This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
.(En) is as in (3.23). By Lemma 3.3, together with the property (35), we have
(:3.29)
.I
lim.f i.p dl1, =
'P dj.l.
En
Hence (328) implies
k fn / lim inf fn as k -----* 00. Thus, letting k and applying the Monotone Convergence Theorem, we have
-----* 00
in (3.25)
lim! (inf fn) dJ.L :s; lim inf! fj dJ.L, ! (lim inf fn) dJ.L = k-+oo n2:k
(3.35)
which gives (3.33). We now tackle the definition of f f dJ.L when f(x) is not necessarily 20 everywhere. For a general measurable function f : X -----* JR, we can write f = f+ - f-, where f+(x) = f(x) when f(x) 2 0, f+(x) = 0 otherwise, and f-(x) = - f(x) where f(x) :s; 0, f-(x) = 0 otherwise. Thus f+, f- E M+(X). We define the integral
(3.36) provided at least one of the terms on the right is finite. If they are both finite or, equivalently, if
! lfl
(3.37) we say
f
dJ.L
f(J;) faT all .1: E X Then, gwen any 8 > O. there eX1 0, FnE
----->
f umformllJ on X \ B.
n E Z+, set
= {.r
E X
for some J 2: n, IfJ(:I:) - I(x)l2: E}
Note that each 'ie't FnE belong& to J The' hypothesi& implie& for each E > 0, so /1(FnE ) -----> 0 as n -----> rx., In paltie-ular, given any 8 such that
(352)
> 0, if we
~ct
E
=
2- k ,
nn>
1 FOE
= 0.
-
there exist~ n
= n(k)
3.
35
Integration on Measure Spaces
Let B
= Uk 2':1 F n (k),2-k,
B E J. We ~ee that IL(B) ::; () and
x tic B, J 2': n(k)
(353)
=?
If)(x) - f(x)1 < 2- k .
This establishes (350). "Ve mention a point which is quite simple but important for applications. Namely. in all the convergeuce theorcms established aboVE'. thc hypothesis that a ~equC'nce of mea~urablc function~ fj converges for all :£ E X can be weakened to the hypothe~is thclt IJ converges ai-most everywhere, i.e., fJ converges for all x E XI. where Xl C X is a measurable subset such that fL(X \ Xd = 0 The proof is ea~y. just restrict attention to the measure spacE' (X 1 ,JI.ltd, where JI = {S E J' S c Xd = {S n Xl: S E J} and JL I i~ J1 restricted to J 1· To close this chapter. we relate the Riemann integral, considered in Chapter 1. to the Lebe~gue integraL which is the integral defined in this chapter whC'n fL i~ Lebe~gue measure The theory of the Lebesgue integral allows us to specify preci~ely which bounded functions on an interval I = [a, b] are Riemalln integrable. Indeed. let ..p : 1-+ lR be a bounded function, and recall the dehnitioll (1.;))-(1 6) ofl( lR is J-measurable, show that there exists fo that is J'measurable, such that fo = f outside a set of /i-measure zero. Hint. First treat the case of simple functions, then positive functions, via the approximation (3.26). 7. Let
f
E
M+(X) and set, for A >"(A) =
(3.61)
E
J',
J
fdfJ, =
J
XAfdfJ,.
A
Show that >.. is a measure. Hint Use the Monotone Convergence Theorem. Note that this result is much stronger than Lemma 3.3.
8 Let f E .cJ(X,fJ,) be given. Show that, for every such that (3.62)
S E J', fJ,(S)
0
E.
S
Hint. Pick cp E 6+(X) such that 0 ::; cp ::; Iff and c/2. Then A = sup cp < 00, so pick (j < c/2A.
J cp dfJ, 2 J If I dfJ, -
9. Give an alternative proof of the Dominated Convergence Theorem, de-ducing it from Egoroff's Theorem and the result of Exercise 8, rather than from Fatou's Lemma. Hznt. Given Ifni::; 9 E .cJ(X,j1), fn ---> f, c > 0, pick a measurable set Y C X such that
fJ,(Y)
0, show that there is a Lebesgue measurable ScI such that m(S) < 0 and 111\S is continuous. Hint. Apply Egoroff's Theorem to the results of Exercise 17. This result is known as LUbin's Theorem 19. If (X, -J', f-L) is a measure space and that there exist Aj E -J' such that
1:X
----+
[0,00] is measurable, show
(3.65) Hint. Set Al
= {x
EX: 1(x)
2: 1} and, inductively, for k 2: 2,
20. Let (X,-J',J-l) be a mea5ure space. Given
1 E M+(X),
t E (0,00), set
(3.66) Show that (3.67)
Then use (3.26) and the l\lonotone Convergence Theorem. Show that, if i.{JJ / 1, then, for each
Hint. First verify this for simple functions
tE(O,oo), S 0, unless v =
o.
The second of these conditions is called the triangle inequality. Given a norm on V, setting d( u, v) = II u - v II defines a distance function on V, making it a metric space. It is easy to see that £l(X,f-L) is a vector space and that IIflip satisfies the first two conditions in (4.3). However, [If[[Ll = 0 if and only if f = 0
-
41
42
LP Spaces
4
almost everywhere. (Recall Exerci&e 4 of Chapter 3.) That is the reason we define L1 (X, fJ,) to consist of equivalence classes defined by (4.2), so L1 (X, fJ,) becomes a normed linear space. Generally speaking, a sequence (v]) in a normed linear space is said to be a Cauchy ~equence if [Iv] - 11k II ----+ 0 as ], k ----+ 00. If every Cauchy sequence ha~ a limit in V, then V is said to be complete, a complete normed linear space is called a Banach space. Theorem 4.1. L1 (X, /1,) is a Banach space.
The proof of completeness of L 1 (X, fJ,) makes use of the following two lemma~, which are essentially re~tatements of the Uonotone Convergence Theorem and the Dominated Convergence Theorem, respectively Lemma 4.2. If fj E £l(X,/L), 0 ::; hex) ::; h(x)::; ", and IlfJllu ::; C < 00, then lim fJ(x) = f(x), with f E L1(X,/L) and Ilfj - fllu ----+ 0 a5 J-'OO
J
----+ 00.
Proof. We know that f E M+(X). The Monotone Convergence Theorem implies J fJ dfJ, /' J f dfJ, Thus J f d/L ::; C. Since Ilf) - fliLl = J f d/L fj d/L in this case, the lemma follows.
J
Lemma 4.3. If fj E £l(X,fJ,),limf](x)
= f(x) /L-a.e., and
~f there is an
£l(X'/L) such that If)(x)1 ::; F(x) 11,-a.e., for all j, then f and Ilfj - fliLl ----+ O.
FE
Proof. Apply the Dominated Convergence Theorem to 9j a.e. Note that 19) I ::; 2F.
E
Ll(X,fJ,),
= IfJ - fl
----+
0
To show L1(X,/L) b complete, suppose (In) is Cauchy in L1. Passing to a subsequence, we can assume Ilfn+l - fnllLl ::; 2- n . Consider the infinite series 00
(4.4)
hex)
+ L[Jn+1(X) -
fn(x)].
n=l
N ow the partial sums are dominated by m
(4.5)
Ih(x)1 + L Ifn+1(x) - fn(x)1 = Ih(x)1 + Gm(x), n=l
and since 0 ::; G 1 ::; G2 ::; ... and llGmll v ::; E2- n ::; 1, we deduce from Lemma 4.2 that G m / ' G fJ,-a.e. and in Ll-norm. Hence the infinite series
4. LP Spaces
43
(4.4) is convergent a.e., to a limit f(x), and via Lemma 4.3 we deduce that fn ---+ f in L1-norm. This proves completeness. Continuing with a description of LP spaces, we define Loo(X, J1,) to consist of bounded measurable functions, L 00 (X, J1,) to consist of equivalence classes of such functions, via (42), and we define Ilfll£o to be the smallest sup of J", f. It i8 easy to show that Loo(X, J1,) is a Banach space. For p E (1. 00 ), we dt'fine LP (X, J1,) to com,ist of measurable functions f such that [/II(x)IP dJ1,(x)]
(46)
lip
x is finitt'. LP (X, It) consists of equivalence classes, via (4.2), and the LP- norm IIIIILP is given by (4.6). This time it takes a little work to verify the triangle inequality. That thi& holds is the content of Minkowski's inequality:
111+ gilLP :S
(4.7)
IIIIILP
+ IlglluJ.
One neat way to t'stablish this is by the following characterization of the LP-norm. Suppose p and q are related by
1 p
We claim that, if
(4.9)
1 q
-+-=1
(48)
I
IIIIILP
E
LP(X, JL),
= ~up{IIIhll[) : h E Lq(X,J1,),
IlhllLq
= 1}.
We can apply (4.9) to 1+ g, which belongs to LP(X, J1,) if I and 9 do, since II + glP :S 2P (IIIP + IgIP) Given this, (4.7) follows easily from the inequality II (f + g)hllv :S IIIhllLl + IIghllv· The identity (4.9) can be regarded as two inequalitit's The ":S" part can be proved by choosing h(x) to be an appropriate multiple Clf(x)IP-l. We leave this as an exercise. The convt'r&e inequality, "2':," is a consequence of Holder's mequality: 1 p
1 q
-+-=l.
(4.10)
Holder's inequality can be proved via the following inequality for positive numbers:
(4.11)
aP
ab < - -p
bq
+ -q
for a, b > 0,
4.
44
LP Spaces
assuming that p E (1, (0) and (4.8) holds. In fact, we claim that, given a,b > 0, 1jp+ 1jq = 1, aP p
bq
'P(t) = - tP + - C q
(4.12)
q
===}
inf 'P(t) = ab,
t>O
which implies (4.11) since the right side of (4.11) is 'P(1). As for (4 12), note that 'P( t) ---> (X) as t "" 0 and as t / ' 00, and the unique critical point occurs for aPtP = bqe q, i.e., for t = blip ja l / q, giving the desired conclusion. Applying (4.11) to the integrand in (4.10) gives
J
1 If(x)g(x)1 dJL(x) :::; -llfll~p
(4.13)
p
1
+ -llglllq· q
This looks weaker than (4.10), but now replace f by tf and g by elg, so that the left side of (4.13) is dominated by (4.14) for all t > O. Another application of (4.12) then gives Holder's inequality. Consequently (4.6) defines a norm on LP(X,{L) Completeness follows as in the p = 1 case discussed above. In detail, given Un) Cauchy in LP(X, {L), we can pass to the case Ilfn+l fnllLP :::; 2- n and define Gm as in (4.5). We have IIGmllLP :::; 1, and hence (via the Monotone Convergence Theorem) deduce that G m / ' G, {L-a.e , and in LP-norm Hence the series (4.4) converges, {L-a.e., to a limit f (x). Since If - fm+11 :::; G - G m , we have by the Dominated Convergence Theorem that
J
J
x
x
if - fm+ll P d{L:::;
as m
---> 00.
(G - Gm)P d{L
--->
0,
Hence LP(X, {L) is complete To summarize, we have
Theorem 4.4. For p E [1, (0), LP( X, {L), with norm given by (4.6), is a Banach space
It is frequently useful to show that a certain linear subspace L of a Banach space V is dense. We give an important case of this here; C(X) denotes the space of continuous functions on X Proposition 4.5. If {L is a finite Borel measure on a compact metric space X, then C(X) is dense in LP(X, JL) for each p E [1, (0).
4.
45
LP Spaces
Proof. First, let K be any compact subset of X. The functions
(4.15)
fK,n(X) = [1
+n
dist(x, K)] -1 E C(X)
are all ::; 1 and decrease monotonically to the characteristic function XK equal to 1 on K, 0 on X \ K. The Monotone Convergence Theorem gives fK,n ----+ XK in LP(X, J-L) for 1 ::; p < 00. Now let A c X be any measurable set. Any Borel measure on a compact metric space is regular, i.e., (416)
J-L(A) = sup{J-L(K) . K
c A, K compact}.
In case X = I = [a, b] and J-L = m is Lebesgue measure, this follows from (2.20) together with the consequence of Theorem 2.11, that all Borel sets in I are Lebesgue measurable. The general case follows from results that will be established in the next chapter; see (5.60). Thus there exists an increasing sequence K j of compact subsets of A such that J-L(A \ Uj K j ) = O. Again, the Monotone Convergence Theorem implies XK J ----+ XA in LP(X, J-L) for 1 ::; p < 00. Thus all simple functions on X are in the closure of C(X) in LP(X, J-L) for p E [1,00). Construction of LP(X, J-L) directly shows that each f E LP(X, J-L) is a norm limit of simple functions, so the result is proved. Using a cut-off, we can easily deduce the following. Let Coo(I~) denote the space of continuous functions on ~ with compact support. Corollary 4.6. For 1 ::; p
< 00, the space
Coo(~)
is dense in
LP(~).
The case L'2(X'll) is special. In addition to the L 2 -norm, there is an mner product, defined by (4.17)
(j, g)u =
.I
f(x)g(x) dJ-L(x).
x
This makes L'2(X, J-L) into a Hzlbert space. It is worthwhile to consider the general notion of Hilbert space in some detail. We devote the next few pages to this and then return to the specific consideration of L2(X, J-L). Generally, a Hilbert space H is a complete inner product space. That is to say, first the space H is a linear space provided with an inner product, denoted (u, v), for u and v in H, satisfying the following defining conditions:
(aul (4.18)
+ U2,V) = a(ul'v) + (U2,V), (u,v) = (v,u), (u, u) > 0 unless u = 0
4.
46
LP Spaces
To such an inner product there is assigned a norm, denoted by
Ilull = V(u,u)
(4.19)
To establish that the triangle inequality holds for Ilu + vii, we can expand jju + vl1 2 = (u + V,u + v) and deduce that this is ::; [[lull + Ilvll]2, as a consequence of Cauchy's inequahty: (4.20)
l(u,v)l::; Ilull·llvll,
a result that can be proved as follows. The fact that (u - v, u - v) 2': 0 implies 2 Re (u, v) ::; Ilul1 2 + Ilv11 2 ; replacing u by eiBu with eiB chosen so that eiB (u, v) is real and positive, we get (4.21) Now in (4.21) we can replace u by tu and v by C 1 v to get (4.22) Minimizing over t gives (4.20). This establishes Cauchy's inequality, so we can deduce the triangle inequality. Thus (4.19) defineb a norm on H. Note the parallel between this argument and the proof of (4.7), via (4.10). The completeness hypothesis on H is that, with thib norm, H is a Banach bpace. The nice properties of Hilbert spaces arise from their similarity with familiar Euclidean bpace, so a great deal of geometrical intuition ib available. For example, we say u and 'V are orthogonal and write u .1 v, provided (u, v) = O. Note that the Pythagorean Theorem holds on a general Hilbert space: (4.23) Thib follows directly from expanding (u
+ v, u + v).
Another useful identity is the following, called the "parallelogram law," valid for all u, v E H (4.24) This also follows directly by expanding (u+v, u+v)+ (u-v, u-v), observing some cancellations. One important application of this simple identity is to the following existence re&ult. Let K be any closed, convex subset of H. Convexity implies x, y E K ::::} (x + y)J2 E K. Given x E H, we define the distance from x to K to be (4.25)
d(x, K)
= inf {llx - yll : y
E K}.
4.
LP Spaces
'l{
Proposition 4.7. If K cHis a nonempty, closed. convex set in a Hilbert space H and if x E H, then there ~s a unique z E K such that d(x, K) =
Ilx - zll· Proof. We can pick Yn E K such that Ilx - Y'lii ---+ d = d(x, K). It will suffice to show that (Yn) mw"t be a Cauchy sequence Use (4.24) with u = Ym - x, 'U = X - Yn, to get
Since K b convex, (Yn
+ Ym)/2
lim sup llYn
E K, so
ilx -
(Yn
- Ymll'2 :s 2d'2 + 2d'2 -
+ Ym)/211 2': d. 4d'2
Therefore
:s 0,
mn-oo
which implies convergence. In particular, this result applies when K i& a closed linear subspace of H. In this case, for x E H, denote by PKX the point in K closest to x. We have (426) We claim that x - PK x belongs to the closed linear space K J.., called the orthogonal complement of K, defined by (4.27)
K J..
= {u
E H . (u, v)
= 0 for all 'U
E K}.
Indeed, take any v E K. Then ~(t) = =
Ilx Ilx -
+ t'Oll'2 2 PKxii + 2t PKx
Re (x - PKX, v)
+ t '2 11'011 '2
is minimal at t = 0, so ~'(O) = 0, i.e., Re (x - PKX, '/') = 0, for all v E K. Replacing v by iv shows that (x - PKX, v) al&o has vanishing imaginary part for any v E K, so our claim i& established The decomposition (4.26) gives (4.28) with Xl = PKx, X2 = X - PKX. Clearly such a decomposition is unique. This implies that H is an orthogonal direct sum of K and K J..; we write (429)
48
4.
£P Spaces
From this it is clear that (430) that (4.31) and that P K and PKl.. are lmear maps on H. We call PK the orthogonal projection of H on K. Note that PKx is uniquely charac-terized by the condition (4.32)
PKX E K, (PKX, v)
=
(x, v) for all v E K
We remark that if K is a linear subspace of H that is not closed, then K ~ coincides with K~, and (4.30) becomes (K ~ ) ~ = K. Using the orthogonal projection discussed above, we can establish the following result. Proposition 4.8. If H is a Hilbert space and
A and p have disjoint supports. Generally, two measures with disjoint supports are said to be mutually smgular. When two measures A and p are mutually singular, we write A .1. p. We have the following result, known as the Lebesgue decomposition of v with respect to J-L. Theorem 4.11. If J-L and v are finite measures on (X, ~), then we can wnte
(455)
v
This decomposztion
tS
= A + p,
A < < J-L,
p ..1
J-L.
unique.
Proof. The measure& A and p are given by (4.51) and (452). The fact that A < < J-L i& contained in (4.52). As we have noted, J-L(X \ Y) = 0, so J-L is supported on y, which is disjoint from Z, on which p is supported, hence p .1. J-L.
-
-
-
If also A and pare mea!'>ures such that v = A + p, A < < J-L, p ..1 I)', we have Z E ~ such that p is supported on Z and J-L(i) = 0. Now J-L(ZUZ) = 0. so A(Z U Z) = 0 and ~(Z U Z) = 0, and, for E E M,
A(E) = A(E \ Z) = v(E \ (Z U Z)), ~(E) = ~(E \ Z) = v(E \ (Z U Z)) This give!'> uniqueness. We say a measure J-L on (X,~) i& (T-finite if we can write X as a countable union UJ~l Xj where Xj E ~ and J-L(Xj ) < 00. A paradigm case is Lebe&gue measure on X = ITt There are routine extensions of Theorems 4.10-4.11 to the case where I)' and v are (T-finite measureh, which we leave to the reader.
Exercises 1. Let V and W be normed linear &paces. Suppose we have linear trans-
formations
(4.56)
4.
LP Spaces
with C independent of j. (We say {Tj } is uniformly bounded.) Suppose also T : V --7 W satisfies this bound. Let L be a dense subspace of V. Then show that (4.57)
Tjv
--7
Tv, V vEL
==}
Tjv
--7
Tv, V v E V.
2. Define
(4.58)
Ts : LP(lR)
~
LP(lR),
Tsf(x)
= f(x - s).
Show that, for p E [1,00), (4.59)
f E LP(lR)
==}
Tsf
--7
f in LP- norm, as [,
--7
Hint. Apply Exercise 1, with V = W = LP(lR) , L Corollary 4.6. Note that IITsfliLl' = IlfllLl'.
O.
= Coo(lR),
as in
One says a metric space is separable if it has a countable dense subset. 3. If I = [a, b] C lR and a :S a < ;3 :S b, define
'Pa;3(x)
=
dist( x,I \ [a,;3]).
Show that the linear span over Q of {'Pa;3 : a,;3 E QnI} is dense in C(I), and deduce that C(1) i& separable. From the denseness and continuity of the inclusion L: C(l) ~ LP(I), prove that £P(l) is separable, for 1 :S p < 00. Then prove that LP(lR) is separable, for 1 :S p < 00. 4 Let X be a compact metric space; X has a countable dense subset {Zj : j ;::=: I}. Given 0 < p < (diam X)/2, set
1/;jp(x)
=
dist (x, X \ Bp(zj))
Show that the algebra generated by {1,Uj,p : j E Z+, P E Q+} and 1 is dense in C(X) and deduce that C(X) is separable Conclude, from Proposition 4.5, that VeX, fJ) i& separable, for p E [1,00), if f.L is a finite measure on the CT-algebra of Borel sets in X. 5. Let {Uj : j ;::=: I} be a countable orthonormal set in a Hilbert space H. Show that 00
(4.60)
~(f,Uj)Uj = Pvf, j=l
4. LP Spaces
54
where V is the closure of the linear span of {11,j}. The Stonc-Weierl:>tra&s Theorem states that, if X is a compact Hausdorff space and A an algcbra of functions in C]R(X) (the space of real-valued continuoul:> functions on X), such that 1 E A, and if A has the property of &eparating points, i.e., for any two distinct p, q E X, there cxists f E A such that f(p) i= f(q). then A i& dense in C]R(X). If A is an algebra in Cc(X) with these propeltie~, plus the property that f E A =} J E A, then A is dense in Cc(X). A proof is given in Appendix A. 6. Use the Stone-Weierstras~ TheOlem to show that, if ek(:c) = eikx , as in (4.40), then the linear span E of {ek : k E Z} is dense in C(Sl), where Sl = JR/27rZ; hence E is dense in LP(SI,dx/27r), for p E [1,00). Hence {ell: k E Z} is an orthonormal basis of L'2(Sl,dx/27r)
7. For f,11, E Crr(JR), the space of smooth in JR, set Kf11,(x) = f
(4.61)
* 11,(x) =
function~
with compact support
J
f(y)11,(x - y) dy
Show that, for 1 ::; p < 00, K f has a unique bounded exten&ion: (4.62) The operation in (461) il:> called convolutwn. H~nt. For f, 'U E Co(JR), if f il:> ~upported in [a, b], show that
f
* 11,(x) = n-+oo lim
b
n
- a n
L J=O
.
f(Lb n
+ (1 -
L)a) TJ'U,(X), n
where Tj11,(X) = 11,(x - Jb/n - (1 - J/n)a) U&e the triangle inequality (4.7) to estimate norml:>. 8. Show that there is a unique extenl:>ion from f E Co(JR) to fELl (JR) of Kf11" with (462) continuing to hold, giving a continuou& linear map K : L1(JR) --+ £(LP(JR»). 9 Let fJ be a sequence of nonnegative functions in L1 (JR) such that (4.63)
/ fJ dx = 1,
supp fJ
C
{:c
E
JR . Ixl < Ijj}
Show that, for 1 ::; p < 00, (4.64)
11, E LP (JR)
====}
fJ
* 11, --+ 11, in
LP norm, as J
--t
00
4.
LP Spaces
55
Derive the same conclusion, upon weakening the second hypothesis in (4.63) to
J
= (3j(c;)
fJ dx
-+
1 as
J -+ 00,
V c; >
o.
Ixl lR+ as follow~. If 5 = U~=l .h is a union of mutually disjoint intervals, set 1),0(5) = I:~=l f-lO(Jk), where f-lO is defined on a single interval J C I as follows:
J),o([x,yJ) = 0, there exist open sets OJ such that A J C OJ and j.t(Oj) :s: j.t(Aj) + 2- j £. It suffices to make such a construction for each EOI.{3 in (5.58), and we need worry only about the case when EOI.{3 is compact; i.e., we need to verify (5.59) when S is compact. For this, keep in mind that X is a metric space. 13. Show that (5.60)
S E 23
===}-
11( S)
= sup
{j.t( K) : K C S, K compact}.
14 If f E £1 (X, j.t) and £ > 0, show that there exists a compact K C X such that j.t( X \ K) < £ and f IK is continuous. Hmt. Using Proposition 45 (or otherwise), produce fv E C(X) such that fv ----7 f, j.t-a.e. Then use Egoroff's Theorem. Then use (5.60). This result is Lusin's Theorem. A special case was stated in Exercise 16 of Chapter 3. 15. Use the Monotone Cla~~ Lemma to give another proof of Proposition 5.5. Furthermore, establish the following variant of Proposition 5.5 Proposition 5.5A. Let A be an algebra of subsets of X, generating the a-algebra M = a(A). Let j.t and v be measures on M. Assume there
exist AJ such that X
=
UA
J,
A J E A,
j.t(AJ )
[0, 001 satisfy
(a) /Lb(0) = 0,
(b) E j E E countable, disjoint, Uj E j = E E E::::} /Lb(E) = 'L- j /Lb(EJ ) Show that /Lb extends to a premeasure /Lo on A, satisfying K
El
, ... ,
EK E E disjoint ::::} ILo(El u· . u EK)
=
L
/Lb(Ej ).
j=l
Show that if
ji,*
/L*(E)
is defined by (5.17), then also, for E eX,
= inf{I:: /Lb(Aj) : A J j~
E
E, E c
UA
J }.
J~
Hint. Start with uniqueness; if also F l , ... , FL E E are disjoint and U~'=l Fc = U~l Ek, write this set as Uk,e(Ek n Fe) to show ILO is well defined. Note. An example is X = [0,1]' E = the collection of intervals in [0,1]' /Lb(J) = t(J), the length. Another family of examples arises at the beginning of Chapter 6.
17. Suppose E is a collection of subsets of X satisfying property (1) Exercise 16 and also satisfying (2') E E E ::::} X \ E is a finite disjoint union of elements of E. Show that E satisfies property (2) in Exercise 16
III
Chapter 6
Product Measures
If (X, M, ft) and (Y,N, v) are measure spaces, we construct a a-algebra M ® N of subsets of X x Y and a product measure ft x v on M ® N, as
follows. We say that a rectangle is a set of the form A x E c X x Y, with A E M, E EN. We define M ®N to be the a-algebra generated by the collection R of rectangles. Note that
(6.1)
(A x E)n(B x F) (A x E) U (B x F)
= (AnB) x (EnF), = A x (E \ F) U (A U B) x (E n F) U
B x (F\E),
the latter decomposition being a disjoint union of rectangles. Also,
(62) Hence the collection of finite disjoint unions of rectangles is an algebra, which we denote M f2J.N, and M ®N is the a-algebra generated by M f2J.N. We define the set function
7r . M
f2J. N ---+
N
(6.3)
[0,00] by
N
7r(U(A
J X
EJ ))
=
j=l
L ft(Aj)v(Ej), j=l
when the rectangles R j = A J x E j are disjoint. We claim on M f2J. N. We need to check that (6.4)
t-t(A)v(E)
7r is a premeasure
= Lft(AJ)v(Ej ) j2:1
when A x E is a countable disjoint union of rectangles Aj x E j . Establishing (6.4) will also show that (6.3) is well defined for an element of M f2J. N
-
71
72
6.
Product Measures
which has several different representations as disjoint unions of rectangles. To prove (6.4), write
(6.5)
XA(X)xE(Y) = XAXE(X,y) = LXAjXEj(X,y) = LXAj(X)xEj(Y) j2:1
Integrating with respect to x and applying Proposition 3 5 yield
M(A)XE(Y) = L
XEj (y)
(6.6) =
J
XAj(X) dM(X)
LIL(Aj)XEj(Y),
VY E Y
Integrating this with respect to Y and using a parallel argument, we get (6.4). We mention that this setup is a special case of that treated in Exercise 16 of Chapter 5. Here [; = n, the collection of rectangles. and A = M ~ N. Using the construction (5.17), we obtain Note that, for SeX X Y,
(6.7) 7l"*(S)
= inf{L M(Ak)V(Ek) : Ak
eW
outer measure
EM, Ek EN, S
k2:1
c
7l"*
on X x Y
U (AkX Ek) } k2:1
By Theorem 5.4, the re~triction of 7l"* to M ®N is a measure, extending the pre measure on M ~ N defined above. We call this measure on M ® N the prod uct measure M x v We note that, if M and v are a-finitc, so is M x v, and, by Proposition 5.5, in this case M x v is the umque measure on M ® N with the property that (IL x v)(A x E) = M(A)v(E), A E M, E EN. If f is a measurablc function on (X x Y, M ® N), we want to see when the iterated mtegral
(68)
.f[.f f(x,y)dv(y)] dM(X)
is well defined and equal to the "product integral" of f with respect to the product measure 11, x v To begin, we define the x-section fx and y-section fY of a function f on X x Y by
(6.9)
fr(Y)
=
r(x) = f(x, y).
Related concepts are those of the x-section Ex and y-section EY of a set E c X x Y, defined by (6.10)
Note that (6.11)
Ex = {y E Y : (x, y) E E},
EY
= {x EX: (x, y)
E E}.
6.
73
Product Measures
Lemma 6.1. If E E M®N, then Ex EN for all x E X and EY EM for all y E Y. More generally, if f is M ®N-measurable, then fx ~s N-measurable for all x E X and fY is M -measurable for all y E Y. Proof. Let
F = {E c X x Y
Ex E N for all x and EY E M for all y}.
Then every rectangle is certainly in F Furthermore,
(U EJ).r = U (EJ)x' j?:l
and (Exr
= (EC)x'
j?:l
Thus F is a cr-algebra, 50 F::J M ® N. Since (jx)-l(S) the lemma follows.
= (j-l(S))x and (jyr1(S) = (j-l(S))Y, the rest of
The next result i5 a basic special case of the forthcoming Theorem 6.3, when f i5 a characteristic function. Proposition 6.2. Assume l.l and v are cr-finite. and take E E M ® N.
Then, cp(x) = v(Ex) and 1.f;(y) = p,(EY) are measurable,
(612)
and (6.13)
(Jt x v)(E) =
J
v(Ex) dJ.l(x)
=
J
p,(EY) dv(y).
Proof. First we assume l.l and v are finite. Let C consist of sets E E M rg;N
such that (6.12) and (6.13) hold, our goal if:> to prove that C = M ®N. We will do this by applying the l\lonotone Cla58 Lemma, Propof:>ition 5.9. Thus, to start, we see that rectanglef:> A x B clearly belong to C, and so do finite disjoint union5 of rectangles, f:>0 M rgj N c C It remains to show that C is a monotone clasf:> Let E j E C, E J / E. Then Ej / EY, so 'Ij}J(Y) = 1.l(E%) / 'tj;(y) = J.l(EY), and the Monotone Convergence Theorem implies (6.14)
J
p,(EY) dv(y) = lim
J
'ljJj(Y) dv(y)
= lim (p,xv)(EJ ) = (p,xv)(E).
Similarly we obain the other identity in (6.13), 50 E E C. On the other hand, if Ej E C and E J ~ E, then EY ~ EY, and hence 'ljJj(Y) ~ 'IjJ(y), provided p, is finite. As long as v is also finite, we can apply
74
6.
Product Measures
the Dominated Convergence Theorem and again get (6.14). Similarly we get the rest of (6.12)-(6.13) for such E, so C has been shown to be a monotone dass, and the proposition is proved, at least for finite measures JI and v. The extem,ion to the CT-finite case is routine. The next two results, known as Tonelli's Theorem and Fubini's Theorem, respectively, are the major result::, of thb chapteI Theorem 6.3. Assume (X, M. jJ,) and (Y,N, v) are CT-finite measure spaces. If f E M+(X x Y), then the funetzons
g(x) =
(615)
J
fx dv,
hey)
=
J
fY dJI
are measurable, hence elements of M+(X) and M+(Y). r'espectwel1j. and
J
(616)
f(x, y) d(II x v)
=
J
g(x) dJL(x)
=
J
hey) dv(y).
Proof. Take a sequence of simple functions fj E 6+(X x Y), fj /' f, such as comMucted in (:320). The results above, for f replaced by fj, are immediate from Proposition 6.2 and additivity of the illtegraL Note that, for each x E X, fJX /' fx and, for each y E y, fi /' p. Thus, with obvious notation, gj /' 9 and h J / ' h, by the l\Ionotone Convergence Theorem. Then (6.16) also follows by the l\Ionotone Convergence Theorem Theorem 6.4. Let X and Y be as m Theorem 6.S. If j E £1 (X X Y, II x v), then
(6.17)
JY
fI E £1 (Y.v) for JL-a e. x E X.
E £1 (X. 11) for v-a.e. y E y,
so g and h, given by (6 15), are defined almost everywhere. We have
(6 18) and (6 16) holds
Proof. If we write f = f+ - f- with f± E M+(X x Y)n£l(X x Y,II x v), then Theorem 6.3 applies to f+ and to f-, yielding g± E M+ (X) and h± E M+(Y) such that
(6.19)
I
f±(x, y) d(JL x v)
=
J
g± dJL
=
J
h± dv
6.
Product Measures
75
The finiteness of the first integral in (6.19) implies that g± and h± are finite a.e. and belong to L1 (X, J.t) and L1 (Y, 1/). respectively. Thus 9 = g+ - gand h = h+ - h- satisfy (6.18), and (6.16) also follows from (6.19). Let us write (616) in thE' form /
f(x, y) d(ll, x 1/) = / [ / f(x. y) dl/(Y)] dp,(x)
(6.20)
= / [/ f(x. y) dtt(x)]
dl/(y).
The first intq!;ral is called a produ("t integral and the other two quantities in the triple identitv (620) are called iterated integrals It is u~eful to have ~ome sufficient ("onditionfl for a function 9 . X x Y to be M ®N-measurable Clearly if 7fx : X x Y --t X and 7fy : X x Y are the standard projectionfl. then (621)
f :X 9 :Y
--4 --4
lR M-measurable
===}
f 0 7fx M ® N-measurable,
N-mea~urable
===}
go 7fy M ® N-measurable
lR
--t --t
lR Y
Also, by Propo~ition 3.1, the algebra 2l ("onsbting of finite sums of products of functions of the type (6.21) are M ® }J-measurable. Also recall that pointwifle limit~ of measurablE' functiom, are measurable. Using these facts together with the StonE'-\Veierstra~s Theorem. we ("an prove the following: Proposition 6.5. Let X and Y be compact Hausdorff spaces and Il3x, ll3y their (J -algebras of Borel sets. Then an'/) contmuous f : X x Y --4 lR is 23 x ® 'By -measurable Proof. The observation~ above imply that f is Il3x ® ll3 y -measurable if f belongs to the algebra C of finite ~um~ of fun("tions of the form g(x)h(y), with 9 and h continuou~. Abo, f is 'B x ® ll3y-mea~urable if f belongs to the clo~ure C of C in C(X x Y). However, the Stone-Weierstrass Theorem implies C = C(X x Y), so we have the re~mlt. If X and Yare compact metric
space~,
we ("an
~ay
more.
Proposition 6.6. If X and Yare compact metric spaces, then any compact K c X x Y belongs to 'Bx ® ll3y. Hence
(6.22)
Il3x x Y = 'Bx ® 'By
6.
76
Product Measures
Proof. In this case, X x Y is a metric space, and, if we denote its distance function by d, then
'lJK(Z) = d(z, K)
(6.23)
is a continuous function on X x Y such that K = {z E X x Y . 'lJK(Z) = O}. According to Proposition 6.5, 'IJ K is SB X ® SBy-measurable, &0 each compact K c X x Y belongs to SBx ® SBy. Thi& implies SBxxy C SBx ® SBy. For the reverse containment, we argue as follows Let Ax = {A eX' A x E E SBxxy, Vcompact E
c Y}.
Then Ax is a cr-algebra, containing all compact A eX, so Ax Hence A E SBx, E c Y compact ====? A x E E SBxxy
~
SB x.
Now, let £y
= {E c Y : A
x E E SBxxy, V A E SBx}
We see that £y is a cr-algebra, containing all compact E ment above, so £y ~ SBy. Hence A E SBx, E E SBy
====?
C
Y, by the argu-
A x E E SBxxy.
It follows that SB x x Y ~ SB x ® SBy. Note that this latter containment holds whenever X and Yare compact Hausdorff 8paces; we do not need them to
be metric spaces. The following i& an immediate consequence of Propo8ition 6.6. Corollary 6.7. If X and Yare cr-compact metric spaces. then (6.22) holds
The product measure space (X x Y, M ® N, ft x v) is almost never complete, but one can construct its completion (X x Y, M ® N, ft x v) via the proces& (3.44). The following i~ a version of the Fubini-Tonelli Theorem for M ® N-measurable function8. Proposition 6.8. Assume (X, M, ft) and (Y,N, v) are complete, cr-finite measure spaces. Let f : X x Y -----t lR be M ® N -measurable and satisfy
(a) f 2: 0
(b) f
or
E
£: 1 (X
X
Y, M ® N)
Then fx is N-measurable . .tor t--t-a.e. x,
(6.24)
fY zs M-measurable, for v-a.e. y.
In case (b), fx and fY are integrable for a e. x (resp., a.e. y). Also, g and h, defined a.e by (6.15), are measurable, and in case (b), integrable, and (6.25)
J
f d(t--t x v)
=
J
g(x) dt--t(x)
=
J
h(Y) dv(y).
6.
77
Product Measures
Proof. Using Exercise 6 of Chapter 3, write f = fo + !I, where fo is M 0Nmeasurable and satisfies either (a) or (b), while !I = 0, f.1 x v-a.e. Theorems 6.3-6.4 apply to fo, so it suffices to show that
(6.26)
!Ix = 0 v-a.e.,
for f.1-a .e x,
fi
= 0 f.1-a.e., for v-a.e.
y.
As in the proof of Theorems 6.3 6.4, it suffices to show that, for E E M 0 (f.1 x v)(E)
(6.27)
= 0 =?
= 0 for f.1-ae. x and f.1(EY) = 0 for v-a.e. v(Ex)
Now E c F with (f.1 x v)(F) = 0; hence Ex C Fx and EY is a consequence of the statement that, for F E M 0 N, (f.1 x v)(F)
(6.28)
N,
C
= 0 =? v(Fx) = 0 for f.1-a.e. x and f.1(FY) = 0 for v-a e
y.
FY. Thus (6.27)
y.
This in turn follows from (6.12)-(6.13), together with (3.59), so the proposition is proved. One can also consider products of K measure spaces (Xj ,Mj ,f.1j), constructing a (I-algebra ~ = Ml 0 ... 0 MK of subsets of Z = TI~l X j , generated by rectangles of the form R = Al X ... x AK, A J E M J. The product measure v = f.11 X ... x f.1K satisfies (6.29) K
v(S)
= inf {Lf.11(Aj1 )· f.1K(Aj K) : AJk
E
Mk, S C
U(II A jk ) }, j2':l k=l
J2':l
for S E ~. Analogues of the results discussed above follow without difficulty; we leave the details to the reader. There is also a notion of product measure for a countable infinite product of measure spaces, (Xj , M J , f.1j), j E Z+, provided f.1j(XJ ) = 1 for all j, i.e, each f.1J is a probability measure. Then one takes rectangles of the form (6.30)
R =
II A j ,
Aj E Mj,
A J = Xj except for finitely many j,
J2':l
and ~ is the (I-algebra generated by such sets. The product measure v is a probability measure on Z = TI j 2':l X j , satisfying
(6.31)
v(S)
= inf
{L: II f.1k(A jk ) : II Ajk E n, S c U(II A jk ) }, j2':l k2':l
k2':l
j2':l k2':l
6
78
Product Measures
for S E ~, where R denotes the class of rectangles of the form (630). Note that, for each "infinite product" Ih>l/1k(A j k) in (6.31), all but finitely many factors are 1 The proof that :"etting vCR) = I1J 2':l/1j(A J ), for sets R as in (6.30), yields a premeasure on the algebra of finite disjoint unions of such sets can be given parallel to the argument involving (6.5)-(66). Thus, buppobe A J E M j, Ej£ E M J' and we have a countable disjoint union Al
X
.
X
Ak
X
X k+ 1
X·
.
=
U Elf x
.. x EfC(f),f
X XfC(£)H
x .
£2':1
so XAr(xd· 'XAk(Xk)
= LXEl£(X1)"
XE"U)f(XK(f»)'
£2':1
One can integrate, first over Xl, then over X2, etc., obtaining, at the kth step, /11(A 1 )
..
/1k(A k )
= LMl(Elf)' . /1k(Ek£)XEk+l f(XkH)'"
XE",U)
f(XK(£))'
£2':1
Continue to integrate over X k + 1 , etc Any given term in the sum above will stabilize eventually, and we obtain
MI(AI) .. /1k(A k )
= Lf1l(Elf)
.. /1K(£) (EfC(f),f) ,
£2':1
as desired. With a little more effort, one can define a product measure on an uncountable product of probability spaces. For a treatment of thi:", we refer to
[HS].
Exercises Let A and B be algebras of bubbets of X and Y, respectively, and let /10 and Vo be pre measures on these algebras, giving rise to measures /1 and v, on M = (l(A) and N = (l(B), respectively. Let us denote by A ~ B the algebra of :"ubbets of X x Y generated by rectangles of the form (6.32)
R
=A
x E,
A E A, E E B.
u.
~
1 UU U{;L
IVH:::i;j,/jUrel>
1. Show that A IZl B consists of finite disjoint unions of rectangles of the form (6.32).
2 Show that o-(A IZl B) = M ® N. Hint. Show that o-(AIZlB) :) MIZlN, the latter being the algebra defined after (6.2). 3. Define a set function N
fO .
A IZl B
---+
[O,ooJ by N
fo(U(A j x E J )) = LMO(A.J)vo(Ej ), J=1
)=1
when the rectangles R j = AJ x BJ E A IZI B are disjoint. Show that is a well-defined premeasure on A IZI B.
fO
4. Show that the measure') on o-(AIZIB) obtained from 'Yo via Theorem 5.4 coincides with the product measure f.L x v, provided f.L and v are (T-finite. Consequently, for S E M ® N,
(6
:~3)
(f.L x v)(S) = inf { L t-to(Aj)vo(Ej) . A J E A B) E B, S
c
j~l
More generally, for any SeX x Y, the outer measure the right side of (6.33). (Compare (6.7) )
U(A
J
x Ej
)}.
j~l
7[*
(S) is given by
In ExerciE>es 5-6, aSf>urne (X, M, II,) and (Y,N, v) are o--finite measure spaces 5 Let Sn con&i&t of simple functions on (X x Y, M ® N) of the form f = I:~:1 aJXRJ , where each Rj is a rectangle, Rj = A) XE J , A J E M, E j E N, such that (f.L x v)(Ry) < 00. Show that Sn is dense in LP(X x Y,M ®N), for 1::; p < 00. 6. If {uJ • j 2: I} is an orthonormal basis of L 2(X,p,) and {Vk : k 2: I} if> an orthonormal basis of L2(y, v), show that {'UJ(X)Vk(Y) . j, k 2: I} is an orthonormal basiE> of the Hilbert space L 2 (X X Y, f.L xv) Hint. Show that the closure of the linear span of thi& orthonormal set contains Sn.
6.
80
Product Measures
In Exercises 7-10, let 00
z=II{O,I}.
(6.34)
k=l
We put on each factor {O, I} the probability measure assigning the measure ~ to {O} and to {I} On Z we put the product iT-alge bra and the product measure, which we denote p. Given a = (aI, a2, a3, .. ), define 00
(635)
P: Z
-----t
P(a) =
[0,1]'
I: ak2-k k=l
Thus P(a) is a number with dyadic expansion 0.ala2a3···.
7. Show that P is measurable Hint. Let a dyadic interval be one with endpoints of the form m2-€ and (m + 1)2-£, with m, f E Z+, m + 1 ::; 2£; a dyadic interval mayor may not contain its endpoints. Show that the iT-algebra generated by the collection of dyadic intervals in I is equal to the Borel iT-algebra 'B 8. Show that P is onto Show that P-1(x) consists of one point unless x has a finite dyadic expansion, in which case p- 1 (x) consists of 2 points. Hint. L~21 2- k = 2- 1
9. Show that S E 'B =? p(p-l(S)) = m(S), where m is Lebesgue measue on I. Hmt. Start with S E Ao, the algebra of finite (disjoint) unions of dyadic intervals. 10. Show that you can take a set Zo C Z of measure zero such that P : Z \ Zo ---+ I is one-to-one and onto and measure preserving. 11 For this exercise, make a slight change of notation; set ex)
Z=II{0,2}. k=l
We use the same sort of product measure as before. Consider the map G . Z ---+ [0, 1] given by 00
G(a)
= Lak3-k. k=l
6.
81
Product Measures
Show that C is a homeomorphism of Z onto the Cantor middle third set K. Show that the measure /1K on [0,1], given by /1K(S) = /1(C-I(S)), coincides with the Lebesgue-Stieltjes measure on [0, 1J given by Exercise 1 of Chapter 5, using the following function cpo On the middle third of [0,1], set cp = 1/2. On the middle third sets of the two intervals remaining, set cp = 1/4 and 3/4, respectively. Continue in the appropriate fashion, to define a continuous monotone function cp : [0, 1J -+ [0, 1J.
12. Let (X,J,/1) be a meabure space and let j ' X Consider the set U Show that U E that
= ((x,t) E X
-+
[0,00) be measurable.
x [0,00): j(x) < t}.
J 0113, where 113 is the class of Borel subsets of lR, and (/1 x
m)(U) =
J
j d/1,
x where m denotes Lebesgue measure on R Compare this result with Exercise 18 of Chapter 3. Hint. First verify this for simple functions, and then consider simple Cpj / ' j, as in (3.26) As an alternative, what can you deduce from Proposition 6.27 13 Here is a variant of Exercise 6. For j E N, let (Xj ,MJ ,/1j) be measure spaces satisfying /1J (XJ ) = 1. Form the infinite product measure space (Z, J, v) as in (6.30)-(6.31), with Z = IT j :2:1 X j . Suppose that for each J E N, {llJk . kEN} is an orthonormal basis of L2(Xj,/1j) and lljl = 1 Consider the collection offunctions on Z, 00
va(z) =
II llkak (Xk),
Z = (Xl, X2, X3, .. ),
k=l
where Q = (QI' Q2, Q3,·· ) runs over the set A of elements of IT~l N such that Qk =I- 1 for only finitely many k. Show that {Va:
Q
E
A}
is an orthonormal basis of L2(Z, v). Hint. For the denseness of the linear span, consult the hint for Exercise 6.
Chapter 7
Lebesgue Measure on ~n and on Manifolds
We constructed Lebm;gue measure on ]R in Chapter 2. The (I-algebra £1 of Lebesgue measurable subsets of]R contains the (I-algebra ~1 of Borel subsets of R The product construction of Chapter 6 applies to ]Rn = ]R x ... x R We have the (I-algebra
(7.1) which, by Corollary 6.7, is the (I-algebra of Borel subsets of ]Rn. We have the larger (I-algebra £1 Q9 .• Q9 £1. As indicated in Chapter 6, the product mea8ure rn x· . x rn (which we will also denote rn) is not complete on this product (I-algebra. Let (]Rn, £n, rn) denote the completion We will also denote the extended meat>ure rn on £n simply by rn. It is easy to &how that thi8 measure space is also the completion of (]Rn, ~n, rn). Frequently, we will omit the subt>cripts from simply ~ and £. respectively.
~n
and
£n,
denoting them
Recall that ~ 1 is the (I-algebra generated by the algebra Al of finite (disjoint) unions of intervals. It follow8 that ~n i8 the (I -algebra generated by the algebra Al ~ . ~ AI, consisting of finite (disjoint) unions of sets in ]Rn which are products of intervals; we call such sets cells. Products of intervals of identical length will be called cubes. It is often convenient to use the formula (6.33) specialized to Lebesgue measure.
(72)
rn(S)
= inf {L: rn(Qj) . QJ j~I
cells, S c
U QJ }, J~l
-
83
84
7.
Lebesgue Measure on ~n and on Manifolds
for S E IB, where, for a cell Q,
(73) In fact, (7.2) hold:;, for all S (7.4)
m(S)
E
£, since (via Exercise 9 of Chapter 5) we have
= inf {rn(E) : E
E IB, E :::) S},
for S E £.
Now ~n has an important structure in addition to its product structure, namely a linear structure The next re:;,ult shows how the measure of a Borel set changes under a linear tramJormation and, more generally, how the Lebesgue integral on ~n behaves under such a transformation. Here, Gl(n,JR.) denoteb the group of invertible linear transformations on JR.n, and det A denotes the determinant of A Proposition 7.1. If A E Gl(n, JR.). and %f f E M+(JR.n) or f E £1 (JR.n,dx), then
(75)
J
f(:r) dx
= Idet AI
J
f(Ax) dx.
Proof. Let 9 be the set of elements A E GI(n,JR.) for which (7.5) is true. Clearly I E g. U:sing det A-I = (det A)-I and det AB = (det A)(det B), it is ea:;,y to show that 9 is a subgroup of Gl(n, JR.). Thus, to prove the propo:;,itioll, it suffices to show that 9 contains all clements of the following three fOllllS. since it is an exerci:;,e in linear algebra to show that these elernmts generate Gl(n,JR.) Here, {ej : 1 ::; j ::; n} denotes the standard ba:;,is of JR.n and (j a permutation of {L. ,n}.
(7.6)
A 1 ej = ea ()), A 2 ej = c)e J ,
A 3e l
c)
= ('1 + C(;2,
i- 0, A 3 (') = c, for J 2: 2
For the cabe A = Al we can work directly If f = XR ib the characteribtic function of a rectangle El x ... x En. the re:;,ult is obvious By the con:;,truction of product meawre the identity (75) then holds for Xs, the characteristic function of a Borel set Then (7.5) follows for f E S+(JR.n ), by additivity, and for f E M+(JR. n ), by (3.25). A similar argument shows that (7.5) holds for transformations of the form A 2 . In such a case, a cell of size f(h), .. ,f(In) is mapped by A2 to a cdl of size !cllf(h), ... , !cnlf(In) , bO it:;, measure is mUltiplied by leI ... cnl = Idet A21· Thus, (75) ho1d& for f = Xs, S E IB, by (7.2)-(7.3), and hmce, as before, for morE' general f
7
Lebesgue Measure on
]Rn
85
and on Manifolds
We will show that (7.5) holds for each of the transformations of the type
A 3 , using the Fubini-Tonelli Theorem. We note that the identity
J
(7.7)
f(x) dx
~
=
J
f(x
+ c) dx,
f
E
~
is elementary. To keep the formulas short, take n we have
J
f(Xl
(7.8)
M+(lR),
J[J = J[J
+ CX2, X2) dx =
f(Xl
= 2. For the ca:;,e A = A 3 ,
+ CX2, X2) dX1]
dX2
f(xl, X2) dXl] dX2,
the second identity by (7.7) Again we get (7.5). It is clear how to extend the analysis of As to n 2: 3, so the proposition is proved.
REMARK 1. One consequence of Propoi:>ition 7 1 is the invariance of Lebesgue measure under rotations. REMARK 2. At the end of this chapter there are exercise sets on determinants and on row reduction, deriving the rei:>ults on matrices used in the proof of Proposition 7.1.
Proposition 7 1 generalizE's to a change of variables formula, which we establish nE'xt, an E'xtE'n&ion to Lebesgue measure of one of the fundamental results of multi-variable calculus. Let F : 0 ---+ n be a C 1 diffeomorphism, where 0 and 0 are open in lRn. Let G = F- 1 , and form the absolute values of the Jacobian dE'terminants:
(7.9)
J(X) = Idet DF(x)1,
K(y) = Idet DG(y)[.
Theorem 7.2. Let F . 0 ---+ 0 be a C l dtfJeomorphism. If either u E M+(O) or u E .c1(O,dx), then
(7.10)
J
u(x) dx
[/
=
J
u(F(x) )J(x) dx.
0
Proof. Take a Borel set Sec O. Fix family of cells {Q j} such that
(711)
E.
> 0, and cover S with a countable
86
7.
Lebesgue Measure on
]Rn
and on Manifolds
Subdividing each of the cells if necessary, we can assume that, if center of Qj and Yj = F(xj),
Xj
is the
(712) and
(7.13) where we set Q~ that
= Qj -
Xj,
centered at 0. From Prop05ition 7.1, we deduce
(7.14) It follows that
(7.15)
Taking
(716)
E --->
0, we have rn(S) 2:
J K(Y)xF(S)(Y) dy or, equivalently,
J
J
(')
[1
cp(x)dx 2:
cp(G(y))K(y)dy,
= XS. By additivity, we have (7.16) for any cp E e+(K), any compact K c O. The l\Ionotone Convergence Theorem then gives (7.16) f01 any cp E M+(O). Setting cp(x) = 'l/J(x)J(x), we have .r~')(x)J(x) dx 2: J IjJ(G(y)) dy or, equivalently, setting v(y) = 'Ip( G(y)), for cp
(7.17)
J
J
[1
(')
1'(Y) dy:S;
v(F(x))J(x) dx,
for all v E M+(n). Reversing the roles of 0 and 0, we hence have (7.10), for U E M+(O), from which the identity for u E £J(O,dx) follows. There are various more refined change of variables formular" some of which we will discuss below. First, we apply the results obtained so far to show how to integrate functions on a Riemannian manifold, i.e., a smooth manifold equipped with a metric tensor. Background material on manifoldr, and metric tensors is given in Appendix B.
7.
Lebesgue Measure on lRn and on Manifolds
87
Let M be a C 1 manifold of dimension n. A continuous metric tensor on M gives a continuous inner product on tangent vectors to M. In a local coordinate system (Xl, ... , x n ), idcnti(ying an open subset of M with an open set 0 C lRn , the metric tensor is given by a positive definite n x n matrix G(x) = (g)k(X)), and the inner product of vectors U and V is given by (718)
(U, V) = U· G(x)V
=
L9)k(X)U)(X)Vk(X), ),k
where U = 2::) uj(x)ej, V = 2::k vk(x)ek, and {el,' ., en} is the standard ba8i8 of }Rn. If we change coordinates by a C 1 diffeomorphism F : 0 ----'> 0, the metric tensor H(y) = (hJk(Y)) in the coordinate system y = F(x) is related to G (.r) by (7.19)
at y
DF(x)U H(y)DF(x)V = (U, V) = U . G(x)V,
= F(x), i.e., G(x) = DF(x)t H(y)DF(x),
(7.20)
or
Now, for a Borel measurable function the integral is given by (7.21)
J
udV =
U
J
u(x)J9dx,
8upported on a coordinate patch,
g(x) = dctG(x).
(One often use~ dS instead of dV when dim Ai = 2 or when one is integrating over the boundary of a manifold of dimension (n + 1).) To see that (7.21) is well defined, note that under the change of coordinates y = F(x) we have by (7.20) that det G(x) = (det DF(x))2 det H(y). Hence
Vh = I det DFI- 1 J9,
(7.22)
h = det H.
so, by (7.10),
J
u(y)Vhdy =
(7.23) =
J
u(F(x)) I det DFI- 1 J91 det DFI dx
J
u(F(x))J9dx.
7.
88
Lebesgue Measure on ]Rn and on Manifolds
More generally f M U dV is defined by writing u as a sum of terms supported on coordinate charts. We see that we have a well-defined measure on the (I-algebra of Borel subsets of a C 1 manifold with a CO metric tensor. In case M is an n-dimensional submanifold of]Rm and a local coordinate chart arises via a C 1 map
U C ]Rn given by cp(r, y) = rcp(y), the Euclidean metric tensor can be written
(f'Jd =
(7.28)
(1
r
2}
1,fm
).
(Compare (724).) To see that the' blank terms vanish, i e., orCP . ox] cp = 0, note that cp(x) cp(x) == 1 ::::} o;rJcp(.r) cp(:r) = 0 Now (7.28) yields (729) We therefore have the following ret>ult for integrating a function in spherical polar coordinates Let us denote by dS the measure on sn-l produced via its Riemannian metric. Proposition 7.3. If f E M+(Il{rt) or f E £l(JRn,dx), then
(730)
J
J [foOO
f(x) dx =
Rn
f(rw)r n - 1 dT] dS(w).
sn-l
See the exercises for applications of Proposition 7.3.
We return to the change of variable formula (7.10), approaching it from another point of view. which yields more general results. In fact, we start with a very general t>etting. Let (0, J', M) and (n, lB, v) be measure spaces and F : 0 --> n a measurable map, i e., S E lB ::::} p-l (S) E J'. If u : n --> R if> measurable and 2: 0, we have
J
J
o
n
u(F(x)) dM(X) =
(7.31)
U(T) dll F- I(;r),
where, for S E lB, (732) Now, if we have the property
v(S) = 0
(7.33)
=::::}
M(p-l(S)) = 0,
we can apply the Radon-Nikodym Theorem, Theorem 4.10, to conclude that there is a lB-measurable function J F-l 2: 0 t>uc-h that (7.34)
J
J
o
n
u(F(x)) dM(X) =
u(X)JF-l (.r) dV(x).
Now we specialize to 0, [2 open in Rn, J', lB the Borel setf> , M = v = Lebesgue measure. ASf>ume P : 0 --> n is continuous, hence' measurable. Suppose P is one-to-one and onto. The property (7.33) certainly holdf> if p-l is Lipf>chitz. If we interchange the roles of 0 and n and of P and p-l, we have the following.
90
7. Lebesgue Measure on
Proposition 7.4. Let F : 0
~n
and on ]Hanifolds
0 be both a Lipschitz map and a homeomorphism. There exists a Borel function JF ~ 0 such that, for any u E M+(O), -+
J
u(x)dx =
(7.35)
o
J
u(F(X))JF(X)dx.
(')
In Theorem 7.2, this result was established for a C 1 diffeomorphism F, and the factor JF(X) was identified with the abbolute value of the Jacobian determinant:
JF(X) = [detDF(x)[.
(7.36)
We will now establish this identity for a larger dass of maps F Note that, if Q ib a cell in 0, then (7.37)
m(F(Q)) =
J
J
o
Q
XF(Q)(x)dx =
Let us assume that 0 is bounded, so m(O) < Extend JF to be 0 on ~n \ 0. For c > 0, set (7.38)
Jc:(x) =
J
m(Q~(x))
,h(x)dx.
00,
JF(Y) dy =
and hence JF E L1(0,dx).
J
'Pc: (x - y)JF(Y) dy,
Q,,(x)
where Qc: is the cube of edge c, centered at x, so 'Pc:(x) have
= c-nXQe;(O)(x). We
(7.39) See Exercise 10 in the first exercise set below. Consequently, by Exercise 11 of Chapter 3, there is a sequence Cv -+ 0 such that
(7.40) Consequently, by (7.37), we have
(7.41)
m(F(Qc:(x))) m(Qc:(x))
----?
JF(X) for a.e. x E 0,
c
= Cv
-+
0
We next establish a lemma which will take the place of (7.14) in our extension of the range of validity of (7.10).
7.
91
Lebesgue Measure on IR n and on Manifolds
Lemma 7.5. If F : 0 Xo E 0, then
---+
n
is a homeomorphism that is differentiable at
(742)
Proof. First absume DF(xo) is invertible. Composing F with a linear tranbformation (and recalling Proposition 71), we can assume DF(xo) = I Thus
(7.43)
F(xo
+ y) = F(xo) + y + R(y),
Fix 6 E (0,1/2). and pkk EO so small that as in (7.13), we have, for E ::; EO,
IR(y)1 = o(IYI).
Iyl ::;
EO ::::::}
IR(y)1 ::;
61yl.
Then,
(7.44) Hence
(7.45) This implie& that the lim sup of the left side of (7.42) is ::; \det DF(xo)1 in this case. We now need to establish the reverse inequality for the lim inf. This is not hard, if one uses borne fundamental results from topology. The same reasoning that gives (744) &hows that the boundary 8Qc(xo) of Qc(xo) is mapped by F onto a set I;, contained in o
(7.46)
C = {x E 0
dist(x,8Qc(xo))::; &} = Q(l+o)c(xo) \ Q(l-o)c(xo).
Furthermore, for each T E [0,1]' we can define FT(xo+Y) = f(xo)+Y+TR(y), and each map FT maps 8Qc('£0) into C Recall that F is assumed to be a o
homeomorphibm. Thus, by degree theory, F(Qc(xo)) contains a point Xl if and only if the map FI&Qe(xo) has degree 1 about Xl. However, the maps o
FT [aQc(xo) all have the same degree about any point Xl E Q(1-o)c(xo), for T E [0,1], so we see that FlaQc(xol has degree 1 about each such point, since Fo obviously has this property. Thus we conclude that (7.47)
hence (7.48)
92
7.
Lebesgue Measure on 1R.n and on Manifolds
This establishes the lemma when DF(xo) is invertible. It remains to treat the case when det DF(xo)
= 0, i.e., A = DF(xo) is
not invertible. In this case, we have
(7.49)
F(xo
+ y) =
F(xo)
+ Ay + R(y),
Consequently, if we fix 0 > 0 and pick olyl, we have
(7.50)
F(Qc:(xo)) C {x E 1R.n
.
EO
IR(y)1
= o(IYI)·
so small that Iyl ::;
EO
:::=}
IR(y)1 ::;
dist(x - F(xo), AQc:(O)) ::; EO},
for E ::; EO. Now AQc:(O) is contained in a set {x E V Ixl::; KE}, where V = R(A) is a linear subspace of 1R.n of dimension::; n - 1. Hence, in this case,
(7.51) so the lim sup of the left side of (7 42) is ::; K 0 in this case. Letting 0 we finish the proof.
---+
0,
Using (7.41), we see that, if F : 0 ---+ n is as in Proposition 7.4 and if also F is differentiable for almost all x E 0, then the identity (7.36) holds for almost all x. We have the following extension of Theorem 7.2. Proposition 7.6. If F : 0 ---+ n is both a Lzpsch%tz map and a homeomorphism and zf F zs differentiable for almost all x E 0, then the conclusion of Theorem 7 2 holds It is a result of Rademacher that every Lipschitz function on 0 it:> differentiable almost everywhere. We will prove this in Chapter 11. To end this chapter, we show that Lebesgue measure on ]Rn is uniquely characterized by translation invariance, together with a simple finiteness condition. Proposition 7.7. Let). be a measure on ~n. Assume that). is translatwn invariant:
(7.52)
8 E
~n ===?
),(8) = ),(8 + y),
Also assume there exists a bounded open U
(7.53) Then).
0< )'(U)
0, where rn is Lebesgue measure on ]Rn.
7.
Lebesgue Measure on
93
and on Manifolds
]Rn
Proof. Let 8 be any bounded Borel set in ]Rn. By (7.52), for each y E ]Rn,
J
xs(x) d'\{x) =
(754)
J
xs{x - y) d'\(x).
Integrate this identity over a ball B with respect to dm(y). Since'\ is finite, one can apply Fubini's Theorem to write (7.55)
m(B)'(S)
~ J~ xs (x - y) dm(Y)]
(T-
d)'(x).
Now m(8) = 0 => the right side of (7.55) is zero, so ,\ «m Hence by the Radon-Nikodym Theorem we have f E M+(]Rn) such that
(756)
'\(8) =
J
f(x) dm(x),
V 8 E \)3n,
s and translation invariance (7.52) implies (7.57)
J
f(x
+ y) dm(x) =
J
f(x) dm(x),
s Note that (7.53) implies Iu f dm
'(U) = m(U) 1 c = m(U)
J
J
u
u
1 f(x) dm(x) = m(U)
f(x
+ y) dm(x).
J"
If we write (7.56)-(7.57) as >'(8) = f(x + y) dm(x) and integrate over U with respect to dm(y), we have by Fubini's Theorem that, for each 8 E \)3n, (7.59)
m(U)>'(8) =
JJ
f(x
8
+ y) dm(y) dm(x) = cm(U)m(8),
u
the last identity by (7.58). This proves that>. = cm.
Exercises 1. Let a
= I~ e- x2 dx. This is called a Gaussian integral. Show that an =
J
e- 1x12 dx.
IR n
94
7.
Lebesgue Measure on
2. Show that a 2 = 27r
fa= e-
r2
~n
and on Manifolds
r dr = 7r.
Hint. Apply Proposition 7.3 and use the fact that 27r is defined to be the length of Sl.
3. Let An denote the n-dimensional measure of the unit sphere sn. Show that
and show that 27rn/2
(7.60)
An - 1 =
f(~) ,
J:'
where f(z) e-ss z - 1 ds, for z > O. The function f(z) is called Euler's gamma function. Hint. Apply Proposition 7.3. 4. Show that zf(z) = f(z + 1). Hint. Integrate by parts. 5 Show that r(1)
=
1 and f(1/2)
6. Let B n be the unit ball in
Hint. Apply (7.30) to
f =
~n,
=
7r 1 / 2 . Deduce that, for k E Z+,
B n = {x E ~n .
Ixl
- Q, if each partition of each interval factor Iv of R involved in the definition of Q is further refined in order to produce the partitions of the factor& Iv, u&ed to define P, via (7.68). It is an exercise to show that any two partitions of R have a common refinement. Note also that, as in (1.4),
(771) Consequently, if Pj are any two partitions of R and Q is a common refinement, we have
(7.72)
7.
98
Lebesgue Measure on lRn and on Manifolds
Now, whenever defined:
1 : R ---+ lR is
(7.73)
=
7(1)
bounded, the following quantities are well
7p (1) ,
inf
1(1) = sup
PElleR)
1p(1) ,
PElleR)
where ll(R) is the set of all partitiont; of R, as defined above. Clearly, by (7.72), 1(1) ::; 7(1). We then say that 1 it; Riemann integrable (on R) provided 7(1) = 1(1), and in such a case, we write 1 E R(R) and bet
(7.74)
1(1)
= 7(1) = 1(1).
The following exercises deal with the Riemann integral and its relation to the Lebesgue integraL 1. Show that
1 E C(R)
=?
j E R(R)
2. Show that j, 9 E R(R) =? j + 9 E R(R) and 1(1 + g) Hint Consult the proof of Proposition 1.1. 3. Show that
1,g E R(R)
=?
= 1(1) + 1(g)
19 E R(R)
4 Given a partition P of R, define maxsize (P) Extend Corollary 1.4 and (1.42) to the multi-diment;ional case. 5. Extend the proof of Propot;ition 3.10 to the following result.
Proposition 7.S. Let R be a cell in JR n , and let cp : R ---+ JR be bounded. Then cp it; Riemann integrable if and only if the set of pointt; x E R at which i.p is discontinuous has Lebet;gue measure zero. If thit; condition holdt;, then the Riemann integral and the Lebesgue integral of i.p coincide. 6. For a bounded set S
c JR n ,
with characterit;tic function Xs, set
as in (1.23), and define S to "have content" if and only if cont+(S) = cont - (S) Show that S hab content if and only if the set of boundary points of S hat; Lebesgue measure zero.
7.
Lebesgue Measure on IRn and on Manifolds
When doing Exercises 7-8, look back at Exercises 10-14 of Chapter 5. 7. If 0
c JRn is open and bounded, show that its Lebesgue measure is given
by m(O) = cont-(O). If K
c JRn is compact, show that its Lebesgue measure is given by m(K)
= cont+(K).
8. If S c JRn is a bounded set, show that its Lebe8gue outer measure is given by m*(S) = inf {m(O) . S C 0, open}.
Remove the hypothesis that S be bounded. Exercises on determinants
Let M (n, JR) denote the space of n x n real matrices. These exercises investigate the existence, uniqueness, and basic properties of the determinant, det : M(n,JR)
(775)
----+
JR,
which will be uniquely specified as a function 1J . }\,f(n, JR)
----+
JR satisfying
(a) 1J is linear in each column a) of A, (b) 1J(A) = -1J(A) if A i8 obtained from A by interchanging two columns, (c) 1J(I) = 1. 1 Let A = (al .... ,(J,n), where a) are column vectors; aj = (al), ... ,anj)t. Show that, if (a) holds, we have the expan8ion detA= La)1 det(ej,a2, ... ,an ) )
(7.76)
= .. =
ajll' . a)"n det (e Jl , e12 ,···, ej,,),
L Jr,
,j"
where {el' .. , en} is the standard basis of JR n,. 2. Show that, if (b) and (c) also hold, then (7.77)
det A
=
L o-ESn
(sgn a-)
ao-(l)1 ao-(2)2
. ao-(n)n,
100
7. Lebesgue Measure on
]Rn
and on Manifolds
where Sn is the set of permutations of {1, ... ,n} and (7.78)
sgn a = det (e u(l),
. .. ,
eu(n») = ±l.
To define sgn a, the "sign" of a permutation a, we note that every permutation a can be written as a product of transpositions: a = 'T1 ••. 'Tv, where a transposition of {1, ... ,n} interchanges two elements and leaves the rest fixed. We say sgn a = 1 if v is even and sgn a = -1 if v is odd. It is necessary to show that sgn a is independent of the choice of such a product representation. (Referring to (7.78) begs the question until we know that det is well defined.) 3. Let a E Sn act on a function of n variables by (7.79) Let P be the polynomial (7.80)
P(X1, ... , xn)
=
II
(Xj - Xk).
l:::;;j 0, there is a path from 1 to Pa in Gl+(n,~). Hint. Use Exercise 5 from the previous exercise set. 4. Show that assertion (a) of Exercise 3 holds. Hint. Evaluate det Ep,v(te). 5. Show that assertion (b) of Exercise 3 holds. Hint. Reduce to the case where each ej = ±1. The number of -l's is even. There is a path from 1 to -1 in Gl+(2,~) given by (
That is, -1 E
COS
t
- sin t )
sin t
Gl+(2,~)
cos t
'
0::; t ::;
is a rotation by
7L
7L
6. Reduce the proof of assertion (c) in Exercise 3 to the following assertion: There is a path from 1 to PaT in Gl+(n, JR) whenever positions in Sn.
0"
and
T
are trans-
In turn, reduce the proof of this assertion to the following two assertions: (a) There is a path from
1=
C
1
J
to A =
G ~) 0 0 1
in Gl+(3, JR).
(b) There is a path from
C
1
1=
1
J
to
1 0
B=
C
0 1
J
in Gl+(4, JR).
7.
Lebesgue Measure on
jRn and
on NlamtOlQS
7. Prove assertion (a) of Exercise 6. Hint. Show that A is a rotation by
21T
-'-vv
/3 about the axis through (1,1, l)t.
8. Prove assertion (b) of Exercise 6. Hint. Show that B is the identity on the linear span of (1,1,0, O)t and (0,0,1, l)t and that it is rotation by 1T on the 2D orthogonal complement of this space.
Exercises on matrix integrals Let M(n, JR.) denote the space of n x n real matrices, M(n, C) the space of n x n complex matrices. Set
(7.96)
O(n)
=
{A
M(n, JR.) : A* A
=
(7.97)
U(n)
=
{A E M(n,C): A*A
=
I}, I},
=
I},
=
I}.
E
where Also set
SO(n) SU(n)
(7.98) (7.99)
= =
{A E O(n) : detA {A E U(n) : detA
1. Use Proposition B.7 from Appendix B to show that each of the four ma-
trix groups (7.96)-(7.99) is a smooth, compact submanifold of M(n, JR.) or M(n, C). 2. For each group G of the form (7.96)-(7.99), define, for 9 E G,
Rg, Lg : G
---7
G,
Rg(x)
=
xg,
Lg(x)
=
gx.
Show that, for each 9 E G, Rg and Lg preserve the metric tensor on G induced from M(n, JR.) or M(n, C). Hint. Show that R g, Lg : M (n, IF) ---7 M (n, IF) are isometries, with IF = JR. or C, as appropriate. 3. Let dV denote the volume element on G associated to the metric tensor discussed above, via (7.21). Show that for each J E Ll(G, dV), (7.100)
J
J
G
G
J(x) dV(x) =
J(gx) dV(x)
=
J
J(xg) dV(x),
G
V9
E
G.
106
7.
Lebesgue Measure on
]Rn
and on Manifolds
We set dg = V(G)-l dV and call this Haar measure on G.
J
4. Show that
f(g-l) dg
J
f(g) dg.
=
G
G
5. Let V be a finite-dimensional Hilbert space and assume 7r : G -----+ GI(V) is a continuous homomorphism such that 7r(g) : V -----+ V is unitary for each g E G. (One says 7r is a unitary representation of G on V.) For v E V, set
Pv =
J
7r(g)v dg.
G
Show that P is the orthogonal projection of V onto
Va
=
Hint. Show that 7r(g)v = v V g
p 2v =
{v
E
==}
J
V: 7r(g)v
Pv = v,
7r(h)Pv dh = Pv,
=
v, Vg
E
G}.
7r(h)Pv = Pv, P*v =
J
V hE G,
7r(g-l)v dg
= Pv.
6. Also let ,\ be a unitary representation of G on a finite-dimensional Hilbert space W. Give £(W, V) the Hilbert-Schmidt inner product, (A, B) = Tr B* A, and define a unitary representation v of G on £(W, V) by v(g)A = 7r(g)A,\(g-l). Define
Q : £(W, V) QA =
Show that
-----+
£(W, V) by
J
J
G
G
v(g)Adg =
7r(g)A,\(g-l) dg.
Q is the orthogonal projection of £(W, V) onto
I(7r,'\) = {A E £(W, V) : 7r(g)A = A'\(g), Vg E G}. 7. Assume the representations 7r and ,\ are irreducible, i.e., V and W have no nontrivial G-invariant linear subspaces. Schur's Lemma (cf. [TIl, Appendix B, §§7-8) implies that I(7r,'\) is either zero- or one-dimensional. Using Exercise 6, what can you conclude about
J
7rij(g)'\kC(g) dg?
G
Here 7rij (g) are the matrix entries of 7r (g) in an orthonormal basis of V, and ,\k£(g) are similarly defined. The resulting formulas are called the Weyl orthogonality relations.
Chapter 8
Signed Measures and Complex Measures
As opposed to the measures we have considered so far, a signed measure is allowed to take on both positive and negative values. To be precise, if M is a IT-algebra of subsets of X, a signed measure von M is a function
(S.l)
v:
M
-----+
[-00,00],
allowed to assume at most one of the values ±oo and satisfying
v(0) = 0
(S.2) and (S.3)
Ej EM disjoint sequence ===> v(UEj ) = 2:v(Ej), j
j
the series L: j v(Ej) being absolutely convergent. A signed measure taking values in [0,00] is what we have dealt with in Chapters 2-7; sometimes we call this a positive measure. If /11 and /12 are positive measures and one of them is finite, then /11 - /12 is a signed measure. The following result is easy to prove but useful. Proposition 8.1. If v is a signed measure on (X, M), then for a sequence
{Ej
} C
(S.4)
M, Ej /
E ===> v(E)
=
lim v(Ej), J->OO
and (S.5)
E j ". E, v(Ed finite ===> v(E) = lim v(Ej ). J->OO
-
107
108
8.
Signed Measures and Complex Measures
Proof. Exercise. If v is a signed measure on (X, M) and E E M, we say v is E-positive if v(F) 2: 0 for every F E M, FeE. Similarly we say v is E-negative if v(F) :S 0 for every such F. We say v is null on E if v(F) = 0 for all such F. Note that, if Pj is a sequence of sets in M, then
(8.6)
v Pj-positive
::::=::}
v P-positive, for P =
UPj. j
To see this, let Qn = Pn \ Uj:Sn-l Pj. Then Qn C Pn , so v is Qn-positive. Now, if FeU Pj, then v(F) = 2: v(F n Qn) 2: 0, as desired. The following key result is known as the Hahn Decomposition Theorem.
Theorem 8.2. If v is a signed measure on (X, M), then there exist P, N E = X, P n N = 0.
M such that v is P-positive and N -negative and PUN Let us assume that v does not take on the value be proved via the following:
+00.
The theorem can
Lemma 8.3. If v(A) > -00, then there exists a measurable PeA such that v is P-positive and v(P) 2: v(A). Given Lemma 8.3, we can prove Theorem 8.2 as follows. Let
s = sup {v(A) : A EM}. By (8.2), s 2: 0. Take Aj EM such that v(Aj) / s, v(Aj) > -00. By the lemma, v is Pj-positive on a sequence of sets Pj C A j , such that v(Pj ) -----) s. Set P = U Pj. By (8.6), v is P-positive. Then we deduce that v(P) = s. In particular, under our hypothesis, s < 00. Set N = X \ P. We claim that v is N-negative. If not, there is a measurable SeN with v(S) > 0, and then v(PUS) = s+v(S) > s, which is not possible. Thus we have the desired partition of X into P and N. To prove Lemma 8.3, it is convenient to start with a weaker result:
Lemma 8.4. If v(A) > -00, then for all c > 0, there exist measurable Be A such that v(B) 2: v(A) and v(E) 2: -c, for all measurable E C B. We show how Lemma 8.4 yields Lemma 8.3. We define a sequence of sets Aj EM inductively. Let Al = A. Given Aj, j :S n-l, take An C A n- 1 such that v(An) 2: v(An-d and v(E) 2: -lin, for all measurable E cAn. Lemma 8.4 says you can do this. Then let P = Aj. Clearly v is P-positive,
n
8.
109
Signed Measures and Complex Measures
and (8.5) implies that I/(P) = liml/(Aj) ~ I/(A).
It remains to prove Lemma 8.4. We use a proof by contradiction. If Lemma 8.4 is false, then (considering B = A), we see that, for some E > 0, there is a measurable El C A such that I/(El) -E. Thus I/(A \ E l ) ~ I/(A) + E. Next, considering B = A \ E l , we have a measurable E2 C A \ El such that I/(E2):S -E, so I/(A\(El nE2)) ~ I/(A)+2E. Continue, producing a disjoint sequence of measurable E j C A with I/(Ej) -E. Then Aj = A \ (El U· .. U E j ) has I/(Aj) 2: I/(A) + jE, and hence, by (8.5), we must have
:s
:s
Aj "" F,
I/(F)
= +00.
But we are working under the hypothesis that 1/ does not take on the value +00, so this gives a contradiction, and the proof is complete.
1/.
We call the pair P, N produced by Theorem 8.2 a "Hahn partition" for It is essentially unique, as the following result shows.
Proposition 8.5. If 1/, P and N are as in Theorem 8.2 and if also P, N is a Hahn partition for 1/, then 1/ is null on P /:;
P= N
/:;
N = (P n N) U (N n P).
Proof. Let E E M, E c P /:; P. Write E = EoUEl, a disjoint union, where we take Eo = En (P n N), El = En (N n P). Then I/(E) = I/(Eo) + I/(El)' But each I/(Ej) is simultaneously 2: and 0, so we have I/(Ej) = 0.
°
:s
If 1/ is a signed measure on (X, M) and we have a Hahn partition of X into PUN, we define two positive measures 1/+ and 1/_ by
(8.7)
I/+(E)
=
I/(P n E),
I/_(E)
= -I/(N n E),
for E E M. We have
(8.8) Now 1/+ is supported on P, i.e., on N, so
(8.9)
1/+
and
1/+
1/_
is null on X \ P. Similarly
1/_
is supported
have disjoint supports.
This is called the Hahn decomposition of the signed measure 1/. By Proposition 8.5, it is unique. We can also form the positive measure
(8.10)
110
8.
Signed Measures and Complex Measures
called the total variation measure of v. Given a signed measure v, we can define J f dv for a suitable class of measurable functions f· We can use the Hahn decomposition (8.8) to do this, setting (8.11) when v satisfies (8.8). In particular this is well defined for
f
E Ll(X,
Ivl).
Let f-l be a positive measure on (X, M). We say that a signed measure von (X, M) is absolutely continuous with respect to f-l (and write v < < f-l) provided that, for E E M, (8.12)
= 0 ==?
f-l(E)
veE)
=
o.
This definition was made in (4.45) in the case when v is also a positive measure. It is useful to have the following. Proposition 8.6. If f-l is a positive measure and v a signed measure on (X, M), with Hahn decomposition (8.8), then (8.13)
v «f-l
==?
v+
«
f-l and v_
«
f-l.
Proof. If E E M and f-l(E) = 0, then f-l(E n P) = f-l(E v+(E) = veE n P) = 0 and v_(E) = -veE n N) = O.
n N)
0, so
We can therefore apply Theorem 4.10 to v+ and v_ to obtain the following extension of the Radon-Nikodym Theorem. Theorem 8.1. Let f-l be a positive finite measure and v a finite signed measure on (X, M), such that v < < f-l. Then there exists h E c8x, f-l) such that
J
(8.14)
Fdv
x for all F E .cl(X,
=
J
Fhdf-l,
x
Ivl).
Proof. Note that v+(X) = v(P) < 00 and v_eX) = -v(N) < 00. Apply Theorem 4.10 to v±, obtaining h± E .cl(X, f-l), and let h = h+ - h_. The only difference between (4.47) and (8.14) is that this time h need not be ~ o.
8.
111
Signed Measures and Complex Measures
It is also useful to consider the concept of a complex measure on (X, M), which is defined to be a function
(8.15)
p :
M
C,
----+
satisfying
p(0) = 0
(8.16) and (8.17)
E EM disjoint sequence j
===}
p(UE
j )
=
j
LP(E
j ).
j
In this case, we do not allow p to assume any "infinite" values. It is clear that we can set
vo(S) = Re p(S),
(8.18)
VI
(S) = 1m p(S),
and Vj are signed measures, which do not take on either value. We have
p(S)
(8.19)
=
vo(S)
+00
or
-00
as a
+ iVl (S),
and we can define
J
(8.20)
F dp =
J
F dvo
+i
J
F dVl,
IF
where in turn dVj are defined as in (8.11). There is an obvious extension of the Radon-Nikodym Theorem: if p, is a finite positive measure and v a complex measure on (X,M), such that v « 11" then (8.14) continues to hold for some complex-valued h.
Exercises 1. 'Write down a proof of Proposition 8.1. In Exercises 2-3, let p, and v be finite positive measures on (X, J). For [0,(0), set i.pT = V - T p,. For each such T, i.pT is a signed measure.
T E
112
8.
Signed Measures and Complex Measures
2. Show that there exists a family P T and a family NT of elements of J such that Po = X, for each T E [0,00), PTlNT is a partition of X,
0, we can take j such that
Then we can pick 1}; E C (lRn) such that 111};('Pj * u) - 'Pj * ullHl,p < E. Of course, 'Pj * u is smooth, so 1};(rpj * u) E COO (lRn). We have established
oo
Proposition 10.1. Forp E [1, (0), the space
Coo (IRn) is dense in H
1 ,P(lRn ).
Sobolev spaces are very useful in analysis, particularly in the study of partial differential equations. We will establish just a few results here, some of which will be useful in Chapter 11. More material can be found in [EG], [Fol], [T1], and [Yo]. The following result is known as a Sobolev Imbedding Theorem. Proposition 10.2. If p > n or if p
= n = 1, then
(10.7) For now we concentrate on the case p E (n, (0). Since dense in H 1 ,P(lRn ), it suffices to establish the estimate (10.8)
COO (IRn)
is then
10.
Sobolev Spaces
131
In turn, it suffices to establish (10.9) o
To get this, it suffices to show that, for a given
p. One technical tool which is useful for our estimates is the following generalized Holder inequality.
10.
134
Lemma 10.5. If Pj E [1,00]' LPjl
Sobolev Spaces
= 1, then
(10.29)
The proof follows by induction from the case m Holder inequality.
= 2, which is the usual
Proposition 10.6. For P E [1, n),
(10.30)
In fact, there is an estimate (10.31)
for u E H1,P(JR n ), with C
= C(p, n).
Proof. It suffices to establish (10.31) for u E Co(JRn ). Clearly
(10.32) where the integrand, written more fully, is 18j u(XI, ... , Yj," ., xn)l. (Note that the right side of (10.32) is independent of Xj.) Hence
(10.33)
lu(x)ln/(n-l)
:s:
D(100-00 n
18j ul dYj
)
I/(n-I)
.
We can integrate (10.33) successively over each variable Xj, j = 1, ... , n, and apply the generalized Holder inequality (10.29) with m = PI = ... = Pm = n - 1 after each integration. We get
(10.34)
This establishes (10.31) in the case P 1, obtaining
= 1. We can apply this to v = lui',
'Y >
(10.35) For P < n, pick 'Y = (n - l)p/(n - p). Then (10.35) gives (10.31) and the nrnnn",it.inn is nroved.
10.
U
Sobolev Spaces
1;5b
There are also Sobolev spaces Hk,P(lR n ), for each k E Z+. By definition E Hk,P(lR n ) provided
(10.36) where [ye" =
(10.37)
lal = al + ... + an,
()~1 ... ()~n,
( - 1)Inl
I
.
()nzp ;:) u dx -uX C'
and, as in (10.2), (10.36) means
J
zpfn d.T,
Given u E Hk,P(lR n ), we can apply Proposition 10.6 to estimate the ()k-llL in terms of II()kuII LP , where we use the notation
Lnp/(n-pL norm of
(10.38)
()kU
= {()nu : lal = k},
II()kuIILP =
L
II()nuII LP ,
Inl=k and proceed inductively, obtaining the following corollary.
Proposition 10.7. For kp
< n,
(10.39)
The next result provides a generalization of Proposition 10.2.
Proposition 10.8. We have
(10.40)
Proof. If p > n, we can apply Proposition 10.2. If p = nand k 2 2, since it suffices to obtain an L oo bound for u E Hk,P(lR n ) with support in the unit ball, just use u E H 2 ,n-c(lRn ) and proceed to the next step of the argument. If p E [1, n), it follows from Proposition 10.6 that
(10.41)
np n-p
Pl = - - .
Thus the hypothesis kp > n implies (k - l)Pl > kp > n. Iterating this argument, we obtain Hk,P(lR n ) c Hf,q(lR n ), for some g 2 1 and q > n, and again we can apply Proposition 10.2.
136
10. Sobolev Spaces
Exercises 1. Write down the details for the proof of the identities in (10.4).
2. Verify the estimates in (10.14). Hint. Write the first integral in (10.13) as l/A times
Jfo1v+(z).'\lU(tz,~t)dtdZ+ Jfo1v_(z).'\lU(tz,1-~t)dtdZ, ~
~
where v±(z) = (±z,1/2). Then calculate an appropriate Jacobian determinant to obtain the second integral in (10.13). 3. Suppose 1 < P < 00. If Tyf(x) = f(x - y), show that f belongs to H 1,P(JRn ) if and only if Tyf is a Lipschitz function of y with values in LP(JRn ), i.e., (10.42)
Hint. Consider the proof of Proposition 10.4. What happens in the case P = I? 4. Show that H n,l(JRn) c C(JR n ) n DXl(JRn ).
Hint. u(x) = J~oo··· J~oo Eh··· 8nu(x + y) dY1··· dYn. 5. If Pj E [1,00] and Uj E LPj, show that U1 U2 E L r provided l/r 1/P1 + 1/P2 and (10.43) Show that this implies (10.29). 6. Given U E L2 (JRn), show that (10.44) 7. Let f E L1(JR), and set g(x) = J~ocJ(Y) dy. Continuity of 9 follows from the Dominated Convergence Theorem. Show that (10.45)
81 g = f·
10.
137
Sobolev Spaces
Hint. Given
O.
Br(x)
Then (11.2)
lim Arf(x) = f(x) a.e.
r--->O
Here and below, m(S) denotes the Lebesgue measure of a set S, and E JRn : Ix - yl ~ r}. From results in previous sections, we know that Arf -----+ f in Ll-norm, as r -----+ O. Hence, for any fixed f E Ll(JRn ), there exists a sequence rv -----+ 0 such that Ar,J(x) -----+ f(x) a.e. However, this does not imply (11.2).
Br(x) = {y
A successful systematic approach to deriving both pointwise a.e. convergence results and additional interesting quantitative information was achieved by Hardy and Littlewood. We look at the "Hardy-Littlewood maximal function," defined by (11.3)
M(f)(x) = sup r>O
m
(~)r
J
If(y)1 dy.
Br(x)
Note that this depends only on the class of fin Ll(JRn ), as does (11.1). The basic estimate on this maximal function is the following.
-
139
140
11. Maximal Functions and A.E. Phenomena
Theorem 11.2. There is a constant C = C(n) such that, for any A > 0, f E Ll(JRn), we have the estimate
m({x
(11.4)
E
JRn: M(f)(x) > A}) ::;
C
"Illfll£1.
We also note that the estimate (11.5)
m({x
1
E
JRn: If(x)1 > A}) ::; ~llfll£1'
known as Tchebychev's inequality, follows by integrating the inequality If I 2: AX8", , where SA is the set {x: If(x)1 > A}. We show how these estimates allow us to prove Theorem 11.1. Indeed, given f E £l(JRn,dx) and c > 0, pick 9 E Co(JR n ) such that Ilf - gll£1 < c. Clearly Arg(x) ---t g(x) uniformly as r ---t 0, so if
(11.6)
EA = {x E JRn : lim sup IArf(x) - f(x)1 > A}, r->O
then EA is unchanged if
f
is replaced by
sup IAr(f - g) - (f r>O
so
EA C {x: M(f - g)(x) > If we apply (11.4)-(11.5) with we get
f
m(EA) ::;
f -
g. Now
g)1 ::; M(f -
~} U {x:
replaced by
C
g)
+ If - gl,
If(x) - g(x)1 >
f - 9
~}.
and A replaced by Aj2,
Cc
"Illf - gll£1 < T·
Since this holds for all c > 0, we deduce that (11.7)
m(EA) =
°
for all A > 0.
This is precisely equivalent to (11.2). We now take up the proof of Theorem 11.2. Let (11.8)
FA = {x E JRn : Mf(x) > A}.
We remark that, for any f E Ll(JRn) and any A > 0, FA is open. Given x E FA, pick r = rx such that Arlfl(x) > A, and let Ex = Brx(x). Thus {Ex: x E FA} is a covering of FA by balls. We will be able to obtain the estimate (11.4) from the following "covering lemma," due to N. Wiener.
11.
141
Maximal Functions and A.E. Phenomena
Lemma 11.3. lfC = {Bet: a E QI.} is a collection of open balls in JRn, with union U, and if ma < m(U), then there is a finite collection of disjoint balls B j E C, 1 :S j :S K, such that
(11.9) We show how the lemma allows us to prove (11.4). In this case, let C = o
0
{Bx : x E F;J. Thus, if ma < m(F>,), there exist disjoint balls B j = Brj(xj) such that m(UBj) > 3-nmo. This implies (11.10)
mo
< 3n
3 L L m(Bj) :S ~ n
J
3 If(x)1 dx :S ~ n
J
If(x)1 dx,
Bj
for all mo < m(F>,), which yields (11.4), with C
=
3n .
We now turn to the proof of Lemma 11.3. We can pick a compact K c U such that m(K) > mo. Then the covering C yields a finite covering of K, say AI' ... ' AN. Let B1 be the ball Aj of the largest radius. Throw out all At' which meet B 1 , and let B2 be the remaining ball of largest radius. Continue until {AI' ... ' AN} is exhausted. One gets disjoint balls B 1 , ... , B K in C. Now each Aj meets some Be, having the property that the radius of Be is 2'" the radius of A j . Thus, if Bj is the ball concentric with B j , with three times the radius, we have K
N
UB ~ UAe ~ K. j
j=l
e=l
This yields (11. 9) . There are more elaborate covering lemmas, due to Vitali and to Besicovitch (amongst others), which arise in measure theory. In appendices to this chapter, we discuss these covering lemmas. Having proved Theorem 11.1, we next establish a slightly stronger result. Proposition 11.4. Given f E .c1(JR n , dx), then
(11.11)
;~ m(~r)
J
If(y) - f(x)1 dy = 0, for a.e. x.
Br(X)
The proof of this goes just as the proof of Theorem 11.1, given the following maximal function estimate. Let (11.12)
-M(J)(x) = sup
r>a m
1 (B) r
J
Br(x)
If(y) - f(x)1 dy.
142
11.
Maximal Functions and A.E. Phenomena
We claim M satisfies an estimate of the form (11.4). Indeed, we have
M(f)(x) :S M(f)(x)
(11.13)
+ If(x)l,
so this is an easy consequence. Clearly, when (11.11) holds, we also have (11.2). Given f E £l(I~n, dx), a point x E ~n for which (11.11) holds is called a Lebesgue point of f. Thus the content of Proposition 11.4 is that almost every x E ~n is a Lebesgue point for any given f E £1 (~n, dx). In analogy with (11.12), define Mp(f)(x) by
~ Mp(f)(x)P = sup
(11.14)
r>O
m
J
1 (B) r
If(y) - f(x)IP dy,
Br(x)
for p E [1,00). Equivalently, (11.15) where dVr is dx/m(B r ), normalized to be a probability measure on Br(x). Using Minkowski's inequality, we obtain (11.16) Note that, if f E (11.17)
LP(~n),
m( {x
E
the estimate (11.4) applied to Ifl P yields
~n : M(lfIP)(x)
> AP}) :S
~ Ilfllir
Now, by (11.16), if Mp(f)(x) > A, then either M(lfIP)(x) If(x)IP> (A/2)P, so we have the maximal function estimate (11.18)
m( {x
~
E ~n : Mp(f)(x)
> (A/2)P or
C > A}) :S AP Ilfll~p.
Thus we have the following extension of Proposition 11.4. Proposition 11.5. Givenp E [1,00), f E (11.19)
:~ m(~r)
J
£P(~n,dx),
then
If(y) - f(x)IP dy = 0, for a.e. x.
Br{X)
The proof is again like that of Theorem 11.1. One uses the denseness of in LP(~n), for p E [1,00). When the limit is 0 in (11.19), we call x an LP-Lebesgue point for f. Co(~n)
11.
143
Maximal Functions and A.E. Phenomena
We next establish some results on differentiability almost everywhere of certain classes of functions. A function f defined on an open set n c lR?n, with values in lR?m, is differentiable at a point x E n provided there is a linear transformation A : lR?n ---+ lR?m such that (11.20)
If(x
+ y)
~
f(x) - AYI =
o(lyl),
as
Iyl
---+
0,
i.e., the ratio of the left side of (11.20) to Iyl tends to 0 as y ---+ O. If such A exists, it must be unique; we denote it by D f (x). If Tn > 1, then f is differentiable at x if and only if each component of f is differentiable at x, as is easily verified. Recall from Chapter 10 the Sobolev spaces Hl,p(lR?n), consisting of f E Vf E LP(lR?n). We showed that
LP(lR?n) with weak derivative (11.21)
We now establish the following.
Proposition 11.6. If f E H1,p(lR?n) and either n < p < CXJ or p = n = 1, then f is differentiable at almost every x E lR?n. More precisely, if x is an LP -Lebesgue point for V f, then fis differentiable at x and D f (x) = V f (x). Proof. Under the hypotheses on
(11.22)
f,
it follows from (10.20) that
1 If(x+y)-f(x)I 0, set (11.28)
R(x, y) = f(y) - f(x) - D f(x)(y - x)
and (11.29)
1]o(x)
= sup {IR(x, y)I/lx - yl : 0 < Ix - yl < 8}.
We know that Tlo(x) -----+ 0 as 8 -----+ 0, for each x E K 1 . By Egoroff's Theorem, there exists compact K C Kl such that m(Kl \ K) < c:/2 and 1]0 -----+ 0 uniformly on K. Thus f = flK' f# = DflK is a pair satisfying the hypotheses of Whitney's Extension Theorem; see Appendix C. That theorem produces a function g E C1(JR n ) satisfying (11.27). We return to a look at averaging operators of the form (11.1), but this time applied to finite (signed) measures in JRn : (11.30)
Using the Lebesgue-Radon-Nikodym Theorem, we can write J-l = fm with f E .c 1 (JRn , dx) and 1I ~ m. Thus we are left with the analysis of 1I ~
(11.31)
+ 1I,
m.
We will establish Proposition 11.10. If 1I is a finite measure on JRn which is purely singular with respect to Lebesgue measure, i. e., 1I ~ m, then lim Arll(X)
(11.32)
r---+O
= 0,
for m-a.e. x.
Proof. There is no loss in assuming 1I is positive. Xc, lI(X) = m(X C ) = O. Given .x > 0, set (11.33)
F>..
=
{x EX: lim sup IArll(X) I > r---+O
Write JRn
=
X U
.x}.
We want to show that m(F>..) = o. We will use an argument not very different from the proof of Theorem 11.2.
146
11.
Maximal Functions and A.E. Phenomena
Given E > 0, there exists an open U ~ X such that v(U) < E. Now, for each x E F).., there is a ball Bx c U, centered at x, such that v(Bx) > Am(Bx). Thus F).. c Bx = V c U.
U
xEF),.
By Wiener's Covering Lemma, if mo < m(V), there exist disjoint B x !, BXJ such that L m(Bxj) > 3-nmo. Thus we have
... ,
(11.34) Hence
(11.35) for all
E
m(F)..) :S m(V) :S
3n
--:\E,
> O. This shows m(F)..) = 0, and the proposition is proved.
The original work of Lebesgue on the one-dimensional case of Theorem 11.1 was done in order to extend the Fundamental Theorem of Calculus beyond the formulation given in Theorems 1.6-1.7 to the two results we now state. The following result extends Theorem 1.6. Proposition 11.11. If f E Ll(JR.), the function
(11.36)
g(x) =
[~f(Y) dy
is differentiable at almost every x E JR., and
(11.37)
Dg(x) = f(x),
a.e. x E R
Proof. It follows from (10.45) that the weak derivative of 9 exists and in fact 01g = f. Then (11.37) is the n = 1 case of Proposition 11.6.
There is also an extension of Theorem 1.7, which we state next. In the generalization, we replace the hypothesis that G be of class C 1 by the hypothesis that G be absolutely continuous. By definition, this means that, for every E > 0, there is a 6 > 0 with the property that, for any finite collection of disjoint intervals (a 1 , bI ), ... , (aN, bN ),
(11.38) Clearly (by Theorem 1.7), if Gis C 1 , with IG'(x)1 :S K, or, more generally, if G is Lipschitz, with Lipschitz constant K, then (11.38) holds, with 6 = ElK.
11.
147
Maximal Functions and A.E. Phenomena
Proposition 11.12. Assume that C : lR -+ lR is absolutely continuous. Then C is differentiable almost everywhere, and DC E Ll (lR). Furthermore, for -00 < a < b < 00, (11.39)
lb
DC(x) dx = C(b) - C(a).
For the proof it is convenient to use material that will be developed in Chapter 13. As will be seen in Exercise 8 of Chapter 13, the hypothesis that C be absolutely continuous implies that alC = f E Ll(lR). Again, by Proposition 11.6, we have DC = f a.e. Hence, by (10.47), for some constant c,
C(x) = c +
lXoo DC(y) dy.
Since both sides are continuous in x, this a.e. identity is valid everywhere. Then (11.39) is an immediate consequence. Another part of Exercise 8 in Chapter 13 will be that, if Ll(lR), then C is absolutely continuous.
al C =
f
E
llA. The Vitali Covering Lemma There is a covering lemma that is a bit more elaborate than the Wiener Covering Lemma (Lemma 11.3), known as the Vitali Covering Lemma, which has widespread use in analysis, and we discuss it in this appendix. It is natural to divide the result into two parts, each of which is occasionally called the Vitali Covering Lemma. The first part does not mention measure. Lemma llA.1. Let C be a collection of closed balls in]Rn (with positive radius) such that diam B < Co < 00 for all BEe. Then there exists a countable family :F of disjoint balls in C with the following property:
(llA.1)
Every ball B in C meets a ball in :F with at least half the radius of B,
and hence
(llA.2)
U B~ U B, BEF
where
B is
BEe
a ball concentric with B with five times the radius.
11.
148
Maximal Functions and A.E. Phenomena
Proof. Set
Let Fl be a maximal disjoint collection of balls in Cl. Inductively, let Fk be a maximal disjoint set of balls in
{B E Ck : B disjoint from all balls in F 1 , ... , Fk- d. Then set F
= UFk. For such F, (llA.l) holds, and then (llA.2) follows.
The second part involves measure.
Lemma llA.2. Let A be a subset ofITf.n, and let C be a collection of closed balls centered at points of A such that for each a E A, E > 0, there is a ball Br (a) E C of radius r < E. Then there exists a family F of disjoint balls in C such that (llA.3)
Proof. Without loss of generality, we can assume all balls in C have radius < 1. Apply Lemma llA.l and consider the resulting family F. We claim (llA.3) holds for this family. It suffices to show that (llA.3) holds with A replaced by AK = An BK(O), for each K < 00. In fact, let FK = {B E F: B C BK+2(0)}. We will show that (llA.4)
m ( AK \
U B) = O. BEJ"K
Write FK
=
{Bj : j E Z+}. Clearly Lj m(Bj)
B, with 5-fold dilation
m(B) :S Cm(B),
L
m(Bj) ~ 0,
as N ~
j>N
Hence it suffices to show that N
(llA.7)
00.
B,
(llA.5)
(llA.6)
N
Bj.
00.
Note that for any ball
11.
14Y
Maximal Functions and A.E. Phenomena
To see this, pick a E AK \ U.f=l Bj. Since the balls B j are closed, we can find some small ball Br(a) E C that does not meet any of the balls B j for j :S N. Now, by Lemma llA.1, Br(a) meets some B E F with radius at least r /2. Furthermore, such B must belong to F K, so B = B j , for some j > N, and hence Br(a) C Hj. Thus (llA.7) holds and Lemma llA.2 is proven. We now use Lemma llA.2 to produce a proof of Theorem 11.1, slightly different from the one given at the beginning of Chapter 11. We start the same way: given f E .c1(lR.n,dx) and E > 0, pick 9 E Co(lR. n ) such that Ilf - gll£1 < E, and note that
EA
(llA.8)
= {x
E
lR. n : lim sup IArf(x) - f(x)1 > A} r---+O
is unchanged if
f
(llA.9)
limsup IArU - g) - U - g)1 :S MoU - g)
is replaced by
f -
g. We also have
+ If -
gl,
r---+O
where we set (llA.lO)
. sup Mof(x) = 11m r---+O
m
1 (B) r
J
If(y)1 dy.
Br{x)
(llA.ll) Assuming this for the moment, we have from (llA.8)-(l1A.9) that (l1A.12)
EA C {x: MoU - g)(x) >
~} U {x : If(x) -
g(x)1 >
~},
so that by (l1A.ll) and Tchebychev's inequality (11.5), (l1A.13)
m(EA) :S
4E
4
:xllf - gll£1 < -:.\'
for all E > 0; hence m(EA) = 0 for all A > 0 and we have (11.2). This gives a second proof of Theorem 11.1, modulo the proof of (llA.11). To prove (llA.ll), we let A = {x E lR. n : Mof(x) > A}, and we let C consist of all balls of radius < 1 such that If(x)1 dx > Am(B). If A -=I 0,
IB
150
11. Maximal Functions and A.E. Phenomena
the hypotheses of Lemma llA.2 are satisfied, so we can take a family of disjoint balls :F = {Bj : j E Z+} C C so that (llA.3) holds. We hence have (llA.14)
m(A)
~ 2: m(Bj) < ~ L J
J
J
If(x)1 dx
~ ~llfllu,
Bj
so (llA.ll) is proven. REMARK
1. We do not elevate the estimate (llA.ll) to the status of a
theorem, parallel to Theorem 11.2, for the simple reason that once the dust clears and Theorem 11.1 is established, we see that Mof(x) = If(x)1 almost everywhere, so (llA.ll) carries no information beyond Tchebychev's inequality (11.5). REMARK 2. The Vitali Covering Lemma can be extended in scope considerably. Indeed, Lemma 11A.1 holds when IRn is replaced by a general metric space. As for Lemma llA.2, Lebesgue measure on IR n can be replaced by any locally finite Borel measure on IRn for which condition (llA.5), called the "doubling condition," holds, and again the setting can be extended to more general metric spaces. For details and further results and references, see the monograph [Rei] by J. Heinonen. We also mention the work of [ehel, extending the scope of Rademacher's Theorem.
lIB. The Besicovitch Covering Lemma The Besicovitch Covering Lemma is a bit more elaborate than the Vitali Covering Lemma, but it permits a generalization of Theorem 11.1 in which Lebesgue measure is replaced by a general locally finite Borel measure on IRn. As in Appendix 11A, the covering lemma comes in two parts, the first part not mentioning measure.
Lemma llB.1. There is an integer K(n) with the following property. If C is a collection of closed balls in IRn with radii in (0, RJ, for some R < 00, and if A is the set of centers of balls in C, then there exist subcollections Ok C C, 1 ~ k ~ K(n), such that each Ok is a countable collection of disjoint balls and K(n)
(llB.1)
Ac
U U B. k=l BE9k
We postpone the proof of Lemma llB.1. Now we show how it leads to its measure-theoretic counterpart.
11.
Maximal Functions and A.E. Phenomena
~iJ~
Lemma IlB.2. Let J.L be a Borel measure on IRn. Assume A c IR n , J.L*(A) < and C is a collection of closed balls (of positive radius), centered at points of A, such that
00,
(11B.2)
inf{r:Br(a) EC}=O,
VaEA.
Then there is a countable collection FcC, consisting of disjoint balls, such that (l1B.3)
Proof. Without loss of generality, we can assume each ball in C has radius :::; 1. Then Lemma 11B.1 applies and we have collections gl,"" gK(n) , each consisting of disjoint balls, such that the union of all these balls covers A. Hence there is one collection, say gk, and disjoint balls B 1,· .. , BL E gk such that (l1B.4)
J.L
* (A n
U B) > K(n) J.L*(A) . . +1 L
J
-
J=l
Since X = Uf=l B j is J.L*-measurable, we have by (5.6) that J.L*(A) = J.L*(An X) + J.L*(A \ X), so (11B.5) Now we can set A2 = A \ Uf=l B j and take C2 to be the collection of balls in C, centered at points in A 2 , disjoint from B 1 , ... , BL, which, by (l1B.2), contains balls with center at each point of A 2 . Then apply the argument above to (A 2 ,C2 ), obtaining disjoint balls B L + 1 , ... ,BL2 E C2 such that (l1B.6) Continuing in this fashion, we obtain F = {Bj holds.
:
j
2 1} such that (11B.3)
We use Lemma 11B.2 to produce a maximal function estimate more general than (l1A.12). To set it up, let J.L be a (positive) locally finite Borel measure on IRn. There is a maximal open set U c IRn such that J.L(U) = 0, namely the union of all balls Br (a) with a E A}, and let C consist of all balls B of radius < 1, centered at some point of A, such that III dJ.L > AJ.L(B). If A # 0, the hypotheses of Lemma llB.2 hold, so we can take a family of disjoint balls F = {Bj : j EN} c C such that (llB.3) holds. Hence
IB
(llB.9)
J.L*(A) ::; LJ.L(Bj )
]Rk such that v = w on K. We consequently deduce (12.47) from (12.45), provided
(12.48) We claim that the right side of (12.48) defines an element of LOO(K), independent of the choice of the Lipschitz extension v of w. To demonstrate this point, it suffices to establish Lemma 12.11. If f E Lip(]Rn), K V' f = 0, £n-a.e. on K.
c
]Rn
compact, and f =
°
on K, then
Proof. Let S be as in (12.43), with v = f. Then (12.45) holds. On the other hand, in this case S = {(x,O) : x E K}, so Hn(s) = £n(K). Hence
J+ (1
(12.49)
IV' f(x)12) 1/2 dx = £n(K).
K
This implies that (1 lemma.
+ IV' f(xW) 1/2 =
1, C~-a.e. on K, which proves the
To pass from E = K, compact, to general Borel E in Proposition 12.10, it suffices to show that, if F = E \ K is a Borel set for which £n(F) is small, then the Hausdorff measure H n ( { (x, w (x)) : x E F}) is small. We establish the following general result. Proposition 12.12. Let X and Y be locally compact metric spaces and F : X ----> Y a Lipschitz map satisfying dy(F(p), F(q)) :S Ldx(p, q). If SeX is a Borel set and HT(S) < 00, then F(S) is HT -measurable and
(12.50) Proof. Since we can pick compact So C S such that HT(S \ So) < since F(So) is compact in Y, it suffices to show that
(12.51) In fact, the hypotheses easily yield (12.52)
REMARK.
If SeX is a Borel set, F(S) need not be a Borel set.
E
and
12.
167
Hausdorff's r-Dimensional Measures
Now that Proposition 12.10 has been proved, it is desirable to consider the Hausdorff measure of images of sets in ]Rn under more general maps into ]Rm = ]Rn+k. For example, suppose 0 C ]Rn is open and u : 0 -----t ]Rm is a C 1 map which is one-to-one and has injective derivative at each point x E O. Then u is said to be an imbedding (loosely speaking), and M = u(O) acquires the structure of a Riemannian manifold, with metric tensor
(12.53)
gij(X)
= Du(x)ei' Du(x)ej =
~ aUg aUg ~ ~~. g
UXi UXj
In other words, we have the matrix identity
(12.54) Hence the volume element of ]\;1 is
(12.55)
dV
= ygdx = (det Du t Du)1/2 dx.
The same argument used to establish Corollary 12.8 establishes the first part of the following. Proposition 12.13. 1fO c then for M = u( 0) we have
]Rn
is open and u : 0 -----t]Rm is a C 1 imbedding,
Hn(M) =
(12.56)
J
Ju(x) dx,
o
where
(12.57) Furthermore, if K
(12.58)
c
0 is a compact set and S
Hn(s) =
= u(K), then
J
Ju(x) dx.
K
The second part of the proposition follows just as in (12.43)-(12.45). Suppose more generally that u : 0 -----t ]Rm is a Lipschitz map that is oneto-one and satisfies det(Du(x)tDu(x)) > 0 for £n-a.e. x E O. Assume 0 is bounded. Given E > 0, there is a compact K C 0 such that £n(o \ K) < E and Dul K is continuous, and also det(Du(x)tDu(x)) 2: 15 > 0 for x E K. If v is a C 1 extension of ul K , then det(Dv(x)tDv(x)) 2: 15/2 > 0 on a neighborhood U of K. We claim that also v : U -----t ]Rm is one-to-one if U is a small enough neighborhood of K, and we pause to establish this.
168
12.
Hausdorff's r-Dimensional Measures
Lemma 12.14. Suppose 0 c jRn is open, v : 0 ---+ jRm is C 1, and K c 0 is compact. If v maps K one-to-one into jRm and Dv(x) is injective for each x E K, then v maps some neighborhood U of K one-to-one into jRm. Proof. Let Uv be a family of neighborhoods shrinking to K. If there are xv, Yv E Uv such that Xv i- Yv and v(xv) = v(Yv), then Xv and Yv must have limit points x, y E K. Furthermore v(x) = v(y), so it is not possible that x i- y. But if x = y E K and Dv(x) is injective, then v must be an imbedding on a neighborhood of x, by the Implicit Function Theorem, and this contradicts the convergence of distinct Xv and Yv to x.
°
Therefore (12.57) applies to S = v(K) = u(K). Letting c ---+ and arguing as in the proof of Proposition 12.9, we conclude that, with M = u( 0), (12.56) continues to hold. We have Proposition 12.15. The conclusions of Proposition 12.13 continue to hold as long as u : 0 ---+ jRm is Lipschitz, one-to-one, and
(12.59) for en-a.e. x E 0, where J u is defined a.e. on 0 by (12.57). It is now within reach to produce a more definitive result. Theorem 12.16. If E c jRn is a Borel set and u : E and Lipschitz, then r = u(E) is Hn-measurable and
(12.60)
Hn(r) =
---+ jRm
is one-to-one
J
Ju(x) dx.
E
°
Proof. Write E = El U E 2 , a disjoint union of Borel sets, such that J u > a.e. on El and Ju = 0 a.e. on E 2 . It suffices to show that (12.60) holds with E replaced by El and r by r 1 = u(Ed and that, for r 2 = u(E2 ), H n (r 2 ) =
o.
Chopping El into a countable number of pieces if necessary, we can assume en(El) < 00. Given c > 0, pick compact K1 eEl such that en(El \ K 1 ) < c, DulKl is continuous, and Ju ~ Ii > 0 on K 1 . Then the argument, involving Lemma 12.14, used to extend (12.54), works here to yield
Hn(u(KI))
=
J
Ju(x) dx.
Kl
Taking c E1.
---+
0 and using Proposition 12.12, we have the desired result for
169
12. Hausdorff's r-Dimensional Measures
To analyze u on E2, we use the following trick. Given e > 0, define -+ jRm+n by vc:(x) = (u(x),eX). Then Vc: is Lipschitz and J Ve = en a.e. on E 2 . By the first part of the argument, we have that
Vc: : E2
1-{n (Vc:(E2)) = en Cn (E2). On the other hand, r 2 = U(E2) is the image of vc:(E2) under a projection, and Proposition 12.12 implies 1-{n(r2) ~ 1-{n (vc:(E2)). Hence 1-{n(r2) ~ en cn(E2) for all e > 0, so 1-{n(r2) = O. This proves the theorem. We have concentrated on the Hausdorff measures 1-{T for r = n E Z+. There are also sets for which it is interesting to consider 1{T for r ~ Z. Such sets are said to have nonintegral Hausdorff dimension, where the Hausdorff dimension of a nonempty subset 5 of a metric space X is Hdim 5 = inf {r 2: 0 : 1{T(5) = O}
(12.61) Note that, if 0
= sup{r 2: 0: 1{T(5) > O}. ~
rl
Sdim S. We obtain (12.71) as follows. Since HT(bSj ) = bT1fT(Sj), the hypothesis of self-similarity implies kb- T1-£T(S) = HT(S). Granted that HT(S) is neither nor 00, this implies that k = bT , or r = (log k) / (log b), as asserted.
°
If HT(S) does not lie in (0,00) for any r 2: 0, we cannot apply (12.71), and the identity Hdim S = Sdim S can fail. To give an example, let Q = Q n [0,1). We can write Q = Ql U Q2where Ql = Q n [0,1/2) and Q2 = Q n [1/2,1). Consequently, Sdim Q = 1, but Hdim Q = 0.
(12.72)
In fact, the Hausdorff dimension of a countable set is always zero. Here is another tool for estimating the Hausdorff dimension of a set. Proposition 12.19. Let KeIRn be compact, and take a E (0,00). Assume there is a positive Borel measure J1, # 0, supported on K, such that
(12.73)
Then Ha(K) = 00, so Hdim K 2: a. Proof. We will assume Ha(K) < 00 and show that the integral in (12.73) must be +00. To do this, consider the set
(12.74)
E
= {x
E K : limsup r- aJ1,(BT(x)) > o}. T->O
Below we will show that if Ha(K) < 00, then
(12.75)
J1,(K \ E)
= 0.
For now we assume this and proceed with the proof.
°
If x E E, there is a sequence rj ~ such that J1,(BTj (x)) 2: Crj, for some C > 0. We can assume J1,( {x}) = 0, since otherwise the integral in (12.73) is dearly infinite. Then there exists qj E (0, rj) such that
(12.76)
12.
173
Hausdorff's r-Dimensional Measures
Passing to a subsequence, we can assume disjoint. Then
rj+l
0, so the integral in (12.73) is
(12.78)
This proves Proposition 12.19, modulo the proof of the next lemma. Lemma 12.20. Given a compact K Borel measure f.l supported on K, (12.79)
F
=
c
]Rn
with 1{a(K)
{x E K: lim sup r-af.l(Br(x))
= O}
0, set
If {Uj } is a countable cover of Fcp by sets of diameter S J S p and each Uj contains a point of Fcp , then there exist balls B j , centered in Fcp , such that B j ~ Uj and diam B j S 2 diam Uj S 2p. Then (12.80) j
j
(12.81 ) This gives (12.79). Proposition 12.19 has the following counterpart.
j
174
12. Hausdorff's r-Dimensional Measures
c lRn and a E (a, 00), if for each nonzero positive Borel measure J-l supported in K one has
Proposition 12.21. Given a compact K
JI
(12.82)
dJ-l(x) dJ-l(Y) = +00
Ix -- yla
'
then 'Jtb(K) = a for all b > a, so Hdim K ~ a. We will not prove Proposition 12.21; see [Fal], p. 78. In Chapter 16, we will use Proposition 12.19 to prove that for almost every Brownian path win lR n , if n 2: 2, T E (a, 00), Hdimw([a, TJ) = 2.
(12.83)
More precisely, Proposition 12.19 will be used to prove Hdimw([a, TJ) 2: 2. The proof that Hdimw([a, TJ) ~ 2 will not use Proposition 12.21, but rather the result of Exercise 9 below. When the Hausdorff dimension of a set S c lRn is not equal to its topological dimension, following B. Mandelbrot, we call S a "fractal." The Cantor sets K(iJ), being totally disconnected, have topological dimension zero, hence they are fractals. By (12.83), Brownian paths in lRn are also fractals, when n 2: 2. Many other examples of fractals, a number of which are self-similar, together with speculations on their role in the description of nature, can be found in [Mdb]. Further measure-theoretic results on fractals can be found in [Fal]. REMARK. More generally than r-dimensional Hausdorff measure, one can take any monotone, continuous function
[a, 00), satisfying a, and set
h~,8 (S) = inf {L 0 lim h'!'r, 8 (S), y,
obtaining a metric outer measure. Then (12.85)
h'P(S) =
h~(S),
for S h~-measurable,
defines a measure. An example is (12.86)
d(x,y) as j
--'> 00,
for each x, y E X. Let these metrics determine Hausdorff r-dimensional measures H d , H d. Show that, for any Borel set SeX, J lim sup HdJ (S) ::; Hd(S), . ]---"=
In particular, if (12.88) is sharpened to
(12.89)
dj(x, y) '\, d(x, y),
then
(12.90)
HdJ (S) '\, Hd(S),
Hint. With A = bja, show that, for any h;,>..6,dj (S) ::; h;,6,d(S)
+ E,
E
> 0 and {) > 0, for j large.
3. Give a detailed demonstration of the first inequality in (12.8), stating that a Steiner symmetrization of a set S C ]Rn does not increase diameter. 4. Let F : ]Rn --'> ]Rn be a C 1 diffeomorphism. Show that Hk and F*H k , defined by F* Hk (S) = 1tk (F- 1 (S) ), are mutually absolutely continuous, but if 0 < k < n, there does not exist a measurable function g such that F*H k = gHk, unless DF(x) is everywhere a scalar multiple of an isometry. How does this relate to the Radon-Nikodym Theorem?
176
12.
Hausdorff's r-Dimensional Measures
5. Give an elementary proof of Lemma 12.11 when f E c1(]Rn). Hint. If Xo E K and \7 f(xo) i= 0, use the Implicit Function Theorem to describe the zero set of f near Xo.
6. With 1{B as in (12.4), show that, for Borel sets A c ]Rn, H'B(A) 1{n(A)
=
£n(A). Hint. Show that the proof of Lemma 12.3 actually yields 1{'B(U) < £n(u), for open U c ]Rn. Note. The isodiametric inequality is not needed in the argument (otherwise parallel to that giving (12.11)) that £n(A) ::; 1{'B(A).
Exercises 7-8 deal with exterior normals. Let 0 c ]Rn be an open set, with closure 0, and let p E 80. We say a unit vector N is a unit exterior normal to 80 at p provided
(12.91)
lim m ( {x EO: (x - p) . N r~O
> 0, j x - p j < r}) =
°
m(Br)
and
(12.92)
. m({xEOc:(x-p)·N 0, where h'P is defined as in (12.84)-(12.85), with
( 1)-1 '
lR.
is positive provided (13.3)
IE C(X), I 2: 0 ==} o:(f) 2:
o.
Clearly, if fL is a (positive) finite Borel measure on X, then (13.4)
o:(f) =
.I I
dfL
is a positive linear functional. We will establish the converse, that every positive linear functional on C(X) is of the form (13.4). It is easy to see that every positive linear functional 0: on C(X) is bounded. In fact, applying 0: to I(x) - a and to b - I(x), we see that, when a and b are real numbers
(13.5)
IE C(X), a ~
I
~ b ==} ao:(l) ~ o:(f) ~ bo:(l),
-
179
um
13.
Radon Measures
so
laU)1 s AIIIII,
(13.6)
A = a(1).
To begin the construction of J-l, we construct a set function J-lo on the collection 0 of open subsets of X by
J-lO(U) = sup {aU) : I --< U},
(13.7) where we say (13.8)
1--< U
~
I
E C(X), 0
sis 1,
Here, supp I is the closure of {x : I (x) we set, for any E c X,
(13.9) Of course, J-l*(U)
i= O}.
and supp
leU.
Clearly J-lo is monotone. Then
J-l*(E) = inf {J-lo(U) : E CUE O}.
= J-lo(U) when U is open.
Lemma 13.1. The set function J-l* is an outer measure.
Proof. By Proposition 5.1, it suffices to show that (13.10) so that we have an analogue of (5.5). Suppose f --< U. We need to show that aU) S L J-lo(Uj ). Now, since supp f = K is compact, we have K C U1 U ... U Ug for some finite C. We claim there are gj --< Uj , 1 S j S C, such that L gj = 1 on K. Granted this, we can set fj = f gj. Then fj --< Uj, so aUj) S J-lo(Uj ). Hence (13.11) as desired. Thus Lemma 13.1 will be proved once we have
Lemma 13.2. If K c X is compact, Uj are open, and K C U1U·· ·UUg then there exist gj --< Uj such that L gj = 1 on K.
= V,
Proof. Set UHl = X \ K. Then {Uj : 1 S j S C+ 1} is an open cover of X. Let {gj : 1 S j S C+ 1} be a partition of unity subordinate to this cover. (See Exercise 9 at the end of this chapter.) Then {gj : 1 S j S C} has the desired properties. Now that we know J-l* is an outer measure, we prepare to apply Caratheodory's Theorem.
181
13. Radon Measures
Lemma 13.3. The outer measure p,* is a metric outer measure. Proof. Let 8 j eX, and assume
(13.12) Take set
~
> O. Given U
:=J 8
Uj
= 81 U 82,
= Un {x
U open, such that p,o(U) ~ p,*(8) +~,
EX: dist(x, 8j )
< c}.
It follows that
(13.13) Now, whenever which implies
I -
0, which, together with subadditivity of p,*, yields the desired identity p,*(8) = p,*(8I) + p,*(82 ). It follows from Proposition 5.8 that every closed set in X is p,* -measurable Hence, by Theorem 5.2, every Borel set in X is p,*-measurable, and the restriction of p,* to Q3(X), which we denote p" is a measure. We make a few useful comments about p,. First, in addition to (13.7), we have
(13.15)
p,(U)
= sup {a(J) : I
~ U},
for U open, where
(13.16)
I
~ U
{:::=}
I
E
C(X), 0 ~
I
~ 1,
1=0 on X \
u.
I
U. Then set
Ij =
To see that (13.7) and (13.15) coincide, take w here ~j (s) is defined by 0 ~j(s)
=
for
2 2s - -:J
8
o~
~
s
1
~-:-,
J
1
2 for -,:= K j , so Lemma 13.4 gives the first inequality of (13.24). Summing (13.23) and (13.24), we have
(13.25)
Hence
(13.26) Letting N
-----t 00,
we have (13.20).
Only the uniqueness of J1 remains to be proved. To see this, let ). be a positive Borel measure on X such that
a(f) =
(13.27) for all
I
E
J
I d)'
C(X). Let K C X be compact, and apply this to
(13.28) By the Monotone Convergence Theorem, we have J Iv dJ1 '\. J XK dJ1 and J Iv d)' '\. J XK d)', so a(fv) '\. J1(K) and a(fv) '\. )'(K). Hence J1(K) = ).(K) for all compact K. Now, by (5.60), for every positive Borel measure ). on a compact metric space X, we have
(13.29)
E
E ~(X) =?
).(E) = sup {)'(K) : K
C
E, K compact}.
This proves uniqueness. Generally, if X is a compact Hausdorff space and), is a finite (positive) measure on ~(X), ). is said to be regular if and only if (13.29) holds. The implication of Exercises 10-13 of Chapter 5 is that every finite Borel measure is regular when X is a compact metric space. If X is compact but not
13. Radon Measures
184
metrizable, a finite measure on IB(X) need not be regular. The generalization of Theorem 13.5 to this case is that, given a positive linear functional a on C(X), there is a unique finite regular Borel measure p, such that (13.20) holds. (Note that if A is any finite measure on IB(X), then (13.27) defines a positive linear functional on C(X), which then gives rise to a regular Borel measure.) For this more general case, the construction of p,* is the same as was done above in (13.7)-(13.9), but the proof that p,* yields a regular measure on IB(X) is a little more elaborate than the proof given above for compact metric spaces. Treatments can be found in [Fol] and [Ru]. We want to extend Theorem 13.5 to the case of a general bounded linear functional
(13.30)
w:
C(X)
-t
R
We start with an analogue of the Hahn decomposition. Lemma 13.6. If w is a bounded (real) linear functional on C(X), then there are positive linear functionals a± on C(X) such that
(13.31)
Proof. We first define a+ on
C+ (X) = {fEe (X) : f 2: O}.
(13.32) For
f
E
C+(X), set a+(f) = sup {w(g) : 9 E C+(X), 0::; 9 ::; f}.
(13.33)
The hypothesis that w is bounded implies Iw(g)1 ::; Kllgll ::; Kllfll when so
o ::; 9 ::; f, (13.34) where K
= Ilwll. Clearly, for c E JR,
(13.35) Now, suppose iI, 12 E C+(X). If 9 E C+(X) and 0 ::; 9 ::; iI + 12, we can writeg = gl+g2 withg E C+(X) and 0 ::; gj::; h· Justtakeg1 = min(g,iI). Hence
(13.36)
185
13. Radon Measures
We claim that a+ has an extension to a linear functional on C(X), which would necessarily be positive. In fact, given f E C(X), write (13.37) A given f E C(X) has many such representations as f = fr - 12; showing that a+(f) is independent of such a representation and defines a linear functional on C(X) is a simple application of (13.36); compare the proof of Proposition 3.7. Note that, if we take fr = f+ = max(f, 0) and 12 = f- = max( - f, 0), we see that (13.38) Finally we set a - = a + - w. It remains only to show that f E C+ (X) ::::} a-(f) :?: 0, i.e., a+(f) :?: w(f). But that is immediate from the definition (13.33), so the lemma is proved. We can combine Lemma 13.6 and Theorem 13.5 to prove the following, known as the Riesz Representation Theorem. Theorem 13.7. If X is a compact metric space and w is a bounded (real) linear functional on C(X), then there is a unique finite signed measure P on ~(X) such that (13.39)
w(f)
=
Jf
dp,
X
for all f E C(X). Furthermore, (13.40)
Ilpll =
Ipl(X) =
Ilwll,
so there is an isometric isomorphism (13.41)
C(X)'
~ 9Jt(X).
Here, 9Jt(X) denotes the linear space of finite signed measures on ~(X), with norm given by the first identity in (13.40). This is also known as the space of finite Radon measures on X. For the proof, write w = a+ - a-, as in (13.31), take finite positive measures J.l± on ~(X) so that a±(f) = f dJ.l±, and set p = J.l+ - J.l-. Thus we have the identity (13.39).
J
186
13.
Radon Measures
We need to prove (13.40). Let p = p+ - p- be the Hahn decomposition of p, so p+ -1 p- and
w(J) =
(13.42) for all
f
E
J
fdp+ -
J
fdp-,
C(X). Consequently
(13.43) Iw(J)I::; Ilfllp+(X) + Ilfllp-(X) = Ilpll ·llfll, so we have Ilwll ::; Ilpll. To prove the reverse inequality, let 15 > O. Suppose p± are supported on disjoint Borel sets X±. Let K± be compact sets in X± such that p±(K±) ::::: p±(X) - 15. We have K+ n K_ = 0, so say dist(K+, K_) = E > O. Let u± = {x : dist(x, K±) < E/4}, so U+ n U_ = 0. Using a simple variant of (13.28), we can construct --< U± such that XK±. Hence, as v --- 00, (13.44)
lR belongs to C* (Y) if, for any 0, there exists a compact KeY such that lu(y)1 < lR be bounded and right-continuous. Show that 1 has the properties (a )-( c) of Exercise 4 if and only if there exists C < 00 such that, for any finite set of real numbers Xo < Xl < ... < Xe,
e
(13.55)
I: If(Xj) -
f(Xj-dl ~ C.
j=l
Hint. To prove one implication, given v E Z+, set fl/(x) :2 v, fl/(x) = fe-v) if X ~ -v, and
f{v) if
X
fl/(x) = f( -v + 2-1/ j) if - v
+ 2-1/ j
~ X
< -v + 2-I/(j + 1), o ~ j < 2v21/.
Show that f 1/ --> f and {ad1/ : v E Z+} is a bounded set of measures on lR+. Consider weak* limits. One says that f has bounded variation on lR if this property holds, and one writes f E BV(lR). 7. Let f E LOO(lR). Show that f is equal a.e. to an element of BV(lR) if and only if you can write f = gl - g2 a.e., with gj E £<Xl(lR) monotone / . In particular, if 9 E LOO(lR) is monotone / , then (hg is a positive measure. Reconsider Exercise 1 of Chapter 5, on Lebesgue-Stieltjes measures, in this light. Hint. Given f E BV(lR) , apply the Hahn decomposition to the signed measure J.L arising in Exerise 4. 8. Let f : lR --> lR be bounded and continuous. One says f is absolutely continuous provided that, for every c > 0, there is a 8 > 0 with the property that, for any finite collection of disjoint intervals, (aI, bl), ... , (aN, bN), (13.56) Show that the following are equivalent: (a) f is absolutely continuous. (b) ad = 9 E Ll{lR). (c) f(x) = r~oo g(y) dy, 9 E Ll(lR). Hint. If (a) holds, first show that 1 has bounded variation and use Exercises 5-6 to get ad = J.L. Then show J.L is absolutely continuous with respect to Lebesgue measure. Compare Exercises 7-9 in Chapter
190
13.
Radon Measures
10 and also Exercise 1 in Chapter 11. 9. Let {Ul,"" U(!+l} be an open cover of a compact metric space X. Show that there exist open sets Vj, j = 1, ... , £ + 1, covering X, such that Vj C Uj. Let hj(x) = dist(x, X \ Vj). -
Show that hj E C(X), supp H j C Vj C Uj, and that h C(X) is > 0 on X. Deduce that
=
£+1
L: j =l hj E
form a partition of unity of X, subordinate to the cover {U1 , . .. ,U(!+d. Hint. Set Hj(x) = dist(x,X \ Uj). Show that H = L:Hj > 0 on X, HE C(X), hence H 2: a > O. Set Vj = {x E Uj : Hj(x) > aj(£+2)}. 10. The countable infinite product Z = I1j~l {O, I} is compact, with the product topology, and metrizable (cf. Appendix A). Let A C C(Z) consist of continuous functions depending on only finitely many variables, so an element j E A has the form j(x) = j(Xl,' .. , Xk), for some k E Z+. For such an j, set
XjE{O,l},l:::;j:::;k
Show that
1(1), where 1(1) is the Riemann integral of j, discussed in Chapter 1 (for n = 1) and in the second exercise set at the end of Chapter 7 (for n > 1). Show that the measure on R produced by Theorem 13.5 coincides with Lebesgue measure (on the Borel subsets of R), as constructed in Chapter 2 (for n = 1) and in Chapter 7 (for n > 1). 13. Let X be a compact Hausdorff space. Show that if C(X) is separable, then X is metrizable. Hint. Define
0, pick 9 E L2(X, JL) such that Ilf - gllLP < E. Then
Hence
(14.11)
limsup IIAnf - AmfilLP ::; 2E,
\IE> O.
m,n---too
This implies the sequence (An!) is Cauchy in LP(X, JL), for each f E V(X, JL). Hence it has a limit; call it Qf· Clearly Qf is linear in f, IIQfllLP ::; IlfIILP, and Q f = P f for f E L 2 (X, JL). Hence Q is the unique continuous extension of P from L2(X, JL) to LP(X, JL) (so we change its name to P). Note that p2 = P on V (X, JL), since it holds on the dense linear subspace L 2 (X, JL). Proposition 14.3 is proven.
Note that P = p*. It follows that P : LP(X, JL) ---> LP(X, JL) for all p E [1,00]. We will show in Proposition 14.7 that (14.9) holds in LP- norm for p < 00. The subject of mean ergodic theorems has been considerably extended and abstracted by K. Yosida, S. Kakutani, W. Eberlein, and others. An account can be found in [Kr]. REMARK.
Such mean ergodic theorems were complemented by pointwise convergence results on Amf(x), first by G. Birkhoff. This can be done via estimates of Yosida and Kakutani on the maximal function
(14.12)
A# f(x)
= sup Amf(x) = sup At[ f(x), m~l
n~l
where
(14.13)
At[ f(x) =
sup
Amf(x).
l::;m::;n
We follow a clean route to such maximal function estimates given in [Gar].
14.
196
Lemma 14.4. With Am given by (14.2)-(14.4) and
(14.14)
I
Ergodic Theory
E Ll(X, J.L), set
En = {x EX: At!: I(x) 2:: O}.
Then
JI
(14.15)
dJ.L 2:: O.
En
Proof. For notational convenience, set
SkI = kAki = 1+ TI
+ ... + T k- 1 I,
l\Ihi = kAt I = sup Sd· 1~f9
For k E {1, ... ,n}, (Mnf)+ 2:: SkI, and hence (because T is positivity preserving) Hence
I 2:: SkI - T(Mnf)+, for 1:S; k :s; n, this holding for k 2:: 2 by the argument above, and trivially for k the max over k E {I, ... , n} yields
=
1. Taking
(14.16)
Integrating (14.16) over En yields
(14.17)
=
J J
En
X
X
(Mnf)+ dJ.L -
x
2::
(Mnf) + dJ.L -
J J
T(Mnf)+ dJ.L
T(Mnf)+ dJ.L = 0,
the first and second identities on the right because Mni 2:: 0 precisely on En, the last inequality because T(Mnf)+ 2:: 0, and the last identity by (14.3). This proves the lemma. Lemma 14.4 leads to the following maximal function estimate.
14.
197
Ergodic Theory
Proposition 14.5. In the setting of Lemma
(14.18)
14·4, one has, for each A> 0,
M({X EX: A;[f(x) 2 A})::::;
1
-:xllfll£1.
Proof. If we set En>. = {x EX: A'/f f(x) 2 A} = {x EX: A;[(f(x) - A) 2 O}, then Lemma 14.4 yields
(14.19)
Thus
Ilfll£1 2
(14.20)
Jf
dM 2 AM(En>.),
En).
as asserted in (14.18). Note that (14.21) so we have M(E>.) ::::; (14.22)
Ilfll£1/A.
Now we introduce the maximal function
A# f(x) = sup IAmf(x)1 ::::; A#lfl(x). m~l
We have (14.23) We are now ready for Birkhoff's Pointwise Ergodic Theorem. Theorem 14.6. If T and Am are given by (14.2)-(14.4), where r.p is a measure-preserving map, then, given f E Ll(X, M),
(14.24)
lim Amf(x)
m-+oo
= Pf(x),
M-a.e.
Proof. Given f E Ll(X, M), c > 0, let us pick Ilf - iI 11£1 : : ; c/2. Then use Lemma 14.1, with H
(14.25)
9 E Ker (I - T),
h
= (I - T)v,
iI E L2(X, M) such that = L2(X, M), to produce
IliI -
(g+h)IIL2::::;
c
2·
14.
198
Ergodic Theory
Here v E £2(X, f-l). It follows that
Ilf -
(14.26)
(g
+ h)ll£1
~
E,
and we have
Amf (14.27)
=
Amg + Amh
= 9 + ~ (v m
Clearly v(x)/m
-----t
0, f-l-a.e., as m
+ Am(f - 9 - h) Tmv) + Am (f - 9 -
-----t 00.
h).
Also
(14.28)
which implies Tmv(x)/m -----t 0, f-l-a.e., as m -----t 00. We deduce that for each A> 0, (14.29) f-l( {x EX: lim sup Amf(x) -lim inf Amf(x) > A})
= f-l ( {x EX: lim sup Am (f -
9 - h) - lim inf Am (f - 9 - h)
~ f-l( {x EX: A#(f -
>
2
~
:xllf -
~
2E -:x.
9-
9 - h)
> A})
~} )
hll£!
Since E can be taken arbitrarily small, this implies that Amf(x) converges as m -----t 00, f-l-a.e. We already know it converges to Pf(x) in £l-norm, so (14.24) follows. We can use the maximal function estimate (14.23) to extend Proposition 14.3, as follows. First, there is the obvious estimate (14.30) Now the Marcinkiewicz Interpolation Theorem (see Appendix D) applied to (14.23) and (14.30) yields (14.31) Using this, we prove the following.
14.
199
Ergodic Theory
Proposition 14.7. In the setting of Proposition 14·3, we have, for all p E [1,00),
fE LP(X,jL) ==::}Amf----tPf,
(14.32)
as m
----t
in LP-norm,
00.
Proof. Take p E (1,00). Given
f
E
LP(X, jL), we have
IAmf(x)l:S A#f(x),
A#f E LP(X,jL).
Since the convergence (14.24) holds pointwise p,-a.e., (14.32) follows from the Dominated Convergence Theorem. That just leaves p = 1, for which we rely on Proposition 14.3.
REMARK.
Since P* = P, it follows from Proposition 14.7 that
f E LP (X, p,) ====> A'!:nf
----t
P f,
weak* in LP(X, p,), for p E (1,00]. More general ergodic theorems, such as can be found in [KrJ, imply one has convergence in LP-norm (and p,-a.e.), for p E [1, 00). Of course if tp is invertible, then such a result is a simple application of the results given above, with tp replaced by tp-1. Having discussed the convergence of Amf, we turn to the second question raised after (14.4), namely whether the limit must be constant. So far we see that the set of limits coincides with Ker (I - T), i.e., with the set of invariant functions, where we say f E LP(X, p,) is invariant if and only if
f(x) = f(tp(x)),
(14.33)
p,-a.e.
We note that the following conditions are equivalent:
f f
f constant (p,-a.e.), E L2(X,p,) invariant::::} f constant (p,-a.e.), S E J invariant ::::} p,(S) = 0 or p,(S) = l.
(a) (14.34)
(b) (c)
Here we say S E (14.35)
E
L1(X, p,) invariant ::::}
J is invariant if and only if p,(tp-1(S)L.S) = 0, J satisfies (14.35), then U tp-k(S) ====> '1'-1(8) = 8 and p,(8L.S) = O.
where AL.B = (A \ B) U (B \ A). Note that if S E (14.36)
8=
n
j~O k~j
To see the equivalence in (14.34), note that if fELl (X, jL) is invariant, then all the sets S),. = {x EX: f(x) > A} are invariant, so (c)::::}(a). Meanwhile clearly (a)::::} (b)::::} (c). A measure-preserving map 'I' : X ----t X satisfying (14.34) is said to be ergodic. Theorem 14.6 and Proposition 14.7 have the following corollary.
14.
Proposition 14.8. If=
Note that the functions in (15.53) are uniformly bounded by V27f. Making use of (15.53), the Fourier transform identity (9.58), and the Dominated Convergence Theorem, we obtain, for each v E S(~),
.I
V
(15.54)
.I (~) d~ .I v(~);yg (~) d~ .I 11 (0 V9k
dl/qk ---->
=
v&)'g.
Since S(~) is dense in C*(~) and all these measures are probability measures, this implies the asserted weak* convergence in (15.47). Chapter 16 is devoted to the construction and study of a very important probability measure, known as Wiener measure, on the space of continuous paths in ~n. There are many naturally occurring Gaussian random variables on this space. We return to the strong law of large numbers and generalize Theorem 15.4 to a setting in which the iJ need not be independent. A sequence {fJ : j E N} of real-valued random variables on (X,~, p,), giving a map F : X ----> R= as in (15.17), is called a statzonary process provided the probability measure I/F on ~= given by (15.18) is invariant under the shift map (15.55) An equivalent condition is that for each k, n E N, the n-tuples {h,···, in} and {h, ... ,ik+n~d are identically distributed ~n-valued random variables. Clearly a sequence of independent, identically distributed random variables is stationary, but there are many other stationary processes. (See the exercises. ) To see what happens to the averages k~l L~=l iJ when one has a stationary process, we can follow the proof of Theorem 15.4. This time, an application of Theorem 14.6 and Proposition 14.7 to the action of on (~=, I/F) gives
e
1
(15.56)
k
k 2..:~j j=l
---->
P6,
I/F-a.e.
and in LP-norm ,
216
15.
Probability Spaces and Random Variables
provided (15.57)
6 E LP(JRoo,VF),
i.e.,
fl E LP(X,J1),
p E [1,00).
Here the map P : L2 (JR oo , VF) ----+ L2 (JRoo, VF) is the orthogonal projection of L2(JR oo , VF) onto the subspace consisting of B-invariant functions, which, by Proposition 14.3, extends uniquely to a continuous projection on V(JR oo , VF) for eachp E [1,00]. Since F: (X,'J,J1) ----+ (JR oo ,23(JROO ),VF) is measure preserving, the result (15.56) yields the following. Proposition 15.6. Let {fJ : ] E N} be a statzonary process, consistmg of fJ ELP(X,J1), withpE [1,00). Then 1
(15.58)
k
k LfJ
----+(P6)oF,
J1-a.e. andm LP-norm.
J=l
The right side of (15.58) can be written as a conddzonal expectatzon. See Exercise 12 in Chapter 17.
Exercises 1. The second computation in (15.39) is equivalent to the identity
Verify this identity. Hmt. Differentiate the identity
1:
e- sx2 dx = J1fs-l/2.
2. Given a probability space (X, 'J, J1) and A J E 'J, we say the sets A J , 1 ::; j ::; K, are independent if and only if their characteristic functions XA J are independent, as defined in (15.15). Show that such a collection of sets is independent if and only if, for any distinct il,··. ,iJ in {I, ... ,K}, (15.59) 3. Let fl, h be random variables on (X, 'J, J1). Show that independent if and only if (15.60)
fl and hare
5.
217
Probability Spaces and Random Variables
Extend this to a criterion for independence of Hznt. Write the left side of (15.60) as
h, ... , fk·
ff e-i(~lxl+6x2)dvh,12(Xl,X2) and the right side as a similar Fourier transform, using dvh x dv 12 . 4. Demonstrate the following partial converse to Lemma 15.1.
Lemma 15.7. Let hand 12 be random variables on (X, p,) such that 6h + 612 zs Gausswn, oj mean zero, Jar each (6,6) E ]R2. Then
E(h12) = 0
====}!l
and 12 are zndependent.
2::7
More generally, zJ ~JJJ are all Gausswn and zJ h, ... , fk are mutually orthogonal zn L2 (X, p,), then Jl' ... ,fk are zndependent. Hznt. Use E(e- i (6h+612)) = e-116h+612112/2,
which follows from (15.41).
Exercises 5--6 deal with results known as the Borel-Cantelli Lemmas. If (X,;g, p,) is a probability space and Ak E ;g, we set
nU 00
A = lim sup Ak = k-.oo
00
Ak,
£=1 k=C
the set of points x E X contained in infinitely many of the sets A k · Equivalently, XA = lim sup XA k • 5. (First Borel-Cantelli Lemma) Show that
L p,(Ak)
1(~) for each ~ E JR.
°
Vk ---> ,
weak* in
Hint. Show that for each E > there exist R, N E (0,00) such that vdJR \ [-R, R]) < E for all k 2: N.
220
15.
Probability Spaces and Random Variables
14. Produce a counterexample to the assertion in Exercise 13 when probability measures but r is not.
Vk
are
15. Establish the following counterpart to Proposition 15.2 and Theorem 15.4. Let {fJ : j E N} be independent, identically distributed random variables on a probability space (X, J, fJ,). Assume fj .2: 0 and fj dJL = +00. Show that, as k ---+ 00,
J
1
k
kL
fJ
----+
+00,
JL-a.e.
J=l
16. Given y E JR, t > 0, show that
C) _
Xt,y ( s > 0 and calculate
(Xs, Xth2('+!) = E(Xs . Xt) (16.19)
= =
J J
p(s, xdp(t -
S,
X2 - Xl) Xl . X2 dXI dX2
p(s, xdp(t - s, y) Xl . (y
+ Xl) dXI dy.
Now Xl . (y + xd = Xl . Y + IX112. The latter contribution is evaluated as in (16.15), and the former contribution is the dot product A(s) . A(t - s), where (16.20)
16.
Wiener Measure and Brownian Motion
So (16.19) is equal to 2ns if t > s > O. Hence, by symmetry,
(16.21)
(Xs, X t )L2('+l)
= 2n
min(s, t).
One can also obtain this by noting IX t - Xsl2 = IXt l2+ IXsl2 - 2Xs . X t and comparing (16.15) and (16.16). Furthermore, comparing (16.21) and (16.15), we see that
(16.22)
t
> s 2' 0
~
(X t
-
X s, Xs)U('+l) = O.
This result is a special case of the following, whose content can be phrased as the statement that if t > s 2' 0, then X t - Xs is independent of Xa for (J :s: S, and also that X t - Xs has the same statistical behavior as X t - s . For more on this independence, see the exercises at the end of this chapter and Chapter 17. Proposition 16.2. Assume 0 < Sl conszder junctzons on ~ oj the jorm
c/viJ
where
(16.33)
J
7/Jn(r) = (47f)-n/2
e-lyI2/4 dy :S a n r n e- r2 /4,
Iyl>r
as r
--* 00.
The relevance of the analysis of E( a, b, c) is that, if we set (16.34) F(k,c,5) = {w E
'+I: oscJ(w) >
410, for some J
c [O,k] n ((JJ+,f(J) :S
~},
16.
227
Wiener Measure and Brownian Motion
where €( J) is the length of the interval J, then
U{E(a, b, 2E) : [a, b]
F(k, E, 15) =
(16.35)
C
[0, kJ,
Ib - al
S;
~}
is an open set, and, via (16.30), we have
W(F(k,E,t5)) S;2k P(Edt5).
(16.36)
Furthermore, with FC(k, E, 0) \,f}o
(16.37)
=
\,f} \ F(k,
E,
0),
= {w: 'Ilk < OO,VE > 0,315 > = FC(k, E, 0)
nn
U
°
such that w E FC(k,E,O)}
k 0=1/1/ 0=1/11
is a Borel set (in fact, an Fao set, i.e., a countable intersection of Fa sets), and we can conclude that W(\,f}o) = 1 from (16.36), given the observation that, for any E > 0,
P(E~O) ~
(16.38)
o
0, as 0 -----+ 0,
which follows immediately from (16.32)-(16.33). Thus, to complete the proof of Proposition 16.3, it remains to establish the estimate (16.30). The next lemma goes most of the way towards that goal. Lemma 16.4. Gwen
such that tl/ - t1
S;
o.
A = {w
(16.39)
0 > 0, take v numbers t J E Q+, Let E,
E \,f} :
IW(t1) - w(tJ)1 >
E,
°
S;
for some J = 1, ... , v}.
Then (J6.40) Proof. Let
B = {w : IW(t1) - w(tl/)I > E/2}, Cj = {w : Iw(tj) - w(tl/)I > E/2}, D J = {w : Iw(td - w(tJ)1 > E and Iw(iI) - w(tk)1 S; E, for all k
(16.41)
Then A
C
B
U
S; j - I}.
U~=l (Cj n Dj), so 1/
(16.42)
W(A) S; W(B)
+L j=l
t1 < ... < tl/'
W(Cj n D j
).
228
16.
Wiener Measure and Brownian Motion
Clearly W(B) :::; p(c/2, J). Furthermore, we have
(16.43)
W(CJ nDJ ) = W(Cj)W(DJ ):::;
p(~,J)W(Dj),
the first identity by Proposition 16.2 (i.e., the independence of CJ and D j ) and the subsequent inequality by the easy estimate W(CJ ) :::; p(c/2, J). Hence (16.44) J
since the D J are mutually disjoint. estimate is independent of v.
This proves (16.40).
Note that this
We now finish the demonstration of (16.30). Given such tj as in the statement of Lemma 16.4, if we set (16.45)
E = {w: Jw(tj) -W(tk)J > 2c, for some j,k E [l,vJ},
it follows that (16.46) since E is a subset of A, given by (16.39). Now, E(a, b, c), given by (16.28), is a countable increasing union of sets of the form (16.45), obtained, e.g., by letting {tl' ... , tv} consist of all t E [a, b] that are rational with denominator :::; K and taking K / +00. Thus we have (16.30), and the proof of Proposition 16.3 is complete. We make the natural identification of paths w E !,po with continuous paths w : [0,00) ---> .!R.n. Note that a function cp on !,po of the form (16.11), with tJ E .!R.+, not necessarily rational, is a pointwise limit on !,po of functions in C#, as long as F is continuous on rr~ JR n , and consequently such cp is measurable. Furthermore, (16.12) continues to hold, by the Dominated Convergence Theorem. An alternative approach to the construction of W would be to replace (16.10) byi1 = rr{JRn: t E .!R.+}. With the product topology, this is compact but not metrizable. The set of continuous paths is a Borel subset of i1, but not a Baire set, so some extra measure-theoretic considerations arise if one takes this route, which was taken in [Nel]. Looking more closely at the estimate (16.36) of the measure of the set F(k, c, J), defined by (16.34), we note that we can take c = K JJ log(l/J), in which case (16.47) Then we obtain the following refinement of Proposition 16.3.
16.
229
Wiener Measure and Brownian Motion
Proposition 16.5. For almost all w E !f}o, we have, for each T < 00,
(16.48)
limsup
(Iw(s) - w(t)l- 8VOIog
Is-tl=8->O
l ):;
0,
s, t
E
[0, T].
Consequently, gwen T < 00, (16.49)
Iw(s) -w(t)l::; C(w,T)VOIogl,
8,t E [O,TJ,
wzth C (w, T) < 00 for almost all w E !f}o. In fact, (16.47) gives W(Sk) = 1 where Sk is the set of paths satisfying (16.48), with 8 replaced by 8 + 11k, since then (16.47) applies with K > 4, so (16.38) holds. Then k Sk is precisely the set of paths satisfying (16.48).
n
The estimate (16.48) is not quite sharp; P. Levy showed that for almost all w E!f}, with fL(O) = 2Jologl/o, · Iw(s) - w(t)1 -_ 1. 1lIn sup Is-tl->o fL(ls - tl)
(16.50) See [McK] for a proof.
Wiener proved that almost all Brownian paths are nowhere differentiable. We refer to [McK] for a proof of this. The following result specifies another respect in which Brownian paths are highly irregular. Proposition 16.6. Assume n 2: 2, and pzck T E (0,00). Then, for almost all w E !f}o, (16.51) w([O, T]) = {w(t) : 0 ::; t ::; T}
C]H.n
has Hausdorff dimenszon 2.
Proof. The fact that Hdimw([O, T]) ::; 2 for W-a.e. w follows from the modulus of continuity estimate (16.48), which implies that for each 0 > 0, w is Holder continuous of order 1I (2 + 0). This implies by Exercise 9 of Chapter 12 that J-i'(w([O, T])) < 00 for T = 2 + O. (Of course this upper bound is trivial in the case n = 2.)
We will obtain the estimate Hdim w( [0, T]) 2: 2 for a.e. w as an application of Proposition 12.19. To get this, we start with the following generalization of (16.16): for 0 < s < t,
E(cp(Xt - Xs)) (16.52)
=
=
JJp(s, xdp(t -
J
8,
X2 - xI)CP(X2 - xd dXl dX2
p(t - s, y)cp(y) dy
= (47f(t - s)) -n/2
J
e- 1y12 /4It-sl cp(y) dy.
16.
230
Wiener Measure and Brownian Motion
We now assume i.p is radial. We switch to spherical polar coordinates. We also allow t < .') and obtain
( 16.53)
E(n(Xt - X )) = r
.
3
An-I. (47Tlt _ SI)n/2
.10roo e-r2/4It-31 (n(r)r r
n- I dr ,
where A n- 1 = Area(Sn-I). We apply this to i.p(r) = r- a to get
E(IX t
-
Xsl- a) = Cnlt - sl- n/ 2
(16.54) =
where C n .a
0, p,#({x EX: j*(x) > A}) ::;
(17.13)
K
-:X'
K
=
sup a
Ilfallu.
Proof. Note that 0: < f3 =* Ilfnllu ::; Ilf611L1. It suffices to demonstrate (17.13) for an arbitrary finite subset {O:J} of A. Thus we can work with Jj = Jaj' 1 ~ j::; N, and fj = E(fkIJJ) for j < k. There is no loss in assuming fN 2:: 0, so all fj 2:: O. Now consider
SA
(17.14)
=
{x EX: j* (x) > A}
=
{x EX: some fj (x) > A}.
There is a pairwise disjoint decomposition N
(17.15)
SA
=
U SAj, j=1
SAj
=
{x: fJ(x) > A but fe(x) ~ A for I! < j}.
236
17.
Conditional Expectation and Martingales
Note that S)..j E ';Sj. Consequently, we have
J
N
fN df-J,N
~
=
fN df-J,N
J- 1s AJ.
SA
(17.16)
J
J
N
=
L
fj df-J,j
J=lS Aj
N
~
L Af-J,j(S)..J)
=
Af-J,N(S>J.
J=l This yields (17.13) in this special case, and the proposition is hence proved.
REMARK
1. If all fn ~ 0, then
REMARK
2. If (fn :
K ::;
0: E
Ilfnll£1 == K.
A) has a final element fELl (X, ';S#, f-J,#), then
Ilfll£1.
The maximal inequality of Proposition 17.2 yields the following (increasing) Martingale Convergence Theorem, which contains Proposition 17.1 as a special case. Proposition 17.3. Let (fn : 0: E A) be a martingale over ';Sn, wIth final element fELl (X, ';Sw, fJl Assume A is countable and totally ordered. Assume ';Sw = CJ(UnEA ';Sn). Then
fn(x)
(17.17)
----t
f(x)
f-J,-a.e.,
as
0: /
w.
Proof. We certainly have (17.17) when f E UnEAL1(X,';Sn,f-J,) = V. But V is dense in L1 (X, ';Sw, f-J,). (See Exercise 6 of Chapter 9.) Thus (17.17) follows for general f E L 1(X,';Sw,f-J,), by the same sort of argument as used in the proof of Theorem 11.1, via the maximal function estimate (17.13). In detail, given f E L 1(X,';Sw,f-J,), pick c > and g E V such that
f = g + h with each A > 0,
Ilhll£1 < E)..
(17.18)
°
c. With obvious notation, fn
=
{x EX: lim sup Ifn(x) - f(x)1 > A}
=
{x EX: lim sup Ihn(x) - h(x)1 > A}.
n n
Noting that (17.19)
= gn + hn and, for
Ihnl ::; E(lhll';Sn), Ihnl ::;
we have
h*,
h* = sup n
E(lhll';Sn),
1 7.
Conditional Expectation and Martingales
and hence (17.20)
JL(E>.)::; JL({x EX: h*(x) > Taking
E
-----7
~}) +JL({X EX: Ih(x)1
>
~})::; ~.
0 gives JL(E>.) = 0, and (17.17) is proven.
In the exercises we explore situations where (fn : 0: E A) does not have a final element. In addition to the martingale convergence result of Proposition 17.3, which is related to convergence results of Chapter 11, there are decreasing martingale convergence theorems, which as we will see are related to results of Chapter 14. For simplicity we consider martingales (fn : 0: E A) with index set A = Z- = {k E Z : k ::; O}. So we have a sequence of cr-algebras of subsets of X:
(17.21) We set
(17.22) Let JL be a probability measure on Jo· Given
f
E V(X,
Jo, JL), we set
(17.23) Note that Pk and pb all have LP-operator norm 1, for each p E [1,00]. Our goal is to produce various results to the effect that Pkf -----7 pb f as k -----7 -00. The next result is somewhat parallel to the L2-mean ergodic theorem contained in Proposition 14.2. Proposition 17.4. If JL zs a probabzlzty measure on Jo, f E L2(X, Jo, JL), and JklJb,Pk, and p b are as m (17.21)--(17.23), then
(17.24) Proof. The key to proving (17.24) is to show that
(17.25)
UKer Pk
is dense in Ker p b , in L2(X, JL).
k::;O
Given this, we set f = fb + g, with fb = pb f, 9 = (I - pb)f E Ker pb Then (17.25) implies Pkg -----7 0 in L2(X, JL), and since Pkf b = fb, we havE (17.24). It remains to establish (17.25). For this, we note that
(17.26)
(Ker Pk)~
= L2(X, Jk, JL),
17.
238
Conditional Expectation and Martingales
so
(17.27)
(U Ker Pk)
-L
k
= n(Ker Pk)-L = n L2(X, Jk,f'L) k k = L2(X, Jb, JL) = (Ker pb)-L,
the third identity by (17.22). This completes the proof. Given that the operators Pk and p b all have operator norm 1 on each LP space, that L2 is dense in LP for p E [1, 2], and that IlfllLP ::; Ilfll£2 for p E [1,2]' the same type of argument used for Proposition 14.3 also establishes the following.
Proposition 17.5. In the settmg of Pmposzizon 17.4. zf p E [1, 2]. (17.28) m LP- norm , as k
-+ -00.
In Proposition 17.7 we will extend this result to p E [1,(0). We next establish a pointwise convergence result, analogous to Theorem 14.6. In analogy with (14.22), we set
(17.29)
A# f(x)
= sup k~O
for
f
IPkf(x)1 ::; sup Pklfl(x), k~O
E L1(X,Jo,JL), and note that Proposition 17.2 gives
(17.30) Proposition 17.6. In the settmg of Propositzon 17.4, (17.31) as k
-+ -00.
Proof. Let us set Vk = Ker Pk n L2(X, Jo, JL), Vb = Ker pb n L2(X, Jo, JL), and recall from the proof of Proposition 17.4 that Uk Vk is dense in Vb, in L2-norm. Now let f E L1(X,Jo,JL) be given and take E > O. Pick h E L2(X, Jo, JL) such that Ilf - h 11£1 < E. Then write h = ff + g, with ff = pb hand g E Vb. Then write g = gK+gE with gK E VK (for sufficiently large negative K) and IIgEIIL2 < E (so also IlgEII£! < c). We hence have (17.32)
17.
Conditional Expectation and Martingales
239
Now
so Hence, for each A > 0,
fL({X EX: lim sup Pkf(x) -liminf Pkf(x) > A}) k-.-CXj
=
(17.33)
k-.-CXj
fLUX EX: lim sup Pkfc(x) -liminf Pkir,:(X) > A})
~ fL ( { x EX: A # fE (x)
>
0})
4c
0, define the map (17.48) Then, given
i.p
bounded and
23= -measurable,
we have
(17.49) The Markov Property gives rise to martingales, via the following construction. Proposition 17.11. Let h(t,x) be smooth m t 2: o,x E IRn, and let it satzsjy Ih(t, :r)1 ::; Cceclxl2 jar all EO> 0 and the backward heat equatwn
ah = -t::.h.
(17.50)
at
Then ~t(w) = h(t,w(t))
Z8
a martmgale over 23 t .
Proof. The hypothesis on h(t, x) implies that, for t, s > 0, (17.51)
h(s, x)
=
.I
p(t, y)h(t
+ s, x
- y) dy,
where p(t, .r) is given by (16.3). Now (17.52)
Ex (~t+sl23s) = Ex (h(t
+ s, w(t + s)) 123 = Ew(s) (h(t + 8,W(t))),
5 )
for Wx-almost all w, by (17.44). This is equal to (17.53)
.I
p(t,y-w(s)) h(t+s,y) dy,
by the characterization (16.12) of expectation, adjusted as in (16.59), and by (17.51) this is equal to h(s,w(s)) = ~s(w).
17.
243
Conditional Expectation and Martingales
Corollary 17.12. For one-dimenszonal Brownian motion, the followmg are martingales over ~t:
(17.54) gwen a > 0. Applying the Martingale Maximal Inequality to 3t(W) = e aw (t)-a 2 t, we obtain the following.
~t(w)
w(t) and to
Proposition 17.13. For one-dimensional Brownian motion, gwen t > O,b> l,a > 0, we have
(17.55)
Wo ({ w E
~o:
sup w(s) > by!4t/1f}) :S
O~s~t
~
b
and (17.56)
W o ( {w E
~o: sup w(s) - as> A}) :S e- aA . O~s~t
Proof. The sets whose measures are estimated in (17.55)-(17.56) are
{WE~o: O~s~t sup and {W E
W(S»A},
~o: sup
e aw (s)-a 2s
A=b
>
Vf4!, -;-
e aA }.
O~s~t
Since paths in ~o are continuous, one can take the sup over [0, t] n Q, which is countable, so Proposition 17.2 applies. Furthermore, we have (17.57)
Eo(l~tl) =
and (17.58)
E O(3t) =
1 Ixl
(4t
00
-00
1:
p(t, x) dx =
eax-a2t
V-;-
p(t, x) dx = 1,
yielding the estimates on the right sides of (17.55) and (17.56). Using more advanced techniques, one can sharpen (17.55) to (17.59)
Wo({w E
~o:
sup w(s) > A})
O~s~t
=2
roo p(t,x)dx.
JA
See, e.g., Proposition 3.8 in Chapter 11 of [Tl]. There is a property more subtle than the Markov Property, called the Strong Markov Property. Material on this can be found in [Dur], [IMc], [McK], and Chapter 11 of [Tl]. We end this chapter with an application of the estimate (17.56) to the following celebrated Law of the Iterated Logarithm. Our treatment follows
[McK].
244
17.
Conditional Expectation and Martingales
Proposition 17.14. For one-dzmensional Browman motion, one has
(17.60)
.
w(t)
hmsup
2y't log log lit
t-+O
= 1,
for Wo-a.e. w.
Proof. We take this in two steps. First we show
(17.61)
lim sup t-+O
w(t)
< 1,
for Wo-a.e. w.
2y't log log lit -
For this, we apply (17.56), with
(17.62) a = a(t) = (1+b)C 1 h(t),
A = A(t) = h(t),
for small positive t. Here, we pick 0 E (0,1). Then aA so e
(17.63)
_).. = (log -1) a
-(1+8)
t
h(t) =
1
t log log-, t
= (1 + 0) log 10g(l/t),
.
Now pick g E (0,1) and set (17.64) Then (17.65) and (17.56) implies (17.66) Since 2: n >1 n-(l+8) < 00, the First Borel-Cantelli Lemma (Exercise 5 of Chapter 15) applies to give
(17.67) Wo(A)
=
1,
A = {w
E
'.Po: sup w(s) - ans ::; An, V large n}.
In particular, for w E A and gn+l
o:s;s:s;e n
< t ::; gn, with n sufficiently large,
w (t) ::; max w (s) ::; an gn s:s;e n
(17.68)
=
+ An
(2+b)h(gn)
2(1 - ()1/2)h(()n) _ 4h(()n+l) (17.76) > 2(1 - 4()1/2)h(()n+l), for infinitely many n. Hence limsuPho w(t)lh(t) 2: 2(1 - 4()1/2), which gives (17.69) in the limit () ---- O. This completes the proof of Proposition 17.14. Since w(t) f---t -w(t) preserves the measure Wo and also w(t) f---t tw(l/t) preserves Wo (cf. Exercise 8, Chapter 16), we have the following.
246
17.
Conditional Expectation and Martingales
Corollary 17.15. For one-dzmensional Brownian mabon,
(17.77)
w ( t) 2Jt log log lit
lim inf 1-.0
= -1
for Wo-a.e. w,
'
and
(17.78)
lim sup h=
w(t) 2Jtloglogt
= 1,
for Wo-a.e. w.
Exercises 1. Suppose (X, J, IL) is a probability space and we have a partition of X into a countable family of disjoint sets X J , with IL( X J ) > O. Let P be the a-algebra generated by {XJ}' Given f E L 1 (X,J,IL), show that EUIP)(x)
=
IL(~J)
J
fdjL,
if x E XJ'
x)
This generalizes (17.10) (17.11). Show that a sequence of progressively finer such partitions of X gives rise to a martingale.
2. Use Proposition 17.9 to show that, for s, t 2 0, G continuous,
Eo(G(w(t
+ s)
- w(s))I23 s )
=
Eo(G(w(t))).
3. Let Jo C JIbe two a-algebras of subsets of X, and let jL be a probability measure on J1. We say a function fELl (X, J1, I},) is zndependent of Jo provided
(17.79)
f
is independent of g,
\/g E L 1 (X,Jo,j1,).
Show that this property is equivalent to
(17.80) Hznt. For (17.79)=?(17.80), note that, for 9 E Ll(X, Jo, jL), E(e iU )
J
gdjL = =
J
e iU gdjL
!
(given (17.79))
E(eiUIJo)g djL.
17.
Conditional Expectation and Martingales
For (17.80)=?(17.79), note that, for such 9 and ~,77 E JR, E(e iU +i1}9) =
J
E(e iU IJo)ei 1)9 dJ-L
= E(c ZU )E(e i 1)9)
4. Deduce for Brownian paths that w (t for s, t :::;. O.
(given (17.80)).
+ s)
- w (s) is independent of SB s,
5. Let X = [0, 1) and let J) be the family of cr-algebras of subsets of X given by (15.4), and let J-L) = where denotes Lebesgue measure. Let 11K be the measure on X, supported on the Cantor middle third set K, constructed in Exercise 11 of Chapter 6. Show that there exist f) E L 1 (X,J),J-L)) such that
ml"" ,
m
l}]
S E Jj
==?
J-LK(S)
=
J
fj dll).
S
Show that the family (f)) is a martingale, without a final element in Ll([O, 1), m).
In Exercises 6-9, we have a set X and cr-algebras Fk C Fk+l, k E Z+. We take A = Fk, F# = cr(A).
U k
We assume given probability measures 11k on Fk, satisfying 11k+ll.h = J-Lk. We aim to produce a probability measure J-L# on F# such that
(17.81) 6. Define J-Lb on S by
Show that J-Lb is a premeasure on A, as defined by (5.15)-(5.16). 7. Use Proposition 5.3, Theorem 5.4, and Proposition 5.5 to obtain a measure J-L# on F# such that (17.81) holds.
248
17.
Conditional Expectation and Martingales
In Exercises 8--9, we retain the setting of Exercises 6-7. Let A be a probability measure on F#, and assume each J-lk < < h ; say
AI.
Also use the Lebesgue-Radon-Nikodym decomposition to write
8. Show that
Note that
f
E
f AIFk « A
fk =
and
vl h
< < A, so
gk+hk' Show that (gk) is a martingale with final element
Ll(X,F#,A).
9. A refinement of the Martingale Convergence Theorem, Proposition 17.3, states that
(17.82)
fk
-----t
f,
A-a.e.
Deduce this from the two results
(17.83)
gk
-----t
f,
A-a.e.,
(17.84)
hk
-----t
0,
A-a.e.
Show that (17.83) follows from Proposition 17.3. Try to prove (17.84). REMARK 1. See [Doo], pp. 631 632, for further discussion ofthis result. REMARK 2. Applying the result of Exercise 9 to the martingale (fj) in Exercise 5, one has fj ---.> 0, m-a.e. Compare this with Proposition 11.10. REMARK :3. For an example of a family of singular martingales, where Exercises 6-9 apply and f = 0, see [RSTj.
10. Let (X, J, J-l) be a probability space and cp : X ---.> X a measure-preserving transformation. Let J c J be the CT-algebra of invariant sets, defined as in (14.35), and let P : LP(X, J-l) ---.> LP(X, J-l) be the projection arising in the Mean Ergodic Theorem, via (14.7). Show that
17.
249
Conditional Expectation and Martingales
11. Assume (X,~, /1) and (Z, 3, v) are probability spaces and F : X ---+ Z is a measure-preserving map, i.e.,
S E
3
=?
F- 1 (S) E ~ and /1(F- 1 (S)) = v(S).
(Do not assume F is bijective.) Let J c 3 be a CT-algebra, and let be the CT-algebra m= {F-l(S) : S E J}.
mc
~
Show that, for all 9 E L1(Z,v),
E(g
0
Flm)
=
E(gIJ)
0
F.
12. Establish the following complement to Proposition 15.6. Let {fj : j E N} be a stationary process, consisting of fJ E LP(X, /1), with p E [1,00). Then 1 k k fJ ---+ E(hlm), /1- a .e. and in LP- norm,
2.:= j=l
where Q( = {F- 1 (S) : S E J}, J consists of elements of !J3(]R=) invariant under the shift in (15.55), and the map F : X ---+ ]R= is given by
e
F(x)
=
(h(x), h(x), h(x), ... ).
13. Let v be a probability measure on]R, and let w be the associated product measure on ]R =. Define
Let Sk = (~o + ... + ~k)/(k + 1) and let 9-k be the CT-algebra generated by {Sk, Sk+l, Sk+2""}' Show that
14. If {fk : k 2': O} are independent, identically distributed random variables on (X,!J3,/1), define F: X ---+]R= by F(x) = (fo(x),h(x),h(x), ... ). Combine the results of Exercises 11 and 13 to establish the identity (17.40).
Appendix A
Metric Spaces, Topological Spaces, and Compactness
A metric space is a set X, together with a distance function d : X x X [0, (0), having the properties that
(A.l)
d(x, y)
=
0
d(x, y)
=
d(y, x),
¢::::::}
d(x, y) «; d(x, z)
x
=
y,
+ d(y, z).
The third of these properties is called the triangle inequality. An exampl of a metric space is the set of rational numbers Q, with d(x, y) = Ix - y Another example is X = ]Rn, with
If (xv) is a sequence in X, indexed by v = 1,2,3, ... , i.e., by v E Z+, or says Xv ---+ Y if d(xv, y) ---+ 0, as v ---+ 00. One says (xv) is a Cauchy sequen< if d(xv, xJ1) ---+ 0 as f-L, v ---+ 00. One says X is a complete metric space every Cauchy sequence converges to a limit in X. Some metric spaces aJ not complete; for example, Q is not complete. You can take a sequence (xJ of rational numbers such that Xv ---+ V2, which is not rational. Then (xv) Cauchy in Q, but it has no limit in Q. If a metric space X is not complete, one can construct its completion as follows. Let an element ~ of X consist of an equzvalence class of Caud
2~
252
Appendix A. Metric Spaces, Topological Spaces, and Compactness
sequences in X, where we say (xv) "-' (Yv) provided d(xv, Yv) -----> O. We write the equivalence class containing (xv) as [xv]. If ~ = [xv] and rJ = [Yv], we can set d(~, rJ);:= limv~oo d(xv, Yv) and verify that this is well defined and that it makes X a complete metric space. If the completion of Ql is constructed by this process, we get JR., the set of real numbers. This construction provides a good way to develop the basic theory of the real numbers.
There are a number of useful concept::; related to the notion of closene::;s. We define some of them here. First, if p is a point in a metric space X and r E (0,00), the set
(A.2)
Br(P) = {x EX: d(x,p) < r}
is called the open ball about p of radius r. Generally, a nezghborhood of p E X is a set containing such a ball, for ::;ome r > O. A set U c X is called open if it contains a neighborhood of each of its points. The complement of an open set is said to be closed. The following result characterizes closed ::;ets.
Proposition A.I. A subset K only ~f
C
X of a metnc space X zs closed
~f
and
(A.3)
Proof. As::;ume K is closed, x J E K, x J -----> p. If p ~ K, then p E X \ K, which is open, ::;0 some Be (p) C X \ K, and d( x)' p) 2: c for all j. This contradiction implies p E K. Conversely, assume (A.3) holds, and let q E U = X \ K. If B1/n(q) is not contained in U for any n, then there exists Xn E K n Bl/n(q); hence Xn -----> q, contradicting (A.3). This complete::; the proof. The following is straightforward.
Proposition A.2. If Uo: ~s a family of open sets zn X, then Uo: Uo: zs open. If Ko: ~s a famzly of closed subsets of X, then no: Ko: is closed. Given 5 c X, we denote by S (the closure of 5) the smallest closed sub::;et of X containing 5, i.e., the intersection of all the closed sets Ko: C X containing S. The following result is also straightforward.
Proposition A.3. Gwen 5 87U:h that x-; -----> 'D.
C
X, pES ~f and only if there exist
Xj E
5
Appendix A. Metric Spaces, Topological Spaces, and Compactness
~;).)
Given SeX, p E X, we say p is an accumulation point of S if and only if, for each E > 0, there exists q E S n BE(p), q =I=- p. It follows that p is an accumulation point of S if and only if each BE(p), E > 0, contains infinitely many points of S. One straightforward observation is that all points of S \ S are accumulation points of S. The znterzor of a set SeX is the largest open set contained in S, i.e., the union of all the open sets contained in S. Note that the complement of the interior of S is equal to the closure of X \ S. We now turn to the notion of compactness. We say a metric space X is compact provided the following property holds: (A)
Each sequence (Xk) in X
has a convergent subsequence.
We will establish various properties of compact metric spaces and providE various equivalent characterizations. For example, it is easily seen that (A: is equivalent to the following: (B)
Each infinite subset SeX has an accumulation point.
The following property is known as total boundedness:
Proposition A.4. If X
(C)
gwen
E
1,S
a compact metrzc space, then
> 0, ::J a fimte set {Xl, ... , XN}
such that Bc(xd, ... , Bc(XN) covers X.
Proof. Take E > 0 and pick :£1 E X. If Bc(Xl) = X, we are done. If no pick .1:2 E X \ Bc(:l;d. If Bc(xJ) U B E (X2) = X, we are done. If not, pic :1:3 E X\[B c (:1:1)UB c (:r2)]. Continue, taking Xk+l E X\[Bc(xd U·· ·UBc(Xk) if Bc(:r;d U··· U Bc(:Dk) =I=- X. Note that, for 1 0, there exists 6 > 0 such that
(A.6)
X, Y E X, d(x, y) ::; 6 ==* If(x) - f(y)1 ::; c.
An equivalent condition is that f have a modulus of contmuzty, i.e., a monotonic function w : [0, 1) -----* [0, (0) such that 6 "" 0 =? w (6) "" 0 and such that
(A.7)
x, y
E
X, d(x, y) ::; 6 ::; 1 ==? If(x) - f(y)1 ::; w(6).
Not all continuous functions are uniformly continuous. For example, if X = (0,1) c JR, then f(x) = sin l/x is continuous, but not uniformly continuous, on X. The following result is useful, for example, in the development of the Riemann integral in Chapter 1.
258
Appendix A. Metric Spaces, Topological Spaces, and Compactness
Proposition A.15. If X is a compact metrzc space and f E C(X), then f is uniformly continuous. Proof. If not, there exist xv,yv E X and E > 0 such that d(xv,Yv):S; 2- V but
(A.8) Taking a convergent subsequence xv] ---) p, we also have YVj ---) p. Now continuity of f at p implies f(xvJ ---) f(p) and f(Yvj) ---) f(p), contradicting (A.8). If X and Yare metric spaces, the space C(X, Y) of continuous maps f : X ---) Y has a natural metric structure, under some additional hypotheses. We use
D(f,g) = sup d(J(x),g(x)).
(A.9)
xEX
This sup exists provided f(X) and g(X) arc bounded subsets of Y, where to say BeY is bounded is to say d : B x B ---) [0, (0) has bounded image. In particular, this supremum exists if X is compact. The following result is frequently useful. Proposition A.16. If X zs a compact metrzc space and Y is a complete metric space, then C(X, Y), wzth the metric (A.9), zs complete. Proof. That D(f, g) satisfies the conditions to define a metric on C(X, Y) is straightforward. We check completeness. Suppose (fv) is a Cauchy sequence in C(X, Y), so, as v ---) 00, sup sup d(Jv+dx),fv(x)):s; Ev ---) O. k?O xEX
Then in particular (f v (x)) is a Cauchy sequence in Y for each x EX, so it converges, say to g(x) E Y. It remains to show that 9 E C(X, Y) and that fv ---) 9 in the metric (A.9). In fact, taking k ---)
00
in the estimate above, we have
sup d(g(x), fv(x)) :S; Ev ---) 0, xEX
i.e., fv ---) 9 uniformly. It remains only to show that 9 is continuous. For this, let Xj ---) x in X and fix E > o. Pick N so that EN < E. Since fN is continuous, there exists J such that j :::=: J::::} d(fN(Xj), fN(X)) < E. Hence j :::=: J::::} d(g(xj),g(x)) :S; d(g(xJ),fN(x J ))
+ d(JN(x J ),fN(X)) + d(JN(X), g(x))
0, set
R(x, y)
(C.1)
=
---+ ]R
and
f(x) - f(y) - f#(y) . (x - y)
and
7](0)
(C.2)
If 7]( 0)
---+
=
yl : x, y E
sup {IR(x, y)l/lx -
0 as 0 ---+ 0, then there exists 9
(C.3)
g(x)
=
f(x),
\1g(x)
E
K, 0 < Ix -
yl < o}.
C 1 (]Rn) such that
=
f#(x),
sup
1\1gl
V x E K.
Furthermore, we can arrange (C.4)
sup
Igl
~ C(n) sup If I, K
~ C(n) sup
If#l·
K
The function 9 will be produced using a carefully constructed partition of unity, described as follows. Suppose K is contained in the interior of a cube Q, of edge £.
-
277
Appendix C. The Whitney Extension Theorem
278
Lemma C.2. There is a partition of unity {] : j 2': O} on IR n that o = 1 for x ~ Q, o = 0 near K,
\
K, such
and j E Co(IR n \ K) for j 2': 1, with the following properties, for some M,C < 00: (1) Each x E IR n is m the support of at most M of the ]. (2) For any 6 E (0, RJ, if x E supp ], j 2': 1, and d(x, K) = 6, then
(C.S)
diam supp ]
1
-s: 26
and
(C.6) Proof. We will write Q\K = U as a countable union of closed cubes, whose interiors are disjoint. We start the process with a collection C consisting of one cube, Q. We successively alter our collection of cubes as follows. If Q is a cube in our collection, let Q denote the concentric cube of three times the linear size (hence 3n times the volume). If Q has nonempty intersection with K, chop Q into 2n smaller cubes, each of half the linear size. Throwaway any of these cubes that are contained in K. Continue this process on each cube with nonempty intersection with U, ad infinitum. We obtain o
0
Q] n Qk
=
0, for j i- k,
Q]
c IR n
\
K, V j 2': 1.
If Qj denotes the cube concentric with Q, with 1.1 times its linear size, then Q] cannot intersect more than 6n other Qk, k i- j. Having such a collection of cubes, we proceed as follows. Let Q be a unit cube, and pick 7/J E C=(Q), supported in the interior, such that 7/J = 1 on Q. Via translation and scaling of coordinates, we obtain 7/J] E C=(Qj), supported in the interior, such that 7/J] = 1 on Qj, for j 2': 1. Clearly IV7/J] I -s: C / (diam Qj). Take 7/)0 E C= (IRn) such that 7/Jo = 1 on IRn \ Q and 7/Jo = 0 on a neighborhood of K. Now set
W(x) =
L 7/Jj(x),
j(x) = 7/Jj(x)/W(x).
j?O
The properties stated for j follow from this construction.
279
Appendix C. The Whitney Extension Theorem
We are now prepared to construct g. For each j, let Yj be a point in K of minimal distance from QJ' Then we set (C.7)
g(x)
=
L
[j(YJ)
+ j#(YJ)(x -
YJ)] j(x),
x E]Rn \ K.
J
Clearly 9 E coo(]Rn \ K). We claim that, as x (C.S)
g(x)
-+
j(x) and \7g(x)
To verify the first part, suppose Ix (C.9)
x E supp J
=?
zl =
Ix - YJI :s; CJ
-+ -+
z E K,
j#(x).
J. Then =?
Ij(Yj) - j(z)1 :s; cJ.
Thus, since there are at most M nonzero terms in the sum (C.7), when evaluated at x, we have
To demonstrate the second part of (C.S), write
We can write the first sum on the right side of (C.1l) as (C.12) as J -+ 0, by the continuity of j# on K. We next examine the last sum in (C.1l). Since I: \7J = \71 = 0, we can write this sum as
By (C.6) plus the estimate assumed on (C.1), the first sum is seen to tend to 0 as J -+ O. Again using I: V'J = 0, we write the second sum in (C.13) as (C.14) and again (C.6) plus continuity of j# on K implies that this tends to 0 as J -+ O. We have now defined 9 on U = ]Rn \ K. Setting 9 = j on K, we see that 9 is continuous. It is clear that 9 is Coo on U, and we have shown that \7g(x) -+ j#(z) as x -+ z E K, x E U. In fact, if we set g#(x) = \7g(x) for x E U, j#(x) for x E K, we have that g# is continuous on ]Rn.
280
Appendix C. The Whitney Extension Theorem
To finish the proof of Theorem C.l, it remains to show that
g(x)
(C.15)
=
J(z)
+ J#(z)(x -
+ o(lx -
z)
zi),
x
E
lR n , z E oK.
If x E K, this is part of the hypothesis. On the other hand, suppose x E lRn \ K, Ix - zl = 6 « 1. Comparing (C.7), we need
where IYj -
zl :s; C6.
Note that
J(z)
(C.17)
=
J(Yj)
J#(z)(x - z)
+ f#(YJ)(z =
J#(YJ )(x -
+ 0(6), z) + 0(6).
Yj)
Multi plying by
0,
+ C2A- qllu2111q·
vTu(2A) ::; VTUI (A)
(D.5)
C 1 A- r
::;
Also,
J
(D.6)
Ifl P dfL = p
x Hence
J
ITul P dfL = p
1
00
x ::; CIP
(D.7)
1
00
Vj(A)A P - 1 dA.
VTu(A)A P - 1 dA
1
00
+C2P
AP -
1
00
l- r (
J lul J
r
dfL )dA
lul>>-
AP -
1-
q(
lulqdfL)dA.
lul9 Now
(D.8)
1
00
AP -
J lul
l- r (
r
dfL) dA = P
~r
Jlul
P
dfL,
lul>>and similarly
(D.9)
1
00
AP-
l-
q(
J
lul q dfL) dA = q
J
~ P lul P dfL·
lul9 Combining these gives the desired estimate on IITull~p. The following complement to Proposition D.1 is applied in the proofs of Proposition 14.7 and Proposition 17.7. Proposition D.2. If 1 ::; r < P < of weak type (r,r) and satzsjies (D.10) then T is bounded on LP.
00
and T is a sublinear operator that is
Appendix D. The Marcinkiewicz Interpolation Theorem
Proof. It suffices to show that T is of weak type (q, q) for each q E (r, 00 ), since then we can take q E (p, 00) and apply Proposition D.1. This time, given A E (0,00), write U = Ul + U2 with Ul(X) = u(x) for lu(x)1 > A/A and U2(X) = u(x) for lu(x)1 ~ A/A, where A E (0,00) is as in (D.lO). It follows that fL({X EX: ITu2(X)1 > A}) = O. It remains to estimate TUl(X). We have
fL({X EX: ITul(X)1 > A}) ~
(D.ll)
=
CA-rllulllLr
J Iulr
CA- r
dfL·
lul>'\/A
Now taking this last integral as the inner product of Holder's inequality, we have
J Iulr ~ (J Iulq dfL
(D.12)
dfL
r/
qfL( {x
Iulr
EX: lu(x)1
and 1 and using
>
~} rq-r)/q
lul>'\/A
~ =
IlullLq (A -q Il ulllq) (q-r)/q A-(q-r) Il ulllq,
the second estimate by Tchebychev's inequality. Plugging the estimate (D.12) into (D.ll) yields the desired estimate (D.3) (with p replaced by q) and completes the proof. There are more general versions of Marcinkiewicz interpolation, involving operators of weak type (p, q). We refer to [FoIl for a treatment.
Appendix E
Sard's Theorem
n ---+
n
pEn
be a C 1 map, with open in ]Rn. If and DF(p) : ]Rn ---+ ]Rn is not surjective, then p is said to be a critzcal pomt and F(p) a crztzcal value. The set C of critical points can be a large subset of n, even all of it, but the set of critical values F( C) must be small in ]Rn. This is part of Sard's Theorem. Let F :
]Rn
Theorem E.1. If F : n ---+ ]Rn is a C 1 map, then the set of critzcal values of F has measure 0 m ]Rn. Proof. If Ken is compact, cover K n C with m-dimensional cubes Qj, with disjoint interiors, of side OJ' Pick PJ E C n QJ' so L j = DF(pj) has rank:::; n - 1. Then, for x E QJ'
where TJj ---+ 0 as OJ ---+ O. Now LJ(QJ) is certainly contained in an (n - 1)dimensional cube of side Co OJ , where Co is an upper bound for vmllDFl1 on K. Since all points of F (QJ) are a distance :::; Pj from (a translate of) LJ(QJ), this implies
oJ + 2PJt- 1 :::; C1TJJO~,
meas F(Qj) :::; 2pJ(Co
provided OJ is sufficiently small that Pj :::; OJ. Now l:j oj is the volume of the cover of K n C. For fixed K this can be assumed to be bounded. Hence meas F(C
n K) :::; C K TJ,
where TJ = max {TJJ}. Picking a cover by small cubes, we make TJ arbitrarily small, so meas F( C n K) = O. Letting K j / ' n, we complete the proof.
-
287
Appendix E. Sard's Theorem
288
Sard's Theorem also treats the more difficult case when n is open in IRm , m > n. Then a more elaborate argument is needed, and one requires more differentiability, namely that F is class C k , with k = m - n + 1. A proof can be found in [Stb]. The main application of Sard's Theorem in this text is to the proof of the change of variable theorem we present in the next appendix. Here, we give another application of Sard's Theorem, to the existence of lots of Morse functions. This application gives the typical flavor of how one uses Sard's Theorem. We begin with a special case: Proposition E.2. Let 0 c IR n be open, f E C=(O). For a E IRn, set fa(x) = f(x) - a· x. Then, for almost every a E IRn, fa is a Morse function, i. e., it has only nondegenemte crit2cal pomts. Proof. Consider F(x) = \1f(x);F: 0 -----t IRn. A point x E 0 is a critical point of fa if and only if F(x) = a, and this critical point is degenerate only if, in addition, a is a critical value of F. Hence the desired conclusion holds for all a E IR n that are not critical values of F. Now for the result on manifolds: Proposition E.3. Let M be an n-dimenswnal mamfold, embedded in IRK. Let f E C=(M), and, for a E IRK, let fa(x) = f(x) -a'x, for x E M C IRK. Then, for almost all a E IRK, fa is a Morse functwn. Proof. Each p E M has a neighborhood Op such that some n of the coordinates Xv on IRK produce coordinates on Op. Let us say Xl, ... ,X n do it. Let (a n +1' ... ,aK) be fixed but arbitrary. Then, by Proposition E.2, for almost every (al,'" ,an) E IR n , fa has only nondegenerate critical points on Op. By Fubini's Theorem, we deduce that, for almost every a E IRK, fa has only nondegenerate critical points on Op. (The set of bad a E IRK is readily seen to be a countable union of closed sets, hence measurable.) Covering M by a countable family of such sets Op, we finish the proof.
Appendix F
A Change of Variable Theorem for Many-to-one Maps
c
Here we present a change of variable theorem for a 1 map defined on an open set in ]Rn that is not assumed to be a diffeomorphism onto its range. Theorem F.l below is hence a generalization of Theorem 7.2. There are further generalizations, involving Lipschitz maps and maps between spaces of different dimensions, that can be found in [EG] and [Fed]. Theorem F.1. Let r2 c ]Rn be open and let F : r2 ---+ ]Rn be a C 1 map. For x E ]Rn set n( x) = card F- 1 (x). Then n zs measurable and for any measurable u 2: 0 on ]Rn,
(F.l)
J
J
n
r
u(F(x)) I det DF(x) I dx =
u(x) n(x) dx.
This is an extension of the standard change of variable formula, in which one assumes that F is a diffeomorphism of r2 onto its image. (Then n(x) is the characteristic function of F(r2).) We will make use of this standard result in the proof of the theorem. We make a sequence of reductions. Let
K
= {x
E
r2 : det DF(x)
P = Fin,
n(x)
= O},
0=
r2 \ K,
= cardp-l(x).
-
289
Appendix F. A Change of Variable Theorem for Many-to-one Maps
290
Sard's Theorem implies F(K) has measure zero. Hence n(x) = n(x) a.e. on ]Rn, so if ~e~can show that n is measurable and (F.l) holds with 0" F, n replaced by 0" F, we will have the desired result. Thus we will henceforth assume that det DF(x) i- 0 for all x E n.
n,
Next, for k E £:+, let
Dk
=
{x En: Ixl :S; k, dist(x, an) 2': 11k}.
Each Dk is a compact subset of n. Furthermore, Dk C Dk+l and 0, = U D k . Let Fk = FIDk and nk(x) = cardFk1(x). Then, for each x E ]Rn, nk(x) /' n(x) as k ---+ 00. Suppose we can show that, for each k, nk is measurable and (F.2)
J
u(F(x)) I det DF(x)1 dx
J
u(x) ndx) dx.
=
Dk
~n
Now the Monotone Convergence Theorem implies that as k ---+ 00 the left side of (F.2) tends to the left side of (F.l) and the right side of (F.2) tends to the right side of (F.l). Hence (F.l) follows from (F.2), so it remains to prove (F.2) (and the measurability of nk). For notational simplicity, drop the index k. We have a compact set F is Cion a neighborhood 0, of D, and det DF(x) is nowhere vanishing. We set nD(x) = cardD n F-1(x) and desire to prove that nD is measurable and
D
C ]Rn,
(F.3)
J
u(F(x)) Idet DF(x)ldx=
D
J
u(x)nD(x)dx.
~n
By the Inverse Function Theorem, each x E 0, has a neighborhood Ox such that F is a diffeomorphism of Ox onto its image. Since D is compact, we can cover D with a finite number of open sets OJ on each of which F is a diffeomorphism. Then there exists 0 such that any subset of D of diameter :S; 0 for all x E n. In such a case, Theorem 7.2 yields
In
Proposition G.1. If F : n --+ 0 zs a C 1 orientation-preserving diffeomorphism and a an mtegmble n-foTm on 0, then
(G.ll)
J J a =
o
F*a.
n
Appendix G. Integration of DiHerential Forms
In Appendix H we will present another proof of the change of variable formula, making direct use of basic results on differential forms developed in this appendix. In addition to the pull-back, there are some other operations on differential forms. The wedge product of dx/s extends to a wedge product on forms as follows. If (3 E Ak ( 0) has the form (G.1) and if
(G.12) define
(G.13)
0: /\ (3 =
L ai(x)bJ(x) dX 21 /\ ... /\ dXic /\ dXJl /\ ... /\ dXjk i,J
in Ak+£(O). We retain the equivalences (G.3). It follows that
(G.14) It is also readily verified that
F* (0: /\ (3) = (F* 0:) /\ (F* (3).
(G.15)
Another important operator on forms is the extenor denvatzve:
(G.16) defined as follows. If (3 E Ak ( 0) is given by (G. 1), then
(G.17)
d(3 =
L J,e
8b
8
J
Xe
dxe /\ dX J1 /\ ... /\ dxJk ·
The antisymmetry dX m /\ dxe = -dxe /\ dx m, together with the identity 82bjj8xe8xm = 82bJj8xm8x£, implies d(d(3) = 0,
(G.18)
for any smooth differential form (3. We also have a product rule:
(G.19)
d(o: /\ (3)
= (do:) /\ (3 + (-1)10: /\ (d(3) ,
0: E AJ(O), (3 E Ak(O).
The exterior derivative has the following important property under pullbacks:
(G.20)
F*(d(3)
= dP* (3,
Appendix G. Integration of Differential Forms
if (3 E Ak(O) and F : n ---; 0 is a smooth map. To see this, extending (C.19) to a formula for d( a /\ (31 /\ ... /\ (3c) and using this to apply d to F* (3, we have (C.21) dF*(3 =
Le fJfJXe (bJ
0
F(x)) dxe /\ (F*dxJI) /\ ... /\ (F*d:c Jk )
],
+ L(±)bJ (F(x) )(F*dx]l)
/\ ... /\ d( F*dx Jv ) /\ ... /\ (F* dXJk)'
.1,1/
Now the definition (G.6)-(G.7) of pull-back gives directly that
(G.22) and hence d(F*dxi) = ddFi = 0, so only the first sum in (G.21) contributes to dF* (3. Meanwhile,
(G.23)
F*d(3 =
L ::: (F(:c))
(F*dxm) /\ (F*dxjl) /\ ... /\ (F*dXjk) ,
j,m
so (G.20) follows from the identity
which in turn follows from the chain rule. Here is another important consequence of the chain rule. Suppose F : n ---; 0 and 1/) : 0 ---; U are smooth maps between open subsets of IRn. We claim that for any form a of any degree,
(G.24) It suffices to check (G.24) for a = dx J • Then (G.7) gives the basic identity 1/;* dXj = ,,£(fJ1/Jj/fJxe) dxe. Consequently,
(G.25)
but the identity of these forms follows from the chain rule:
(G.26)
Dip = (D1/;)(DF) =* ;ipj = Xm
Le ~1/;jXc ;Fc . Xm
Appendix G. Integration of Differential Forms
297
One can define a k-form on an n-dimensional manifold M as follows. Say M is covered by open sets OJ and there are coordinate charts F j : D j ----7 OJ, with fl J C ]Rn open. A collection of forms (3J E Ak (D J ) is said to define a k-form on M provided the following compatibility condition holds. If O 2 n OJ cI- 0 and we consider n~J = F i- 1 (0 2 n OJ) and diffeomorphisms (G.27) we reqmre (G.28) The fact that this is a consistent definition is a consequence of (G.24). For example, if G : ]1.1 ----7 Rm is a smooth map and, is a k-form on ]Rm, then there is a well-defined k-form (3 = G*, on .!vI, represented in such coordinate charts by (3J = (G 0 FJ ) * T Similarly, if {3 is a k- form on M as defined above and G : U ----7 .!vI is smooth, with U C ]Rm open, then G* (3 is a well-defined k-form on U. We give an intrinsic definition of fM a when a is an n-form on M, provided !'vI is oTzented, i.e., there is a coordinate cover as above such that det DtpJk > O. The object called an "orientation" on M can be identified as an equivalence class of nowhere vanishing n-forms on M, two such forms being equivalent if one is a multiple of another by a positive function in COO(D). A member of this equivalence class, say w, defines the orientation. The standard orientation on ]Rn is determined by dXl/\ ... /\ dx n . The equivalence class of positive multiples a( x)w is said to consist of "positive" forms. A smooth map '1/) : S ----7 .!vI between oriented n-dimensional manifolds preserves orientation provided '1/)*(1 is positive on S whenever (1 E An(M) is positive. \Ve mention that there exist surfaces that cannot be oriented, such as the famous "l\Iobius strip." We define the integral of an n-form over an oriented n-dimensional manifold as follows. First, if Ct is an n-form supported on an open set 0 C ]Rn, given by (G.4), then we define fa a by (G.5). More generally, if !'vI is an n-dimensional manifold with an orientation, say the image of an open set 0 C ]Rn by tp : 0 ----7 M, carrying the natural orientation of 0, we can set (G.29)
for an n-form a on M. If it takes several coordinate patches to cover M, define f M a by writing a as a sum of forms, each supported on one patch.
298
Appendix G. Integration of DiHerential Forms
We need to show that this definition of JM a is independent of the choice of coordinate system on M (as long as the orientation of M is respected). Thus, suppose cp : 0 -----> U c M and 1j; : n -----> U C M are both coordinate patches, so that F = 1j;-1 0 cp : 0 -----> n is an orientation-preserving diffeomorphism. We need to check that, if a is an n-form on M, supported on U, then (G.30)
J J cp*a =
(')
1j;*a.
n
To establish this, we use (G.24). This implies that the left side of (G.30) is equal to (G.31)
J
F*(1j;*a),
(')
which is equal to the right side of (G .30), by (G .11) (with slightly altered notation). Thus the integral of an n-form over an oriented n-dimensional manifold is well defined. We turn now to the Gauss-Green-Stokes formula for differential forms, commonly called simply the Stokes formula. This involves integrating a k-form over a k-dimensional manifold with boundary. We first define that concept. Let S be a smooth k-dimensional manifold, and let M be an open subset of S, such that its closure M (in ]RN) is contained in S. Its boundary is aM = M \ M. We say M is a smooth surface with boundary if also aM is a smooth (k - I)-dimensional surface. In such a case, any p E aM has a neighborhood U C S with a coordinate chart cp : 0 -----> U, where 0 is an open neighborhood of 0 in ]Rk, such that cp(O) = p and cp maps {x EO: Xl = O} onto Un aM. If S is oriented, then M is oriented, and aM inherits an orientation, uniquely determined by the following requirement: if
(G.32) then aM = {(X2' ... ,Xk)} has the orientation determined by dX2 /\ ... /\ dXk. We can now state the Stokes formula. Proposition G.2. Given a compactly supported (k -I)-form {3 of class c l on an oriented k-dzmenswnal surface M (of class C 2 ) with boundary aM, with its natural onentation, (G.33)
Appendix G. Integration of Differential Forms
299
Proof. Using a partition of unity and invariance of the integral and the exterior derivative under coordinate transformations, it suffices to prove this when M has the form (G.32). In that case, we will be able to deduce (G.33) from the Fundamental Theorem of Calculus. Indeed, if
(G.34) with bj(x) of bounded support, we have
(G.35)
d(3 = (-1)1
_1
8bJ dXl 1\ ... 1\ dXk· 8xj
-
If ] > 1, we have
(G.36)
and also K* (3 = 0, where for] = I, we have
K :
(G.37)
= =
8M
-----t
A1 is the inclusion. On the other hand,
J J
b1 (0, x') dx'
(3.
8M
This proves Stokes' formula (G.33). The reason we required !vI to be a surface of class C 2 in Proposition G.2 is the following. Due to the formulas (G.6)-(G.7) for a pull-back, if (3 is of class cj and F is of class C€, then F* (3 is generally of class CP" with fJ = min(j,£ - 1). Thus, if j = £ = I, F*(3 might be only of class Co, so there is not a well-defined notion of a differential form of class CIon a C 1 surface, though such a notion is well defined on a C 2 surface. This problem can be overcome, and one can extend Proposition G.2 to the case where M is a C 1 surface and (3 is a (k -I)-form with the property that both (3 and d(3 are continuous. One can go further and formulate (G.33) for a (k - I)-form (3 with the property that
(G.38)
300
Appendix G. Integration of Differential Forms
where L : oM ---+ M is the natural inclusion, a class of forms that can be shown to be invariant under bi-Lipschitz maps. (It can be shown that the first two conditions in (G.38) imply L*j3 E H1,l(oM)'.) We will not go into the details. However, in Appendix I we will present an elementary treatment of (G.33), stated in a more classical language, when AI is an open domain in IRk whose boundary is locally the graph of a Lipschitz function. A far reaching extension, due to H. Federer, can be found in [Fed]; see also [EG]. The calculus of differential forms has many applications to differential equations, differential geometry, and topology. 1\lore on this can be found in [Spi] and also in [TI] (particularly Chapters 1, 5, and 10). To end this appendix, we make use of the calculus of differential forms to provide simple proofs of some important topological results of Brouwer. The first two results concern retmctwns. If Y is a subset of X, by definition a retraction of X onto Y is a map i.p : X ---+ Y such that i.p(:r;) = :1: for all :1: E Y.
Proposition G.3. There 'is no smooth r-etmctwn closed umt ball B zn IR n onto ,tis bO'lmdary 5,,-1.
i.p :
B
---+
5n-
1
of the
In fact, it is just as easy to prove the following more general result. The approach we use is adapted from [Kan].
Proposition G.4. If AI 'is a compact orIented n-dzmenswnal manifold 'Wl,th nonempty boundary oAI, there 'is no smooth retmctwn i.p : AI ---+ oAI. Proof. You can pick W E An-1 (oM) to be an (n -I)-form on fJM such that J8M W > O. Now apply Stokes' Theorem to j3 = i.p*w. If 'P is a retraction, then i.p 0 J(x) = x, where J : oAI "---+ AI is the natural inclusion. Hence j*i.p*w = W, so we have
(G.39)
J .I W =
8M
di.p*w.
M
But di.p*w = i.p*dw = 0, so the integral (G.39) is zero. This is a contradiction, so there can be no retraction. A simple consequence of this is the famous Brouwer Fixed-Point Theorem.
Theorem G.5. If F : B ---+ B zs a continuous map on the closed umt ball zn IRn, then F has a fixed poznt. Proof. First, an approximation argument shows that if there is a continuous such F without a fixed point, then there is a smooth one, so assume F : B ---+ B is smooth. We are claiming that F(x) = x for some x E B. If not,
Appendix G. Integration of Differential Forms
301
then for each x E B define rp(x) to be the endpoint of the ray from F(x) to x, continued until it hits DB = sn-l. It is clear that rp would be a smooth retraction, contradicting Proposition G .3.
REMARK. Typical proofs of the Brouwer Fixed-Point Theorem use concepts of algebraic topology; cf. [S pa]. In fact, the proof of Proposition G.4 contains a germ of de Rham cohomology. See [TIl, Chapter 1, §19 for more on this. An integral calculus proof of the Brouwer Fixed-Point Theorem that does not involve differential forms is given in [DSl, Vol. 1, pp. 467-470. One might compare it with the proof given above.
Appendix H
Change of Variables Revisited
As indicated in Exercise 10 of Chapter 1, the change of variable formula for a one-variable integral,
j
(H.I)
'P(b)
'P( a)
J(x) dx
= Ib
J(c.p(x))c.p'(x) dx,
a
given J continuous and c.p of class C 1 , satisfying c.p' > 0, can be established via the Fundamental Theorem of Calculus and the chain rule. By comparison, let us recall the change of variable formula for multiple integrals, as given in Theorem 7.2.
Theorem H.I. Let 0, 0 be open sets on ]Rn and let c.p : 0 -----) 0 be a C 1 dzffeomorphzsm. Gwen J E M+(O) or J E £1(0, dx), we have
(H.2)
.I o
J(c.p(x)) Idet Dc.p(x) I dx
=
.I
J(x)dx.
0
The proof of Theorem 7.2 given in Chapter 7 was very different from the argument indicated above, and it is our goal here to provide a proof of Theorem H.I more closely parallel to that of (H.I). Our proof is based on one of P. Lax [La], who found a fresh approach to the proof of the multidimensional change of variable formula. More precisely, [La] established the following result, from which Theorem H.I can be deduced. Our proof, adapted from [T2], will differ from that of [La] in that we make use of basic results on differential forms, from Appendix G.
-
303
Appendix H. Change of Variables Revisited
304
Theorem H.2. Let 'P : lRn --+ ][{n be a C I map. Assume 'P(x) = x for Ixi 2': R. Let f be a continuous function on lRn wzth compact support. Then
J
(H.3)
f('P(x)) detD'P(x)dx
=
J
f(x)dx.
Proof. Via standard approximation arguments, it suffices to prove this when 'P is C 2 and f E CJ(lRn ), which we will assume from here on.
To begin, pick A> 0 such that f(x - Aed is supported in {x : Ixl > R}, where el = (1,0, ... ,0). Also take A large enough that the image of {x : Ixl ::; R} under 'P does not intersect the support of f(.1: - Ael). We can set fJ1jJ F(x) = f(x) - f(x - Ael) = ~(x), UXI
(H.4) where
(H.5) Then we have the following identities involving n-forms:
(H.6)
fJ1jJ a = F(x) dXI /\ ... /\ dX n = dXI /\ ... /\ dX n fJxI = d1jJ /\ dX2 /\ ... /\ dX n
= d( 1jJ dX2
/\ ... /\ dx n ),
i.e., a = d(3, with (3 = 1jJ dX2/\' .. /\ dX n a compactly supported (n - 1)-form of class C I . Now the pull-back of a under 'P is given by (H.7)
'P*a = F('P(x)) detD'P(x)dxI/\" ·/\dx n .
Furthermore, the right side of (H.7) is equal to
(H.8)
f('P(x)) detD'P(x) dXI/\"'/\ dX n - f(x - Aed dXI/\"'/\ dx n .
Hence we have
J
J J
f('P(x)) detD'P(x)dxI· .. dxn -
(H.9)
=
J J 'P*a =
'P*d(3 =
where we use the identity (H.10)
'P* d(3 = d( 'P* (3),
f(X)dxI .. ·dxn
d('P*(3),
Appendix H. Change of Variables Revisited
305
established in Apendix G. On the other hand, a very special case of Stokes' Theorem applies to
(H.ll)
tp* (3
= I =
2.:.: I) (2:) dXl /\ ... /\ dx; /\ ... /\ dx n , J
with I} E
cd (lRll).
Namely "(
) -1
dl=~-lJ
(H.12)
J
al] dX1/\···/\dxn, ax ]
and hence, by the Fundamental Theorem of Calculus,
J
(H.13)
dl = O.
This gives the desired identity (H.3), from (H.9). We make some remarks on Theorem H.2. Note that 'P is not assumed to be one-to-one or onto. In fact, as noted in [La], the identity (H.3) implies that such 'P must be onto, and this has important topological implications. In more detail, it is readily seen that the range of cp must be closed, so if cp is not onto, it must omit some nonempty open set U. One can take a function J supported in U and with a positive integral and obtain a contradiction in (H.3), since the left side would have to vanish. One then easily observes that, if there is a smooth retraction of the ball B onto its boundary, such a cp would exist that is not onto. This contradiction then gives another proof of the Brouwer no-retraction result, Proposition G.3. We recall that, if one puts absolute values around det Dcp(x) in (H.3), the appropriate formula is (H.14)
J
J (cp(x))
Idet Dcp(x) Idx =
J
J(x) n(x) dx,
where n(x) = #{y: cp(y) = x}. This result was proven in Appendix F. We want to extend Theorem H.2 to treat more general maps cp and functions J. The fact that Theorem H.2 holds for maps that do not have to be diffeomorphisms simplifies the task of making the extension to more singular maps. We begin with the following: Proposition H.3. Let'P: lR n Ixl 2:: R. Furthermore, assume
(H.15)
----+
lR n be contmuous. Assume cp(x) = x for
Dcp E Lr~c(lRn),
where Dcp is the weak derzvatzve of cp. Then (H.3) holds Jor all continuous J with compact support.
306
Appendix H. Change of Variables Revisited
Proof. Using a mollifier, we can produce Coo maps CPv : ]Rn ---+ ]Rn, such that CPv(x) = x for Ixl ~ R + 1, such that CPv ---+ cp uniformly, and such that (H.I6) for all compact K
c ]Rn. By Theorem H.2 we have
(H.I7) for each v. Under our hypotheses, f 0 CPv ---+ f 0 cp uniformly on ]Rn, and we can pick a compact K c ]Rn containing the support of f 0 CPv for all v. By (H.I6) we have (H.I8)
det Dcpv
---+
det Dcp
CPv det Dcpv
---+
f
in L 1 (K),
and hence (H.I9)
f
0
0
cp det Dcp
in L 1 (]Rn).
Thus we can pass to the limit in (H.I7) to obtain (H.3) in this setting. We mention that, if the hypothesis (H.I5) is strengthened to Dcp E Lfoc(]Rn), for some p > n, then, by Proposition Il.6, cp is differentiable a.e. and its pointwise derivative coincides a.e. with the weak derivative. In light of this, one can compare Proposition H.3 (and Proposition H.4 below) with Propositions 7.4 and 7.6. Now we establish a result valid for more singular
f.
Proposition H.4. Retam the hypotheses of Propositzon H.3. Furthermore, assume (H.20)
det Dcp(x) ~ 0
for a.e. x.
Then (H.3) holds for all posztwe measurable f. Proof. We have (H.3) for compactly supported continuous f. An application of the Monotone Convergence Theorem then gives (H.3) for all characteristic functions f = XK, K c ]Rn compact. A second application of monotone convergence gives (H.3) for all characteristic functions f = Xu, U C ]Rn open and bounded. Now if S c ]Rn is a bounded measurable set, we can write Kv / So, Uv '" Sl, So eSc Sl, with meas(So) = meas(SI) = A, say, and monotone convergence gives
J J
XSo (cp(x)) det Dcp(x) dx
(H.2I)
=
A,
XSl(CP(X)) detDcp(x)dx=A,
Appendix H. Change of Variables Revisited
,JUI
so (H.3) holds for f = XS. Hence (H.3) holds for positive simple functions f· By one more application of monotone convergence it hence holds for all positive measurable functions (with the standard convention that +00 . 0 = 0). We now discuss how Theorem H.1 can be deduced from Theorem H.2 (or more precisely from the extension to allow arbitrary f E M+(O), as in Proposition H.4). It suffices to obtain (H.2) for f E M+(O) supported in some compact K c 0, by the Monotone Convergence Theorem. Also, to prove Theorem H.1, it suffices to assume det D O. We will use the following lemma.
Lemma H.5. In the settmg of Theorem H.1 (and wzth detD 0), gwen p E 0, there zs a nezghborhood U of p and a C 1 map : lR n ---+ lRn such that =