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0 there is a 8 > 0 with the property that JL(r) < E whenever v(r) < 8. 3.3. 16 Exercise: Let f be a non-negative, measurable function on the measure space (E, B, JL) . If f is integrable, show that (3.3. 17) 3.3. 15 Exercise:
Next, produce a non-negative measurable f on ( [0, 1] , B[o , l ] , A [o , I] ) (.X[o, l ] is used here to denote the restriction of AR to B[o, 1 ] ) such that (3.3. 17) holds but f fails to be integrable. Finally, show that if f is a non-negative measurable function on the finite measure space (E, B, JL), then f is integrable if and only if CX) JL(f > n ) < oo. n =l 3.3. 18 Exercise: Let J be a closed rectangle in � N and f J � � a continuous function. Show that the Riemann integral (R) JJ f(x) dx of f over J is equal to the Lebesgue integral JJ J (x) AJRN (dx) . Next, suppose that f E L 1 (AJRN ) is continuous, and use the preceding to show that
L
:
J f(x) >.RN (dx)
J /limR N (R) JJ{ f(x) dx , where the limit means that, for any E > 0, there exists a rectangle Jf. such that =
j f(x) >.RN (dx) - (R) 1 f(x) dx < f
whenever J is a rectangle containing Jf. . For this reason, even when f is not continuous, it is conventional to use J f(x) dx instead of J f dAR N to denote the Lebesgue of f. 3.3. 19 Exercise: Let (E, B, JL) be a measure space and let {fn}1 be a se quence of measurable functions on (E, B) . Next, suppose that {gn}1 C L 1 (JL) and that gn � g E L 1 (JL) in L 1 (JL) . The following variants of Fatou ' s Lemma and Lebesgue ' s Dominated Convergence Theorem are often useful. ( i ) If fn < gn (a.e., JL) for each n > 1, show that
j
j
nlim ---+ CX) fn dJL. ---+ CX) fn djL < nlim ( ii ) If fn � f either in JL-measure or JL-almost everywhere and if I fn i < gn ( a.e., JL) for each n > 1, show that II fn - f II £ 1 (J.L) � 0 and therefore that limn ---+ CX) J fn dJL == J f dJL.
3.3
Convergence of Integrals
61
Let (E, B, JL) be a measure space. A family JC of measurable functions f on (E, B, JL) is said to be uniformly JL-absolutely continuous if for each E > 0 there is a 8 > 0 such that fr I f I dJL < E for all f E JC whenever E B and JL( ) < 8; and it is said to be uniformly JL-integrable if for each E > 0 there is an R < oo such that � / > R l f l dJL < E for all f E JC . I ( i ) Show that JC is uniformly JL-integrable if it is uniformly JL-absolutely con tinuous and supIC l l f ii £ 1 (JL ) < oo . /E Conversely, suppose that JC is uniformly JL-integrable and show that it is then necessarily uniformly JL-absolutely continuous and, when JL(E) < oo , that sup / E IC l l f ii £ 1 (JL ) < oo . ( ii ) If sup / E IC J 1!1 1 + 6 dJL < oo for some 8 > 0, show that JC is uniformly JL-integrable. ( iii ) Let {/n}1 C L 1 (JL) be given. If fn � f in L 1 (JL), show that {fn}! U {f} is uniformly JL-absolutely continuous and uniformly JL-integrable. Conversely, assuming that JL(E ) < oo , show that fn � f in L 1 (JL ) if fn � f in JL measure and {fn}1 is uniformly JL-integrable. ( iv ) Assume that JL(E) == oo . We say that a family JC of measurable func tions f on (E, B, JL) is tight if for each E > 0 there is a E B such that JL( ) < oo and sup / E IC frc I! I dJL < E. Assuming that JC is tight, show that JC is uniformly JL-integrable if and only if it is uniformly JL-absolutely contin uous and sup / E IC ll f ii £ 1 (JL ) < oo . Next, show that JC is uniformly JL-integrable if JC is tight and sup / E IC J 1! 1 1 + 6 dJL < oo for some 8 > 0. Finally, suppose that {fn}1 C L 1 (JL) is tight and that fn � f in JL-measure. Show that l l fn - f i i£ 1 (JL ) � 0 if and only if {fn} is uniformly JL-integrable. 3.3. 20 Exercise:
r
r
r
r
Let (E, B, JL) be a finite measure space. Show that fn � f in JL-measure if and only if J l fn - ! I 1\ 1 dJL � 0.
3.3. 21 Exercise:
Let (E, p) be a metric space and {En}! a non-decreasing sequence of open subsets of E such that En / E. Let Jl and v be two measures on (E, BE ) with the properties that JL(En) V v(En) < oo for every n > 1 and J cp dJL == J cp dv whenever cp is a bounded p-uniformly continuous cp for which there is an n > 1 such that cp 0 off En. Show that Jl == v on BE . 3.3. 22 Exercise:
Although almost everywhere convergence does not follow from convergence in measure, it nearly does. Indeed, suppose that {fn}1 is a sequence of measurable, �-valued functions on (E, B, JL) . Given an �-valued, measurable function J, show that (3.3.8) holds, and therefore that fn � f both (a.e., JL) and in JL-measure, if 3.3. 23 Exercise:
00
L1 JL (Ifn - f l
> E)
0.
62
III Lebesgue Integration
In particular, conclude that fn � f (a.e., J-t ) and in J.-t- measure if
L1 l l fn - J IIL 1 (JL)
0, a 8 == 8 ( x , E) > 0 such that 0
�
r ! J (t) - f( x ) l dt < E whenever i 3 is an open interval with I I I < 8. };_ I I I it X
In other words, except on a set of Lebesgue measure 0, an integrable function can be recovered by differentiating its indefinite integral.
In order to understand the strategy behind our proof, first note that (3.4.1) is completely obvious when f is continuous. Hence, since ( cf. Corollary 3.3.14 ) the continuous elements of L 1 ( �) L 1 (,\R) are dense in L 1 (�), it suffices for us to show that the set g of f E L 1 (�) for which (3.4. 1) holds is closed in L 1 (�) . To this end, we introduce the Hardy-Littlewood maximal function _
(3.4.2)
M f( x )
sup
{};_I I I }{I i f(t) i dt : j } 3
x ,
xE
�'
for f E L 1 (�). Next, for each f E L 1 ( � ) and E > 0 set b,. ( f, E)
==
{x : 1'\.{)imx} };_I II Jt� i f(t) - f(x) i dt } >
t:
.
Clearly, (3.4.1) holds if and only if I D,.(j, E) I e == 0 for every E > 0. Moreover, for any E > 0 and any J, g E L 1 (�): D,. ( j, 3 E ) C { x : M (f - g) > E } U b,. (g, E) U { x : l g( x ) - f( x) ! > E } .
63
3.4 Lebesgue 's Differentiation Theorem
In particular, this means that if g E g, then
where, in the passage to the last line we have used l � (g, E) l e == 0 and Markov ' s inequality. Finally, suppose that { gn }1 C g and that 9n � f in £ 1 (� ) . Then, the preceding line of reasoning leads us to the conclusion that � � ( / , 3E) I e
< nlim ! {M(J - gn) > E} l e ' ---+ CX)
and so we would be done if we knew that ( 3.4.3) With the preceding in mind, we now turn our attention to the analysis of the Hardy-Littlewood maximal function. To begin with, we first note that M f is measurable, and therefore that we can drop the subscript "e" in ( 3.4.3). To see this, observe that an alternative expression for M f is 1 { sup Mf(x) = ,a b E(O , CX)) a + b }( -a ,b) I J(x + t) l dt and that, for each a, b E (0, oo ) and x < y ,
f I J( x + t) l dt - r I J( y + t) l dt < r I J(t) l dt . r ,b) r ,b) J( x -a , y-a]U[x + b, y+ b) J(-a J(-a Hence, by Exercise 3.3. 15, we know that, for each a , b E (0, oo ) , 1 r 1 1 ( x + t) 1 dt E R xER a + b J( -a ,b ) f---+
is uniformly continuous and, therefore, for each a E � ' that {x : Mf(x) > a} ==
{
i
1 : x l f(x + t) l dt > o: U b a + ( -a ,b ) a , b E (O , CX))
}
is open. We next observe that control on the size of M f will follow from control on the one-sided maximal functions M+ f(x) sup � { I J(t) l dt. I J(t) l dt and M_f(x) sup � { h>O J( x -h, x ] h>O J[x , x +h)
64
III Lebesgue Integration
Indeed, for any I 3 x , let h+ and h_ be the lengths of [x, oo ) ni and ( -oo, x] n i , respectively, and note that 0
0
1
0
� r I J(t) i dt = � I J( t ) l dt I J (t) i dt + � r I I I }J I I I J[ x,x + h+) I I I ( x - h_ , x ] < M+ f( x ) + M_ f(x) < M+ f(x) V M_ f( x) ,
�A
�il
and therefore, since it is essentially obvious that M+ f V M_ f < MJ, we have (3.4.4) as
Moreover, by precisely the same argument we used to prove that { Mf > a} is open, we know that the same is true of both {M+ f > a} and {M_ f > a}. Finally, note that M_ f( x) == M+ /( -x) , where /(t) == f( - t) , which means that we really need to learn how to control only M+ f · For this purpose, let f E £ 1 ( �) and E > 0 be given, and consider the function F€ (x) = { i f(t) i dt - EX , X E JR. }( - CX) , X] By Exercise 3.3. 15, Ff_ is uniformly continuous, and, obviously, lim x � ± CX) Ff_(x) == =f OO . Moreover, it is an elementary matter to see that as
Hence, we will see shortly, all that we need is the following wonderfully simple observation. 3.4.5 Lemma (Sunrise Lemma) . * Let F : � � � be a continuous func tion with the property that limx � ± CX) F (x) == =fOO . Set G == {x : 3y > x F( y ) > F(x) }. Then G is an open, and each non-empty, open, connected component of G is a bounded interval (a, {3) with F(a) < F({3) .
PROOF: Clearly G is open. Next, suppose that (a, {3 ) is a non-empty, con
nected, open component of G, and take 1 E (a, {3). If either {3 == oo or {3 < oo and F({3) < F(1) , then there exists a unique x E [1, {3) such that F(x) == F(1) and F( y ) < F(1) for all y > x . But this would mean that, on the one hand, x E G and, on the other hand, F( y ) < F(x) for all y > x , which is impossible. Hence, we now know that {3 < oo and that F( 1) < F ( {3 ) for all 1 E (a, {3) . Finally, from this it is clear that a > -oo and that F(a) < F({3). D * The name derives from the following picture. The sun is rising infinitely far to the right in mountainous (one-dimensional) terrain, F(x ) is the elevation at x, and G is the region in shadow at the instant when the sun comes over the horizon.
65
3.4 Lebesgue 's Differentiation Theorem
Applying Lemma 3.4.5 to the function Ff. , we see that G€ {M+ f > t } is either empty or (cf. the first part of Lemma 2. 1.9) is the union of countably many disjoint, bounded, open intervals ( an , f3n ) satisfying 0 < F€ (f3n ) - F€ ( an ) =
1(O:n ,/3n ) i f(t) i dt - E (f3n - an )
for each n. Hence, either IG € 1 == 0 or, after summing the preceding over n , we arrive at In other words, we have now proved first that and then, after taking left limits with respect to t > 0,
v
In fact, because M_ f( x) == M+ f( ) we also know that (3.4.6) continues to hold when M+ f is replaced by M_ f throughout. Finally, in conjunction with (3.4.4), these leads to -x ,
which means that we have now proved the renowned
Hardy-Littlewood
maximal inequality:
( 3.4.7) Since (3.4.7) certainly implies (3.4.3) , and (3.4.3) was the only missing in gredient in the program with which this section began, the derivation of the following statement is complete. 3.4.8 Theorem ( Lebesgue's Differentiation Theorem ) . For any Lebes gue integrable function f on �' (3.4. 1) holds. In particular, (3.4.9)
lim _..;_ { j ( t ) = j(x) }
i�{x} I I I
;
for almost every X
E JR.
Before closing this section, there are several comments which should be made. First, one should notice that the conclusions drawn in Theorem 3.4.8 remain true for any Lebesgue measurable f which is integrable on each compact
66
III Lebesgue Integration
subset of JR. Indeed, all the assertions there are completely local and therefore follow by replacing f with f l ( - R , R) , restricting ones attention to x E ( -R, R) , and then letting R / oo. Second, one should notice that (3.4. 7) would be a trivial consequence of Markov ' s inequality if we had the estimate l i M f i i £ 1 ( R ) < 2 II J II£ 1 (R ) · Thus, it is reasonable to ask whether such an estimate is true. That the answer is a resounding no can be most easily seen from the observation that, if ll f ii £ 1 ( R) f= 0, then a f( -r, r ) I J(t) l dt > 0 for some r > 0 and therefore Mf(x) > l xN-r for all x E JR. That is, if f E £ 1 (JR) does not vanish almost everywhere, then Mf is not integrable. (To see that the situation is even worse and that, in general, M f need not be integrable over bounded sets, see Exercise 3.4. 12 below.) Thus, in a very real sense, (3.4. 7) is about as well as one can do. (See Exercise 6.2.27 for an interesting continuation of these considerations.) Because this sort of situation arises quite often, inequalities of the form in ( 3.4. 7) have been given a special name: they are called weak-type inequalities to distinguish them from inequalities of the form li M J II £ 1 (R ) < C I I J I I£ 1 (R ) , which would be called a strong-type inequality. Finally, it should be clear that, except for the derivation of (3.4. 7), the arguments given here would work equally well in JR N . Thus, we would know that, for each Lebesgue integrable f on JR N , 1 I for almost every ) ( X E IR. N f(x) = 0 d J ( 3.4. 10) B lim y y l � { x} I B }B if we knew that c (3.4.1 1) I {Mf > E } l < -E IIJ IIL 1 ( AR N ) ' where M f is the Hardy-Littlewood maximal function Mf(x) �� � I IJ( y ) l dy I and B denotes a generic open ball in JR N . It turns out that (3.4. 11), and therefore (3.4.10), are both true. However, the proof of ( 3.4. 11 ) for N > 2 is somewhat more involved than the one which we have given of (3.4.7)*.
1r
L
Exercises
Define f : JR � [O, oo) so that f(x) == (x( log x) 2 ) - 1 if x E ( o, - 1 ) and f(x) == 0 if x � ( o, 1 ). Using Exercise 3.3. 18 and the Fundamental Theorem of Calculus, check that f E £ 1 (JR) and that - 1 , X E ( 0, - l ) . f ( t) dt = log X 3.4.12 Exercise: e
e
1(O,x)
-
e
In particular, conclude that f( o , r ) M f(x) dx == oo for every r > 0.
* See, for example, E . M . Stein's Singular Integrals and Differentiability Properties of Func tions, published by Princeton Univ . Press ( 1 970)
67
3.4 Lebesgue 's Differentiation Theorem
Given f E L 1 (1R) , define the Lebesgue set of f to be the set Leb(f) of those x E � for which the limit in (3.4. 1) is 0. Clearly, (3.4. 1) is the statement that Leb(j)C has Lebesgue measure 0, and clearly Leb(f) is the set on which f is well behaved in the sense that the averages m f; f (t) dt converge to f(x) as j \. {x} for X E Leb(f) . The purpose of this exercise is to show that, in the same sense, other averaging procedures converge to f on Leb(f). To be precise, let p be a bounded continuous function on � having one bounded, continuous derivative p'. Further, assume that p E £ 1 (JR) , J p(t) dt == 1, J l tp'(t) I dt < oo , and lim l t i � CX) p(t) == limt � CX) tp'(t) == 0. Next, for each t > 0, set pf. (t) == t - 1 p ( t - 1 t ) and define f€ (x) = p€ (x - t)f(t) dt, x E lR and f E L 1 (1R). 3.4. 13 Exercise:
J
The purpose of this exercise is to show that (3.4.14) ft: (x) � f(x) as t � 0 for each x E Leb(f) . (i) Show that, for any f E £ 1 (JR) and x E Leb(j),
8lim �0 ; j(x-8,x] f(t) dt. �0 ; J[rx ,x+8) f(t) dt f(x) 8lim v
(ii)
that
=
=
v
Assuming that f is continuous and vanishes off a compact set, first show f€ (x) =
rJ[o, CX) ) p( - t)f(x + d) dt + J[ro ,CX)) p(t)f(x - d) dt,
and then (using Exercise 3.3. 18 and Theorem 1.2.7) verify the following equal ities: (3.4.16) and
rJ[o, CX)) p( -t) f(x + Et) dt Jr(o, CX) ) tp' ( -t) ( ..!..tt J[rx ,x+t:t ) ! ( s) ds) dt =
rJ[o, CX)) p(t)f(x - d) dt
(j
)
f(s) ds dt. tp' (t) ..!..tt J( o , CX) ) (x-t:t , x] Next (using Corollary 3.3. 14) argue that ( 3.4.16) and (3.4. 16 ' ) continue to hold for every f E L 1 (1R) and x E Leb(J). (iii) Combining part (i) with (3.4. 16) and (3.4. 16 ' ), conclude that ( 3.4.16 ' )
=
-
r
' J f. (t) dt, for x Leb(J), tp (x) f -f(x) f.lim �O E
==
and, after another application of Exercise 3.3.18 and Theorem 2. 1.7, note that
- J tp' (t) dt J p(t) dt =
=
1.
Chapter IV Products of Measures
4. 1
Fubini's Theorem.
Just before Lemma 3.2.2, we introduced the product (E1 E2 , B1 B2 ) of two measurable spaces (E 1 , B 1 ) and (E2 , B2 ). We now want to show that if Jl i , i E { 1, 2}, is a measure on ( Ei , Bi ), then, under reasonable conditions, there is a unique measure v on (E1 E2 , B1 B2 ) with the property that v(rl X r2 ) == Il l (rl ) JL(r2 ) for all ri E Bi . The key to the construction of v is found in the following function analog of ?T- and A-systems ( cf. Lemma 3. 1.3). Namely, given a space E, we will say that a collection £, of functions f : E � ( ] is a semi-lattice if both j + and f - are in £, whenever f E .C . A sub collection /C C £, will be called an £-system if: ( a ) 1 E /C; ( b ) if f, g E /C and { ! == oo} n {g == oo} == 0, then g - f E /C whenever either f < g or g - f E .C; ( c ) if a , f3 E [0, ) and J , g E /C, then af + (3g E /C; (d) if {fn}1 C /C and fn / f, then f E /C whenever f is bounded or f E .C . x
x
x
x
- oo, oo
oo
The analog of Lemma 3. 1.3 in this context is the following. 4. 1 . 1
Lemma. Let C be a ?T-system which generates the a-algebra
and let £, be a semi-lattice of functions f : E ( ] £-system and l r E /C for every r E C, then /C contains every f E measurable on ( E, B) .
E,
�
- oo, oo .
B
over If /C is an £, which is
PROOF: First note that {r C E : 1 r E /C} is a A-system which contains C. Hence, by Lemma 3. 1.3, 1 r E /C for every r E B. Combined with ( c ) above, this means that /C contains every non-negative, measurable, simple function
on (E, B) . Next, suppose that f E £, is measurable on (E, B). Then both j + and f - have the same properties, and, by ( b ) above, it is enough to show that j+ , f- E /C in order to know that f E /C. Thus, without loss in generality, we assume that f E £, is a non-negative measurable function on (E, B). But in that case f is the non-decreasing limit of non-negative measurable simple functions ; and so f E /C by (d) . D
4.1
Fubini 's Theorem
69
The power of Lemma 4. 1 . 1 to handle questions involving products is already apparent in the following. 4.1.2 Lemma. Let (E1 , 8 1 ) and (E2 , 82 ) be measurable spaces, and suppose that f is an �-valued measurable function on (E 1 x E2 , 8 1 x 82 ). Then for each x 1 E E1 and x 2 E E2 , j(x 1 , ) and f( x 2 ) are measurable functions on (E2 , 82 ) and (E1 , 8 1 ), respectively. Next, suppose that Jli , i E {1, 2}, is a finite measure on ( Ei, 8i ). Then for every measurable function f on ( E1 x E2 , 8 1 x 82 ) ·
·
,
which is either bounded or non-negative, the functions
(E1 , 81 ) and (E2, 82 ), respectively. PROOF: Clearly it is enough to check all these assertions when f is non negative. Let £, be the collection of all non-negative functions on E1 x E2 , and define /C to be those elements of £, which have all the asserted properties. It is clear that 1 r1 xr2 E /C for all ri E 8i . Moreover, it is easy to check that /C is an £-system. Hence, by Lemma 4. 1.1, we are done. D
are measurable on
(E 1 , 8 1 , Jl i ) and (E2 , 82 , Jl 2 ) of finite measure spaces, there exists a unique measure v on (E1 x E2 , 8 1 x 82 ) such that 4.1.3
Lemma. Given a pair
Moreover, for every non-negative function
rE
J (4.1.4)
1 xE
f on (E1 x E2 , 8 1 x 82 ),
v(dx ) x ) x , f(x dx l l 2 2 2
=
L2 (L1 j(x 1 , x2 ) M 1 (dx 1 ) ) M2 (dx2 ) L1 (L2 J(x 1 , x2 ) 1-t2 (dx2 ) ) /-ll (dx l ) · =
PROOF: The uniqueness of v is guaranteed by Exercise 3. 1.8. To prove the
existence of v, define
and
70
IV Products of Measures
for r E B l X 82 . using the Monotone Convergence Theorem, one sees that both v1 , 2 and v2 , 1 are finite measures on (E1 E2 , 8 1 82 ). Moreover, by the same sort of argument as was used to prove Lemma 4. 1.2, for every non-negative measurable function on (E1 E2 , 81 82 ): x
x
and
x
x
j j dv2 , 1 l2 (l1 f(x 1 , x2 ) /-l2 (dx2 ) ) /-l 1 (dx 1 ). =
Finally, since VI , 2 (rl X r2 ) == JL(rl ) JL(r2 ) == v2 , 1 (rl X r2 ) for all ri E Bi we see that both v1 , 2 and v2 , 1 fulfill the requirements placed on v. Hence, not only does v exist, but it is also equal to both v1 , 2 and v2 , 1 ; and so the preceding equalities lead to (4.1.4). D In order to extend the preceding construction to measures which need not be finite, we must ( cf. Exercise 4.1. 12) introduce a qualified notion of finiteness. Namely, we will say that the measure JL on (E, B) is a- fi nite and will call (E, B, JL) a a-finite measure space if E can be written as the union of a countable number of sets r E B for each of which JL(r) < oo . Thus, for example, (� N , BRN , AR N ) is a a-finite measure space. 4.1.5 Theorem ( Tonelli's Theorem ) . Let (E1 , 8 1 , JL I ) and (E2 , 82 , JL 2 ) be a-finite measure spaces. Then there is a unique measure v on (E1 E2 , 8 1 82 ) such that v(rl X r2 ) == JL I (rl ) JL(r2 ) for all ri E Bi · In addition, for every non negative measurable function f on ( E1 E2 , 8 1 82 ), J f( · , x 2 ) JL2 (dx 2 ) and J j(x 1 , · ) JL I (dx l ) are measurable on (E1 , B 1 ) and (E2 , B2 ) , respectively, and (4.1 .4) continues to hold. PROOF: Choose sequences {Ei, n}� 1 C Bi for i E {1, 2} so that JLi (Ei,n) < oo for each n > 1 and Ei == U� 1 Ei,n · Without loss in generality, we assume that Ei,m n Ei, n == 0 for m f= n . For each n E z + , define JLi , n(ri) == JLi (ri n Ei, n), ri E Bi ; and, for ( m , n ) E z + 2 ' let V(m , n ) on (El X E2 , B l X 82 ) be the measure constructed from JL I,m and JL 2 , n as in Lemma 4. 1.3. Clearly, by Lemma 4.1.2, for any non-negative measurable function f on ( El X E2 , Bl X 82 ), '
x
x
x
x
is measurable on (E 1 , 8 1 ) ; and, similarly, JE 1 j(x 1 , · ) JL I (dx 1 ) is measurable on (E2 , 82 ). Finally, the map r E 8 1 B2 r-----+ E:, n =l v( m , n ) (r) defines a measure vo on (E 1 E2 , 8 1 82 ), and it is easy to check that vo has all the required properties. At the same time, if v is any other measure on (E1 E2 , 81 82 ) for which v(rl X r 2 ) == /L l (rl ) JL 2 (r2 ), ri E Bi , then, by the uniqueness assertion x
x
x
x
x
4. 1
71
Fubini 's Theorem
in Lemma 4.2 1.3, v coincides with v(m , n ) on 81 82 [El,m E2 , n ] for each ( m , n) E z + and is therefore equal to Vo on 8 1 82 . D The measure v constructed in Theorem 4. 1.5 is called the product of Jll times JL2 and is denoted by Jl l Jl2 · x
x
X
x
4.1.6
Theorem (Fubini's Theorem) . Let (EI , 8 1 , Jll ) and (E2 , 82 , JL2 ) be
a-finite measure spaces and f a measurable function on (E1 Then the f is Jll x Jl2-integrable if and only if
x
E2 , 8 1
x
82 ) .
if and only if
Next, set
and
and define fi on Ei , i
E {1, 2},
by XI
if E Al otherwise and
Then fi is an �-valued, measurable function on ( Ei , 8i ) . Finally, if f is Jll x Jl2integrable, then Jli (Ai C) == 0, fi E L 1 (J-Li ), and
( 4. 1.7) for i
E {1, 2}.
72
IV Products of Measures
PROOF: The first assertion is an immediate consequence of Theorem 4. 1.5.
Moreover, since Ai E Bi , it is easy to check (from Lemma 4. 1.2) that fi is an �-valued, measurable function on (Ei , Bi)· Finally, if f is Jl l x JL 2 -integrable, then, by the first assertion, Jli(Ai C) == 0 and fi E L 1 ( J-Li )· Hence, by Theorem 4.1.5 applied to j + and f- , we see that
r 1
E 1 x E2
j(x 1 , X 2 ) ( J.Ll X J.L2 ) (dx 1 X dx 2 )
- r
}A l
=
X E2
� - (x 1 , X 2 )
( J.Ll X J.L2 ) ( dx 1 X dx 2 )
1 1 (L2 j+ (x 1 , x2 ) J.L 1 (dx2 ) ) J.Ll (dxt ) - 1 1 (L2 f (x 1 , x2 ) J.L2 (dx2 ) ) J.L 1 (dx 1 ) -
=
r ft (X ! ) J.L 1 ( dx ! ) i
jA
1
and the same line of reasoning applies to f2 .
D
Exercises
Let (E, B, JL) be a a-finite measure space. Given a non negative measurable function f on (E, B) , define r(j) == { (x, t) E E x [O, oo) : t < f (x) } 4.1.8
Exercise:
and f (J )
{ (x, t) E E x [O, oo) : t < f (x ) }. -Show that both r(j) and r(j) are elements of B X B[o ,CX)) and, in addition, that ==
( 4.1.9) In proving measurability, consider the function (x , t ) E E x [0, oo) r-----+ f(x ) - t E ( - oo, oo] ; and get (4. 1.9) as an application of Tonelli ' s Theorem. Clearly (4. 1.9) can be interpreted as the statement that the integral of a non Hint :
negative function is the area under its graph.
4. 1 Fubini 's Theorem
73
Let (E1 , 81 , J.-t 1 ) and (E2 , 82 , J.-t 2 ) be a-finite measure spaces and assume that, for i E { 1, 2} , 8i == a(Ei; Ci), where Ci is a 1r-system con taining a sequence {Ei, n}� 1 such that Ei == U� 1 Ei , n and J-t i(Ei , n) < oo, n > 1. Show that if v is a measure on (E 1 E2 , 81 82 ) with the property that v(r1 X r2 ) == J.-l 1 (r1 ) J.-l 2 (r2 ) for all ri E Ci , then V == J.-l 1 X J.-l 2 . Use this fact to show that, for any M, N E z + , 4. 1 . 10 Exercise:
x
x
Let (E1 , 8 1 , J.-t 1 ) and (E2 , 82 , J.-t2 ) be a-finite measure spaces. Given r E 8 1 X 82 ' define r(l ) (x2 ) { Xi E El : (x l , x2 ) E r } E B l for X 2 E E2
4. 1 . 1 1 Exercise:
and r(2 ) (x l )
{ X2 E E2
:
(x l , x 2 ) E r } for Xi E El .
Check both that r( i ) (xj) E 8i for each Xj E Ej and that Xj E Ej r-----+ J.-ti ( r( i ) (xj) ) E [O, oo] is measurable on (Ej, 8j) ({i, j} == {1, 2}). Finally, show that JL 1 JL2 (r) = 0 if and only if JLi (r( i ) ( xi ) ) = 0 for ttralmost every xi E Ei ; and, conclude that JL 1 (r(l ) ( x2 ) ) = 0 for tt2 -almost every x 2 E E2 if and only if /L 2 (r(2) ( xt) ) = 0 for IL l -almost every Xi E El . In other words, r E Bl X 82 has J.-t 1 J.-t 2 -measure 0 if and only if �-t 1 -almost every vertical slice (J.-t 2 -almost every horizontal slice) has �-t 2 -measure (J.-t 1 -measure) 0. 4. 1 . 12 Exercise: The condition that the measure spaces of which one is taking a product be a-finite is essential if one wants to carry out the program in this section. To see this, let E1 == E2 == (0, 1 ) and 8 1 == 82 == 8( o, I ) . Define J.-l 1 on ( E1 , 8 1 ) so that �-t 1 (r) is the number of elements in r ( oo if r is not a finite set) and show that J.-li is a measure on (E1 , 8 1 ) . Next, take �-t 2 to be Lebesgue ' s measure A (o, I ) on (E2 , 82 ). Show that there is a set r E 81 82 such that x
x
x
r l (x l , X 2 ) JL 2 (dx 2 ) = 0 for every X l E El
JE2
r
but ( Hint : Try r == {(x, x) : 0 < X < 1}.) In particular, there is no way that the second equality in ( 4.1.4 ) can be made to hold. Notice that what fails here is really the uniqueness and not the existence in Lemma 4. 1.3.
IV Products
74
of Measures
< a < b < oo and a function f : (a, b) E � � with the properties that /( · , x) E C((a, b)) for every x E E and f ( t, ) is measurable on ( E, B) for every t E (a, b), show that f is measurable on ( (a , b) E, B( a, b) B) . Next, suppose that f ( , x) E C 1 ((a , b)) for each x E E, set f'(t, x) == d� f(t, x) , x E E, and show that f' is measurable on ( (a, b) E, B( a , b ) B) . Finally, suppose that Jl is a measure on (E, B) and that there is a g E L 1 (JL) such that I f( t, x ) I V I f' ( t, x ) I < g(x ) for all (t, x ) E (a , b) E. Show not only that JE f( x) JL(dx) E C 1 ((a, b)) but also that f(t, x) J.L (dx) = f ' (t, x) J.L (dx) . 4. 1.13 Exercise: Let (E, B) be a measurable space. Given
-oo
x
·
x
x
x
x
x
·
�l
4.2
·
,
l
Steiner Symmetrization and the Isodiametric Inequality.
In order to provide an example which displays the power of Fubini ' s The orem, we will prove in this section an elementary but important inequality about Lebesgue ' s measure. Namely, we will show that, for any bounded sub set r C �N , (4.2. 1) where n N denotes the volume of the unit ball B(O, 1 ) in �N and rad(r) sup { ly; x l : x, y E r } _
is the radius (i.e., half the di amet er) of r. Notice (cf. ( ii ) in Exercise 2.2.3) that what (4.2.1 ) says is that, among all the subsets of �N with a given diameter, the ball of that diameter has the largest volume; it is for this reason that (4.2. 1) is called the isodiametric inequality. At first glance one might be inclined to think that there is nothing to (4.2.1). Indeed, one might carelessly suppose that every r is a subset of a closed ball of radius rad(r) and therefore that (4.2.1) is trivial. This is true when N == 1. However, after a moment ' s thought, one realizes that, for N > 1, although r is always contained in a closed ball whose radius is equal to the diameter of r, it is not necessarily contained in one with the same radius as r. ( For example, consider an equilateral triangle in �2 .) Thus, the inequality 1 r 1e < n N ( 2rad ( r )) N is trivial, but the inequality in ( 4.2. 1) is not! On the other hand, there are many r ' s for which (4.2.1) is easy. In particular, if r is symmetric in the sense that r == - r { -X : X E r}, then it is clear that x E r ===} 2 l x l == l x + x l < 2rad(r) and therefore r c B (O, rad(r) ) . Hence ( 4.2.1) is trivial when r is symmetric, and so all that we have to do is devise a procedure to reduce the general case to the symmetric one.
75
4.2 Steiner Symmetrization and the Isodiametric Inequality
The method with which we will perform this reduction is based on a famous construction known as the Steiner symmetrization procedure. To describe Steiner ' s procedure, we must first introduce a little notation. Given v from the unit (.LV - 1 ) -sphere s N - 1 {x E �N : l x l == 1 }, let L(v) denote the line {tv : t E �}, P(v) the (N - I ) -dimensional subspace {x E �N : x ..l v}, and define S(r; v) {x + tv : x E P(v) and l t l < �f(r; v, x) } , where f(r; v, x) l {t E � : x + tv E r} l e is the length of the intersection of the line + L(v) with r. Notice that, in the creation S(r; v) from r, we have taken the intersection of r with x + L(v) , squashed it to remove all gaps, and then slid the resulting inter val along x + L(v) until its center point falls on x. In particular, S(r; v) is the symmetrization of r with respect to the subspace P(v) in the sense that, for each x E P(v), X
(4.2.2)
x + tv E S(r; v)
¢:::::;>
x - tv E S(r; v);
what is only slightly less obvious is that S(r; v) possesses the properties proved in the next lemma. 4.2.3 Lemma. Let r be a bounded element of BRN . Then, for each v E s N - 1 , S(r; v) is also a bounded element of �N , rad (S( r; v) ) < rad ( r ) , and I S(r; v) I == 1 r 1 . Finally, if R : �N � �N is rotation for which L(v) and r are invariant (i.e., R ( L(v) ) == L(v) and R(r) == r), then RS(r; v) == S(r; v) . PROOF: We begin with the observations that there is nothing to do when N == 1 and that, because none of the quantities under consideration depends on the particular choice of coordinate axes, we may and will assume not only that N > 1 but also that v == eN _ (0, . . . , 0, 1 ) . In particular, this means, by Lemma 4. 1 .2, that a
� E IRN - l f(�) � l lr((�, t) ) dt E [0, oo) �
is BR N- 1 -measurable; and therefore, by Exercise 4. 1 .8, because S(r; e N ) is equal to { ( � , t) E �N - 1 X [O, oo) : t < ( ) } U { ( � , t) E �N - 1 X ( - oo, O] : - t < ( ) } ,
f�
we know both that S(; eN ) is an element of BRN and that where, in the final step, we have applied Tonelli ' s Theorem.
f�
76
IV Products of Measures
We next turn to the proof that rad (S (r ; eN ) ) cannot be larger than rad(r) ; and, in doing so, we will, without loss in generality, add the assumption that r is compact. Now suppose that x , y E S(r; eN ) are given, and choose � ' 'fJ E �N - l and s, t E � so that x == ( � , s) and y == ( TJ , t) . Next, set M ± (x) == ± sup{ T : ( � , ± 7 ) E r} and M ± ( y ) == ± sup{ T : ( TJ , ±T ) E r}, and note that, because r is compact, all four of the points x ± (� ' M ± (x)) and y ± == ( TJ , M ± ( y )) are elements of r. Moreover, 2 l s l < M+ (x) - M - (x) and 2 l t 1 < M + ( y ) - M - ( y ) ; and therefore
In particular, this means that 2 2 2 2 2 < == t s t x s IY - l ITJ - �1 + I - l I TJ - �1 + ( l i + 1 l ) 2 2 + + < 111 - �1 + ( ( M ( y ) - M - (x)) V ( M (x) - M - ( y ) ) ) < (I Y + - x - 1 v IX + - y - 1) 2 < 4 rad(r) 2 . Finally, let R be a rotation. It is then an easy matter to check that P(Rv) == R ( P(v) ) and that f(Rr ; Rv, Rx) == f(r ; v, x) for all x E P(v). Hence, S(Rr, Rv) == RS(r, v) . In particular, if r == Rr and L(v) == R (L (v)) , then Rv == ± v, and so the preceding ( together with (4.2.2)) leads to RS( r, v ) == S(r, v). D Theorem. The inequality in (4.2.1) holds for every bounded r C �N . PROOF: Clearly it suffices to treat the case when r is compact and therefore Borel measurable. Thus, let a compact r be given, choose an orthonormal basis {e l , . . . , eN } for �N , set r o == r, and define rn == S(rn - l ; en) for 1 < n < N. By repeated application of (4.2.2) and Lemma 4.2.3, we know that 1 rn1 == 1 r 1 , rad(rN ) < rad(r) , and that Rm rn == rn , 1 < m < n < N, where Rm is the rotation given by Rm x == x - 2(x , em )RNem for each x E �N . In particular, this means that Rm rN == Rm rN for all 1 < m < N, hence - rN == rN , and therefore ( cf. the discussion preceding the introduction of Steiner ' s procedure) 4.2.4
We will now use ( 4.2.1) to give a description, due to F. Hausdorff, of Lebes gue ' s measure which, as distinguished from the one given at the beginning of
4.2 Steiner Symmetrization and the Isodiametric Inequality
r
77
Section 2. 1, is completely coordinate free. Namely, we are going to show that, for all C �N , (4.2.5)
1 r 1 e HN (r) inf { L ON rad(C) N ==
:
C a countable cover of
CEC
r} .
H (r)
We emphasize that we have placed no restriction on the sets C making up the cover C. On the other hand, it should be clear that N would be unchanged if we were to restrict ourselves to coverings by closed sets or, for that matter, to coverings by open sets. Directly from its definition, one sees that H N is monotone and subadditive in the sense that and
HN (Qrn) < �HN(rn)
rn
for all { } f' C P (R N ) .
Indeed, the first of these is completely trivial, and the second follows by choos ing, for a given E > 0, {Cm }1 so that c U cm and n N (rad ( C )) N < H N m + T m f
rm
(r )
L
CEC m
and noting that
Moreover, because
1r 1 e < L C i e for any contable cover C, the inequality 1 r 1 e < H N (r) is an essentially trivial consequence of ( 4.2.1). In order to prove the opposite inequality, we will use the following lemma. I
CEC
sequence {Bn}! property that
�N
with IGI < oo , there exists a of mutually disjoint closed balls contained in G with the
4.2.6 Lemma. For any open set G in 00
(4.2.7)
PROOF: If G == 0 , there is nothing to do. Thus, assume that G =I= 0 , and set
Go
==
G.
Using Lemma 2. 1.9, choose a countable, exact cover C0 of G0 by
78
IV Products of Measures
non-overlapping cubes Q. Next, given Q E C0 , choose x E JRN and 8 E [0, oo ) so that N i Q == Il [x - 8, xi + 8] and set BQ == B (x , � ) . 1 Clearly, the BQ ' s are mutually disjoint closed balls. At the same time, there is a dimensional constant a N E (0, 1) for which I BQ I > a N I Q I ; and therefore we can choose a finite subset { Bo, I . . . , Bo, no } C { BQ : Q E Co } in such a way that no Go \ U Bo,m < ,BN I Go l where ,BN 1 - ¥- E (0, 1). m= 1 Now set G 1 == G0 \ U�o Bo , m · Noting that G 1 is again non-empty and open, we can repeat the preceding argument to find a finite collection of mutually disjoint closed balls B1 , m C G 1 , 1 < m < n 1 , in such a way that n1 G 1 \ U B 1 ,m < ,BN I G 1 I · m= 1 More generally, we can use induction on f E z + to construct open sets Gt C Gt - 1 and finite collections Bt , 1 , , Bt , n�. of mutually disjoint closed balls B c Gt so that nt I Gt + l l < ,BN I Gt l where Gt+ 1 == Gt \ U Bt ,m ; m= 1 clearly the collection •
•
•
:
{ Bt ,m f E N and 1 < m < n�_} has the required properties. D RN . PROOF: As we have already pointed out, the inequality 1r1e < H N ( r ) is an immediate consequence of ( 4.2. 1 ) . To get the opposite inequality, first observe that HN ( r ) == 0 if 1r1e == 0. Indeed, if 1r1e == 0, then, for each E > 0 we can first find an open G � r with I G I < E and then, by Lemma 2.1.9, a countable, exact cover C of G by non-overlapping cubes Q, which means that 4.2.8 Theorem. The equality in (4.2.5) holds for any set r c
Next, because H N is countably subadditive, it suffices to prove that H N ( r ) < l r le for bounded sets r. Finally, suppose that r is a bounded set, and let G be
4.2 Steiner Symmetrization and the Isodiametric Inequality
79
any open superset of r with I G I < oo . By Lemma 4.2. 7, we can find a sequence {Bn}! of mutually disjoint closed balls B C G for which CX)
I G \ A I == 0 where A == U Bn. 1
Hence, because H N (r) that H N (f)
1. Finally, since cp is continuous ( a. e., v) , cp
and so, not only is cp ( (a, b] , B (a , b] ) , but also -
v
== nlim ----+- CX) -cpn == nlim ----+- CX) 'Pn ( a.e., v); ==
limn CX) cpn ( a.e., v) and therefore measurable on ---+-
r - t) AIR ( dt) . J( , ) JE Hence, either JL(f > 8) == oo for some 8 > 0, in which case both sides of ( 5.1.5) are infinite, or for each 0 < 8 < r < oo the map t E [8, r] � cp '(t)JL(f > t) is (5. 1.5)
Riemann integrable and
o f(x) J-L( dx) = lim (R) f r.p ' (t ) J-L(f > t) dt. f r.p }E r/8 �oo0 }[8,r] PROOF: It is clear that t E (0, oo) � JL(f > t) is right-continuous and non increasing. Hence, if 8 == sup{ t E (0, oo) : JL(f > t) == oo } , then JL (f > t) == oo for t E (0, 8) and t E (8, oo) � JL(f > t) has at most a countable number of
discontinuities. Furthermore, if
h (t ) == Jl ( f > ktl ) for t E ( � , ktl J , k > 0, and n > 1, n
then each hn is clearly measurable on ( (O, oo), B( o , oo ) ) and h (t ) � JL(f > t) for each t E (O, oo). Hence, t E (O, oo) � JL(f > t) is measurable on ( ( 0, oo), B(o , oo ) ) . We turn next to the proof of ( 5.1.5). Since (cf. Exercise 3.3. 18) n
limo f o:�
}( o: , 8]
limo (R) {6 r.p' ( t) dt = r.p( 8) r.p' ( t) dt = o:� l
o:
and therefore
(l
'P
0
f dfJ,
) (lco, oo ) r.p' (t) J-L(f 1\
>
t) AJR (dt )
)
>
r.p (8 ) J-L(f > 8) ,
it is clear that both sides of (5. 1.5) are infinite when JL(f > 8) == oo for some 8 > 0. Thus we will assume that JL(f > 8) < oo for every 8 > 0. Then the restriction of Jl J to B[8 , oo ) is a finite measure for every 8 > 0. Given 8 > 0, set
84
V Changes
of Variable
==
J.t J ((8, t]) for t E [8, oo) and apply Theorem 5.1.2 and Theorem 1.2.7 to see that r dJ.L J = (R) r
8 } , and so the required convergence follows by the Monotone Con vergence Theorem. Finally, to prove the last part of the theorem when J.t(f > 8) < oo for every 8 > 0, simply note that I
1
r
J(O
cp ' (t)�-t(f j r/ oo (li, r]
lim 1 (t) J.L (f > t) AR(dt) = 6)0
t) AR(dt) ,
5. 2 Polar
Coordinates
85
Exercises
5 . 1 . 6 Exercise:
Here are two familiar applications of the ideas discussed in
this section. (i) Let 1/J be a continuous, non-decreasing function on the compact interval [a, b] . Show that (5. 1 . 7)
(R) { f o '1/J ( s) d'lj;( ) = (R) { s
1[a , b]
1[1/J (a),'lj;(b)]
f(t) dt, f E C ( [a, bl ) .
Suppose that Jl is a measure on ( �, BR) with the properties that JL(l) < oo for each compact interval I and JL( { t}) = 0 for each t E � - Let q, : � � � be a function satisfying JL( [a, b] ) = q,(b) - q,( a ) for all - oo < a < b < oo . Note that q, is necessarily continuous and non-decreasing, and show that q,*Jl coincides with the restriction of AR to BR [q,(�) J . (ii)
A particularly important case of Theorem cp(t) = tP for some p E (0, oo ), in which case (5. 1 . 5) yields 5 . 1 . 8 Exercise:
5. 1 .4
1s when
(5. 1 .9)
Use (5. 1 .9) to show that I J I P is JL-integrable if and only if CX)
CX)
Ln= 1 nP+ l J-l ( l f l > ! ) + nL= 1 nP- 1 1-l ( l f l > n) 1
< oo .
Compare this result to the one obtained in the last part of Exercise 3.3. 16.
5.2 Polar Coordinates.
From now on, at least whenever the meaning is clear from the context, we will use the notation "dx " instead of the more cumbersome ",\R N ( dx ) " when doing Lebesgue integration with respect to Lebesgue ' s measure on �N . In this section, we examine a change of variables which plays an extremely important role in the evaluation of many Lebesgue integrals over �N . Let s N - 1 denote the unit (N - 1 )-sphere { x E �N : lx l = 1 } in � N , and define : � N \ {0} -----+ s N - 1 by q,(x) = 1�1 · Clearly q, is continuous. Next, define the surface measure AsN on s N - 1 to be the image under q, of N AR N restricted to Bn ( o , 1 )\{ 0 } · Noting that q,(rx) = q,(x) for all r > 0 and x E �N \ {0} , we conclude from Exercise 5.0.3 that { j o if!(x) dx = r N { j o if! (x) dx 1B ( O , r)\{ 0 } 1B (0, 1 )\{ 0 }
-1
V Changes of Variable
86 and therefore that
N r {n j o if!(x) dx = N {sN- 1 j(w) >. gN- 1 (dw) (5.2.1) J J ( O , r)\{ 0 } for every non-negative measurable f on ( S N - 1 , B8N-1 ). In particular, using N--11 - AsN - 1 ( s N - 1 ) to denote the surface area of s N - 1 ' we have that W� is the volume n N of the unit ball B(O, 1) in �N . Next, define w : (O, oo) X s N - 1 � �N \ {0} by w(r, w) = rw. Note that w is one-to-one and onto; the pair (r, w) w - 1 (x) = (lx l , �(x)) are called the polar coordinates of the point x E � N \ {0}. Finally, define the measure RN on ( (0, oo) , B( o , CX) ) ) by RN (r) = fr r N - 1 dr. The importance of these considerations is contained in the following result.
w
5.2.2 Theorem. Referring to the preceding, one has that
ARN = w* ( RN X AsN- 1 ) a
on
BR N \{ 0 } ·
In particular, if f is non-negative, measurable function on
(
(�N , BR N ) ,
then
)
r N j(x) dx = r r N - 1 rsN- j(rw) ).gN- 1 (dw) dr l 1 JR J(O , CX) ) (5.2.3) = rsN- 1 r f(rw)r N - 1 dr AsN- 1 (dw) . J( o , CX)) l PROOF: By Exercise 3.3.22 and Theorem 4.1.5, all that we have to do is check that the first equation in ( 5.2.3 ) holds for every f E Cc ( �N ) {! E C ( �N ) : f 0 off of some compact set}. To this end, let f E Cc( �N ) be given and set F(r) = fn ( o ,r) f(x) dx for r > 0. Then, by (5.2.1), for all r, h > 0:
(
)
=
f(x) dx F(r + h) - F(r) = f jB ( O , r + h)\B (O , r) f o \II ( r, if!(x)) dx B ( O , r + h)\B ( O , r) ( f(x) - f w(r, �(x)) ) dx + B ( O , r + h)\B ( O , r) (r + h) N - r N o(h), sN-l (dw) + o w) \II ( r, f >. N gN- 1 where "o ( h)" denotes a function which tends to 0 faster than h. Hence, F is continuously differentiable on (0, oo) and its derivative at r E (0, oo) is
J
=
i
J
0
5. 2 Polar Coordinates
87
given by r N - 1 fsN- 1 f o 'll (r, w) AsN-1 (dw). Since F(r) -----+ 0 as r '\. 0, the desired result now follows from Theorem 5. 1.2 and the Fundamental Theorem of Calculus. D Exercises
5.2.4 Exercise:
In this exercise we discuss a few elementary properties of
AgN- 1 . ( i ) Show that if r =I= 0 is an open subset of s N - 1 , then AgN- 1 (r) > 0. Next, show that AsN- 1 is rotation invariant . That is, show that if 0 is an N N orthogonal matrix and To is the associated transformation on �N ( cf. the paragraph preceding Theorem 2.2.2), then ( To ) * AsN- 1 == AsN- 1 . Finally, use this fact to show that (5.2.5 ( i )) (e, w)RN AsN- 1 (dw) = o sN- 1 and (5.2.5 ( ii)) (e , w ) R N (TJ , w) JRN AsN- 1 ( dw) = n N (e , TJ ) RN S N- 1 for any e , TJ E �N . Hint: In proving these, let e E � N \ {0} be given and consider the rotation Oe which sends e to - e but acts as the identity on the orthogonal complement of e . x
I
I
( ii ) Define
:
[ ��::] ,
and set J.L = * A [o, 2 ,.1 . Given S 1 by (O) = any rotation invariant finite measure v on (S 1 , B81 ), show that v = "�!1 ) J.L· In particular, conclude that ,\8 1 == M · ( Cf. Exercise 5.3. 15 below. ) Hint: Define 0 cos () sin () for () E [0, 21r] , == - sin 0 cos 0 and note that v = _.!._ f f f d f f o To8 (w) v (dw) dO. 27r s1 s1 J[o, 21r) l l ( iii ) For N E z + ' define 3 : [ - 1, 1] X s N - 1 -----+ s N by 1 2 S(p, w) = ( 1 - � ) 2 w ' and let l 2 1 (r) = � p ( - ) AIR A8N- 1 (dp dw) J.L N for r E B[ - 1 , 1 ) X BsN- 1 . Show that AsN == 3* M N · [0, 21r]
0
-----+
[
]
(
£
)
]
[
x
x
88
V Changes of Variable
( Hint: Consider
(
)
{ rN { r 3 (p, w)) J-L N (dp X dw ) dr f ( sN1 1[ - 1 , 1 ] x 1( o, oo ) for continuous f : �N + 1 � � with compact support.) Finally, use this result to show that rsN J ( ( O , w)JRN ) AsN (dw) = WN - 1 r f(p) ( l - p2 ) � - 1 dp 1 1[ - 1 , 1 ] for all (} E s N and all measurable f on ( [- 1, 1] , B[ - 1 , 1 1 ) which are either bounded or non-negative. Perform the calculations outlined in the following. (i ) Justify Gauss ' s trick: 2 r � � r e - dx == rJR2 e - 2 dx == 27r r re - 2 dr == 27r 1( o, oo ) 1R 1 and conclude that for any N E z + and symmetric N N-matrix A which is strictly positive definite ( i.e., all the eigenvalues of A are strictly positive), 5 . 2 .6 Exercise:
(
2
2)
x
(5.2.7) try the change of variable �(x) == TA _ ! x. ( ii ) D efine r ( 1' ) == f( o, oo ) fY - 1 e - t d for 1' E (0, oo). Show that, for any 1' E (0, oo ) , r(1' + 1) == 1'r( 1') · Also, show that r ( � ) == 1r � . The function r( · ) is called the Euler ' s Gamma function. Notice that it provides an extension of the factorial function in the sense that r( n + 1) == n! for integers n > 0. ( iii ) Show that N 2 2{ 7r) WN - 1 = r ( � ) ,
Hint :
t
and conclude that the volume O N of the N-dimensional unit ball is given by N 2 7r f! N == r ( -N + 1 ) ( iv ) Given a , E (O, oo) , show that
/3
. 2
1(O, oo ) t-! exp [ 2 t 2
-a
-
/3-t2 ]
5. 3
89
Jacobi 's Transformation and Surface Measure
Finally, use the preceding to show that
1
[
]
a2 7r ! e - 2 o ,B 2 � fJ t- exp - a t - t dt == a . fJ ( O ,CX)) Hint: Define 'lj;( s ) for s E � to be the unique t E (0, oo ) satisfying s at ! - (3t - ! , and use part ( i ) of Exercise 5. 1 .6 to show that 2 o ,8 2 e (3 2 - s ds. e t - 1 exp - a t - - dt == t a R
1
( O ,CX))
2
2
[
-
]
1
2
5.3 Jacobi's Transformation and Surface Measure.
We begin this section by deriving Jacobi ' s famous generalization to non linear maps of the result in Theorem 2.2.2. We will then apply Jacobi ' s result to show that Lebesgue 's measure can be differentiated across a smooth surface. Given an open set G C �N and a continuously differentiable map* x E G r-----+ 4l(x) ==
�(x)
E �N '
we define the Jacobian matrix J4l(x) == �! (x) of 4l at x to be the N N matrix whose jth column is the vector x
84>1 8x · J
In addition, we call 8 4l( x ) l det(J4l(x)) l the Jacobian of 4l at x. 5 .3. 1 Lemma. Let G be an open set in � N and 4l an element of C ( G ; � N )
1
whose Jacobian never vanishes on G. Then tit maps open (or BRN -measurable) subsets of G into open or (BRN -measurable) sets in � N . In addition, if r C G with 1 r 1 e == 0, then 1 4l (r) l e == 0; and if r c 4l( G ) with 1 r 1 e == 0, then (r) l e == 0. In particular,
I tlt - 1
* Because we will be dealing with balls in different dimensional Euclidean spaces here , we will use the notation BJR( a , r) to emphasize that the ball is in � N .
V Changes
90
of Variable
PROOF: By the Inverse Function Theorem,t for each x
E
G there is an open neighborhood u c G of X such that 4l r u is invertible and its inverse has first derivatives which are bounded and continuous. Hence, G can be written as the union of a countable number of open sets on each of which 4l admits an inverse having bounded continuous first order derivatives; and so, without loss in generality, we may and will assume that � admits such an inverse on G itself. But, in that case, it is obvious that both 4l and 4l - 1 are continuous. In addition, by Lemma 2.2. 1 , both take sets of Lebesgue measure 0 into sets of Lebesgue measure 0. Hence, by that same lemma, we now know that both 4l and 4l - 1 take Lebesgue measurable sets into Lebesgue measurable sets. D A continuously differentiable map 4l on an open set U C �N into �N is called a diffeomorphism if it is injective (i.e., one-to-one ) and 8clt never vanishes. If 4l is a diffeomorphism on the open set U and if W == 4l( U ), then we say that 4l is diffeomorphic from U onto W . In what follows, for any given set r C �N and 8 > 0, we use r < 6) - { X E �N : I Y - x l < 8 for some y E r} to denote the open 8-hull of r. 5.3.2 Theorem (Jacobi's Transformation Formula) . Let
set in � N and 4l an element of C2 (G; �N ). ishes, then, for every measurable function f measurable on ( G, BR N [G] ) and (5.3.3)
r
J( y ) dy
1 and Gn / G. Clearly, the result for 4l on G follows from the result for 4l r Gn on Gn for every n > 1. Thus, from now on, we assume that G is bounded, the first and second derivatives of 4l are bounded, and 8clt is uniformly positive on G. Let Q == Q ( c; r ) == flf [ci - r, ci + r] C G. Then, by Taylor 's Theorem, there is an L E [0, oo ) (depending only on the bound on the second derivatives of 4l) such that (Cf. Section 2.2 for the notation here.) At the same time, there is an M < oo (depending only on L, the lower bound on 8 4l, and the upper bounds on the first derivatives of 4l) such that
Hence, by Theorem 2.2.2, 1 4l(Q) I < 84l(c) I Q ( O, r + Mr 2 ) I == (1 + Mr) N 84l(c) I Q I . Now define JL(r) == fr 84l(x) dx for r E BRN [G] , and set v == 4l*JL• Given an open set H C 4l (G), use Lemma 2.1.9 to choose, for each m E z + , a countable, non-overlapping, exact cover Cm of 4l - 1 (H) by cubes Q with diam( Q) < ! . Then, by the preceding paragraph,
where CQ denotes the center of the cube Q. After letting m � oo in the preceding, we conclude that I H I < v(H) for open H C 4l(G) ; and so, by Exercise 3. 1.9, it follows that (5.3.5)
l r l < v(r) for all r E BR N [4l(G)] .
Starting from (5.3.5) , working first with simple functions, and then passing to monotone limits, we now conclude that ( 5.3.3 ) holds for all non-negative, measurable functions f on (4l(G) , BR N [4l(G)]) . D As an essentially immediate consequence of Theorem 5.3.2, we have the following.
V Changes of Variable
92 5.3.6 Corollary. Let
morphism. Set
G be an open set in JRN and cit E C 2 ( G; JRN )
M� (r) =
l 8+(x) dx
for r E
BR N [ G ] .
Then clt*Jlif! coincides with the restriction Aif!(G) of AR N to particular,
f E L 1 (cit ( G), BRN [cit ( G)] , Aif!(G) )
�
a diffeo
BR N [clt ( G )] .
In
f � E L 1 ( G, BR N [G] , Jlif!) ' 0
in which case (5.3.4) holds.
As a mnemonic device, it is useful to represent the conclusion of Corollary 5.3.6 as the statement f( y ) dy == f o �(x) 8�(x) dx when y == clt(x). We now want to apply Jacobi ' s formula to show how to differentiate Lebes gue 's measure across a smooth surface. To be precise, assume that N > 2 and say that M c RN is a hypersurface if, for each p E M, there exists an r > 0 and a three times continuously differentiable F : BR N (p, r) � lR with the properties that N (p, r) n M == { y E BR N (p, r) : F(y) == o } B R (5.3.7) and I VF( y ) l -1- 0 for any y E BR N (p, r) , where \7 F(y ) [F , 1 ( y ), . . . , F , N ( Y ) ] is the gradient of F at y and, once again, we have used the notation F ,j to denote the partial derivative in the direction ej . Given p E M, the tangent space Tp ( M ) to M at p is the set of v E JR N for which there exists an E > 0 and a twice continuously differentiable curve 1 on ( - E , E ) into M such that 1(0) == p and i' (O) == v . ( If 1 is a differentiable curve, we use i' (t) to denote its derivative � at t.) Canonical Example: The unit sphere s N - 1 is a hypersurface in ]R N . In fact, at every point p E s N - 1 we can use the function F(y) == IYI 2 - 1 and can identify Tp (s N - 1 ) with the subspace of v E JRN for which v - p is orthogonal to p. 5 .3.8 Lemma. Every hypersurface M can be written as the countable union of compact sets and is therefore Borel measurable. In addition, for each p E M, Tp ( M ) is an (N - I)-dimensional subspace of R N . In fact, if r > 0 and F E C3 ( BR N (p, r) ; R) satisfy (5.3.7) , then, for every y E BR N (p, r)nM, T y ( M ) coincides with the space of the vectors v E JRN which are orthogonal to \7 F(y) . Finally, if, for r c M and p > 0, (5.3.9) r(p) { y E ]RN : :3p E M (y - p ) l_ Tp ( M ) and I Y - PI < p} } ,
5.3 Jacobi 's Transformation and Surface Measure
93
PROOF: To see that M is the countable union of compacts, choose, for each p E M, an r(p) > 0 and a function Fp so that (5.3.7) holds. Next, select a countable subset {Pn}! from M so that 00
M C U BR N (Pn, � ) , with rn 1
==
r(pn)·
Clearly, for each n E z+ ' Kn { y E BR N (Pn, � ) : F( y ) = 0 } is compact ; and M == U� Kn. Next, let p E M be given, choose associated r and F, and let y E BR N (p , r) n M. To see that VF( y ) l_ Ty (M) , let v E Ty (M) be given and choose t > 0 and 1 accordingly. Then, because 1 : ( - t , t ) � M, Conversely, if v E �N satisfying (v , VF( y )) R N == 0 is given, set V(z) = v -
( v , VF(z)) R N
F(z) V' 2 z F IV' ( ) l for z E BR N (p, r) . By the basic existence theory for ordinary differential equations,* we can then find an t > 0 and a twice continuously differentiable curve 1 : ( - t , t ) � BR N (p, r) such that 1(0) == y and i' (t) == V (1(t)) for all t E ( - t , t ) . Clearly i' (O) == v , and it is an easy matter to check that Hence, v E Ty (M) . To prove the final assertion, note that a covering argument, just like the one given at the beginning of this proof, allows us to reduce to the case when there is an r > 0 and an a three times continuously differentiable F : M (2r) � � such that I VF I is uniformly positive and M == {x E M (2r ) : F(x) == 0} . But in that case, we see that : x E r and r(p) = x + t VF(x) lt p , l < F(x) l i V' and so the desired measurability follows as an application of Lemma 2.2 . 1 to the Lipschitz function
{
}
D * See Chapter 1 of E. Coddington and N . Levinson's Theory of Ordinary Differential Equa tions, McGraw Hill ( 1 955) .
V Changes of Variable
94
We are, at last, in a position to say where we are going. Namely, we want to show that there is a unique measure AM on (M, BR N [M] ) with the property that (cf. (5.3.9)) ( 5.3. 10)
1 - AR N (r(p) ) AM (r) == plim �O 2p for bounded r E BR N with r c M.
Notice that, aside from the obvious question about whether the limit exists at all, there is a serious question about the additivity of the resulting map r � AM (r) . Indeed, just because r1 and r2 are disjoint subsets of M' it will not be true, in general, that r1 (p) and r2 (p) will be disjoint. For example, when M == s N - 1 and p > 1 ' r1 (p) and r2 (p) will intersect as soon as both are non-empty. On the other hand, at least in this example, everything will be all right when p < 1; and, in fact, we already know that (5.3. 10 ) defines a measure when M == S N - 1 . To see this, observe that, when p E ( 0, 1) and M == S N - 1 , r(p) =
{
y :
1 - p < IYI < 1 + p and , : ,
E
}
r '
and apply (5.2.3) to see that
where the measure AsN- 1 is the one described in Section 4.2. Hence, after letting p � 0, we see not only that the required limit exists but also gives the measure AsN - 1 . In other words, the program works in the case M == s N - 1 , and, perhaps less important, the notation used here is consistent with the notation used in Section 4.2. In order to handle the problems raised in the preceding paragraph for general hypersurfaces, we are going to have to reduce, at least locally, to the essentially trivial case when M == �N- 1 {0}. In that case, it is clear that we can identify M with �N- 1 and r(p) with r X ( -p, p) . Hence, even before passing to a limit, we see that x
In the lemmata which follow, we will develop the requisite machinery with which to make the reduction. p E
M there is an open neighborhood U of 0 in �N- 1 and a three times continuously differentiable injection (i.e. , one-to-one) \}1 U -----+ M with the properties that p == w(O) and, for each E U, the set { \}1 , 1 ( ) , . . . , \}1 , N 1 ( ) } forms a basis in T w ( u) ( M) . 5.3. 1 1 Lemma. For each :
u
_
u
u
5.3 Jacobi 's
Transformation and Surface Measure
95
PROOF: Choose r and F so that (5.3.7) holds . After renumbering the coor
dinates if necessary, we may and will assume that F , N (P) =I= 0. Now consider the map Y1 - P1 E �N . YN - 1 - P N - 1 F( y ) Clearly, cit is three times continuously differentiable. In addition,
[ 1
J cit (p) == IRVN-
]
0 F , N (P ) '
where v == [F , 1 (p) , . . . , F , N _ 1 (p) J . In particular, 8 clt (p) =I= 0, and so the In verse Function Theorem guarantees the existence of a p E (0, r] such that cit r BR N (p, p) is diffeomorphic and clt - 1 has three continuous derivatives on the open set W cit ( BR N (p, p)) . Thus, if U
{u E � N - 1 : (u, O) E W },
then U is an open neighborhood of 0 in �N - 1 , and is one-to-one and has three continuous derivatives. Finally, because \}1 takes its values in M, it is obvious that
At the same time, as the first (N - 1) columns in the non-degenerate matrix Jclt - 1 (u) , the \}1 ,j (u) ' s must be linearly independent. Hence, they form a basis in Tw (u) ( M) . D Given a non-empty, open set U in �N - 1 and a twice continuously differen tiable injection \}1 U � M with the property that :
{ w , 1 , . . . , w , N - 1 } forms a basis in Tw(u) ( M ) for every u E U, we say that the pair ( \}1 , U) is a coordinate chart for M. 5.3. 12 Lemma. Suppose that (w, U) is a coordinate chart for M, and define (5.3. 13)
V Changes of Variable
96
Then 8 '}1 never vanishes, and there exists a unique twice continuously differ entiable n : U ------+ s N - 1 with the properties that n(u) l_ Tw ( u ) (M) and *
det [ W , 1 . . . W , N - 1 n(u) T ] = 8W(u)
for every u
E
U.
Finally, define
�(u, t) == W(u) + tn(u) T
for (u, t)
E
U X �-
Then
8 w(u, 0) == 8 w(u) , u E U, and there exists an open set U in �N such that � r U is a diffeomorphism, u == { u E �N - 1 : (u, 0) E U} , and w(U) == �(U) n M. In particular, if x and y are distinct elements of \}1 ( U), then { x} (p) n \}1 ( U) is disjoint from { y } (p) n w(U) for all p > 0. PROOF: Given a u E U and an n E s N - 1 which is orthogonal to Tw ( u ) (M) , { \}1 1 ( u), . . . , \}1 , N - 1 ( u) , nT } is a basis for �N and therefore det [ w , 1 (u) . . . W , N - 1 (u) nT ] =f- 0. Hence, for each u E U there is precisely one n(u) E s N - 1 n Tw ( u ) (M) with the property that det [ W , 1 (u) . . . W , N - 1 (u) n(u) T ] > 0. To see that u E U � n(u) E s N - 1 is twice continuously differentiable, set p == W(u) and choose r and F for p so that (5.3.7) holds. Then n(u) = ± 1 ���=� 1 , and so, by continuity, we know that, with the same sign throughout, n( w ) = ± \7 F(w(w)) I V' F(W(w)) l for every w in a neighborhood of u. Turning to the function \}1 , note that 8�(u, 0) 2 = (det [ W , 1 (u) . . . W , N - 1 (u) n(u) T J r W , 1 (u)T w , 1 ( u) . . . w , N - 1 ( u) n ( u) T ] == det [ \}1 , N - 1 ( ) T n(u) = 8W(u) 2 . = det (( ( W , i (u) , : ,j (u)) R N (5.3. 14)
,
U
[
)) �]
* Below, and throughout , we use A T to denote the transpose of a matrix A. In particular , if A is a row vector, then A T is the corresponding column vector, and vice versa.
5. 3 Jacobi 's Transformation and Surface Measure
97
Hence, (5.3. 14) is proved. In particular, this means, by the Inverse Function Theorem, that, for each u E U, there is a neighborhood of (u, 0) in RN+ 1 on which \}1 is a diffeomorphism. In fact, given u E U, choose r and F for p == w(u), and take p > 0 so that
'VF ( W ( w )) Then, because n (w) = ± j F ( w ( )) j for all w E BR N- 1 (u, p) , w V'
F ( �(w, t) ) == ± t ! VF ( w(w)) ! + E(w, t) for (w, t) E BR N- 1 (u, p) ( - � , � ) , where ! E(w, t) l < Ct2 for some C E (0, oo ) . Hence, by readjusting the choice of p > 0, we can guarantee that \}1 r BIRN - 1 ( u, p) X ( - p, p) is both diffeomorphic and satisfies x
In order to -prove we must find an open U in �N such - the final -assertion, that w(U) == w(U) n M and w r U is a diffeomorphism. To this end, for each u E U, we use the preceding to choose p( u) > 0 so that: BJRN - 1 ( u, p( u) ) C U and � r BJRN - 1 ( u, p( u) ) X ( - p( u) ' p( u) ) is both a diffeomorphism and satisfies Next, choose a countable set { un}1 C U so that CX)
U == U BJRN- 1 (un , if- ) where 1 and set
Pn
== p (un) ;
n
Un == U BJRN -1 (urn , e; ) and Rn == P 1 1\ . . . 1\ Pn · 1 To construct the open set U, we proceed inductively as follows. Namely, set E 1 == '¥- and k1 == u 1 X [ - E 1 , E 1 ] · Next given Kn , define En + 1 by
( {I
2En+ 1 = Rn + l 1\ En 1\ inf W ( ) - � ( t) and take
U
W,
I
:
U
W,
E Un + l \ Un , ( t) E kn ,
)
and l u - w l > R';+ 1 } ,
V Changes of Variable
98
Clearly, for each n E z +-' Kn is compact, Kn c Kn + 1 ' Un X ( - En, En) c K-n , and 8 '}1 never vanishes on Kn. Thus, if we can show that En > 0 and that \}1 f Kn is one to one for each n E z + ' then we can take [; to be the interior of u � Kn. To this end, first- observe - that there is nothing to do when n == 1. Furthermore, if En > 0 and \}1 f Kn is one to one, then En + 1 == 0 is possible only if there exists a u E BR N-1 (Un + 1 , Pn + 1 ) and -a (w, t) E BRN- 1 (Um , Pm ) X ( - pm , Pm ) for some 1 < m < n such that w(u) == w(w, t) and l u - w l > R�+ 1 • But, because this would mean that \}1 ( w) == \}1 ( u) and therefore, since \}1 is one to one, would R�+ 1 • Hence, En + 1 > 0. Finally, lead to the contradiction that 0 == ! w - u l > - to see that \}1 f Kn + 1 is one to one, we need only check that (u, s) E (Un + 1 \ Un) X [ - En + 1 , En + 1 ] and (w, t) E Kn ===} �(u, s) =/= �(w, t) . But, if l u - w l < R�+ 1 , then both (u, s) and ( t) are in BR N- 1 (un+ 1 , Pn + 1 ) ( -Pn + 1 , Pn + 1 ) and � is one to one there. On the other hand, if l u - w l > R�+ 1 , then l �(u, s) - � ( , t) l > l w(u) - �( , t) l - l sl > 2En + 1 - En + 1 == En + 1 > 0. D w,
x
w
w
5 . 3 . 1 5 Lemma. If (w, U) is a coordinate chart for M and
ment of BR N with
r c w(U) ,
then (cf.
(5.3. 13))
r a bounded ele
(5.3. 16) in Lemma 5.3.12. Since r is a compact subset of the open set G == �(U), E == dist ( r, GC) > 0. Hence, for p E (0, E), PROOF: Choose U and \}1
as
which, by (5.3.4 ) and Tonelli ' s Theorem, means that
Since w - 1 (r ) == � - 1 (r ) is compact, 8� f w - 1 ( r ) is uniformly bounded and continuous. In particular, by Lebesgue ' s Dominated Convergence Theorem and (5.3. 14) ,
f _!_ 2p }( - , ) 8�(u, t) dt � 8�(u, 0) = 8W(u) p p
5.3 Jacobi 's
Transformation and Surface Measure
99
boundedly and point-wise for u E U. Hence, after a second application of Lebesgue ' s Dominated Convergence Theorem, we arrive at (5.3. 16). D 5.3. 1 7 Theorem. Let M be a hypersurface in � N . Then there exists a unique
measure AM on (M, BRN [M] ) for which (5.3. 10) holds. In fact, AM (K) < oo for every compact subset of M. Finally, if ( \}1 , U) is a coordinate system for M and f is a non-negative, BRN [M]-measurable function, then (cf. (5.4. 13))
(5.3.18)
r
lw ( U)
f(x) AM (dx) = r f W(u)8W(u) )..R N - 1 (du). lu 0
PROOF: For each p E M, use Lemma 5.3. 11 to produce an r(p) > 0 and a coordinate chart ( \}1 Up ) for M such that BRN (p, 3r(p) ) is contained in ( cf. Lemma 5.3. 12 ) w(Up ) · Next, select a countable set {Pn}! C M so that P'
CX) M C U BRN (pn , rn), where rn r(pn), 1
n- 1 Mn == (BRN (pn , rn) n M) \ u Mm for n > 2. 1 Finally, for each n E z + , define the finite measure Jln on (M, BRN [M] ) by
and set (5.3.19) Given a compact K c M, choose n E z + so that K c u� BRN (pm , rm ), and set r == r 1 1\ · · · 1\ rn and E == j . It is then an easy matter to check that, for any pair of distinct elements x and y from K, either l x - Y l > r, in which case it is obvious that {x}(E) - -n { y }(E) == 0 , or l x - Y l < r, in which case both {x}(E) and { y }(E) lie in W m (U m ) for some 1 < m < n and, therefore, the last part of Lemma 5. 3.12 applies and says that { x} (E) n { y } (E) == 0 . In particular, if r E BRN is a subset of K and rm == r n Mm , then, for each p E [0, E), {r1 (p ) , . . . , rn (P) } is a cover of r( p ) by mutually disjoint measurable sets. Hence, for each 0 < p < E: n ARN ( r (p ) ) == ARN ( rm (P) ) . 1 P
L
V Changes of Variable
100
At the same time, by Lemma 5.3. 15,
In particular, we have now proved that the measure defined in (5.3.19) satisfies (5.3. 10) and that AM is finite on compacts. Moreover, since (cf. Lemma 5.3.8) M is a countable union of compacts, it is clear that there can be only one measure satisfying (5.3. 10) . Finally, if ( \}1' U) is a coordinate chart for M and r E BR N with r c \}1 ( U) is bounded, then (5.3.18) with f == l r is an immediate consequence of (5.3.16). Hence, (5.3. 18) follows in general by taking linear combinations and monotone limits. D The measure AM produced in Theorem 5.3.1 7 is called the surface measure on M. Exercises
5.3.20 Exercise: In the final assertion of part ( ii ) in Exercise 5.2.4 and again in ( i ) of Exercise 5.2.6, we tacitly accepted the equality of 1r , the volume 02 of
the unit ball BR 2 (0, 1 ) in �2 , with the half-period of the sin and cos functions. We are now in a position to justify this identification. To this end, define 4l : G (0, 1) (0, 27r) ----+ �2 (the here is the half-period of sin and cos) by 4l(r, 8) == (r cos O, r sin O)T. Note that 4l(G) == B(O, 1) \ {x E B(O, 1) : x 1 == 0} and therefore that 1 4l(G) I == 0 2 . Now use Jacobi ' s Transformation Formula to compute 1 4l(G) I . 1r ,
x
1r
Let M a hypersurface in �N . Show that, for each p E M, the tangent space Tp (M) coincides with the set of v E � N such that 5.3.21 Exercise:
1. lmo dist(p +�2�v, M) e�
0, set s N - l (r) == { X E �N : l x l == r } , and observe that s N - l (r) is a smooth region. Next, define � r : s N - l � s N - l (r) by � r (w) == rw ; and show that AsN-l (r) == r N - l (� r )*AsN- 1 . 5 .3.22 Exercise:
5.3.23 Exercise:
then AM (r) > 0.
Show that if r =I= 0 is an open subset of a hypersurface M,
5. 3 Jacobi 's Transformation and Surface Measure
10 1
In this exercise we introduce a function which is intimately related to Euler ' s Gamma function. ( i ) For (a, {3) E ( 0 , oo ) 2 , define
5.3.24 Exercise:
B (a, /3) = f u - 1 ( 1 - u) f1 - 1 du. 1(0, 1 ) 0
Show that
(a)r(/3) B ( a, {3) = rr(a + {3) where r( · ) is the Gamma function described in ( ii ) of Exercise 5.2.6. ( See part ( iv ) of Exercise 6.3.18 for another derivation.) The function B is called the Beta function. Clearly it provides an extension of the binomial coefficients in the sense that
(
m+n+ 1 m+n B (m + 1 , n + 1 ) m for all non-negative integers m and n.
)
Think of r(a) r({3) an integral in (s, t) over (O, oo) 2 , and consider the map uv 2• E oo) ( u , v ) E (0 , oo) (0 , 1 ) (0 , u( 1 v) as
Hint:
f--+
x
( ii ) For ,\ > � show that
[
]
where {cf. part ( ii ) of Exercise 5.2.6) W N _ 1 is the surface area of s N - 1 . In particular, conclude that
Use polar coordinates and then try the change of variable 4l{r) == 1 � r2 • ( iii ) For ,\ E { 0 oo ), show that 2
Hint :
,
and conclude that, for any N E z + ,
e ) � - 1 d�
WN . WN - 1 (- 1,1) Finally, check that this last result is consistent with part ( iii ) of Exercise 5.2.4.
1
(1 -
=
102
V Changes of Variable
Let U be a non-empty, open subset of JR N - 1 and f E C3 (U ; JR) be given, take 5.3.25 Exercise:
M = { (u, f(u)) : u E U} and W(u) =
[!�)] ,
u E U.
That is, M is the graph of f. (i) Check that M is a hypersurface and that (U, w) is coordinate chart for M which is global in the sense that M == w(U ) . (ii) Show that
(((w w ) IRN )) 1 - 0 so that BR N (p, 2 t) c 4- (U) , set U == � - 1 ( BR N -(p, t)n8G) , and take \}1 and \}1 to be, respectively, the restrictions of � and � to U and U == U X (-t, t). Turning to the second part of the lemma, set n == \}1 ( U) , and define p( y ) = W ( --J, - 1 ( y h , . . . , --J, - 1 ( y ) N - 1 ) and n( y ) = n(p( y ) ) for y E n. Given w E s N - 1 ' set
Wn (Y ) == (w, n( y )) R N and Wn ( Y ) == W0 ( y ) n (y ) , y E fl ; and, given f E C(�N ; �) with compact support K c n, set r ==dist ( K, nC ) . Then, by the translation invariance of Lebesgue ' s measure, for every � E �: �( � )
fa f(x + �w) dx - fa f(x) dx = � 1 (�) + � 2 (�),
where and
�2 (� )
L ( 1 c (x - �wn (x) ) - 1 a (x) ) f(x) dx.
In order to prove that (5.4.3) holds, we will show that
( ) � � 1 0 lim = ---+ 0 � e � 2 (� ) = f(x) wn (x) A (dx); lim aa e ---+ O �
(5.4.7)
1
G
and we will begin with the second, and easier part of ( 5.4. 7) . To this end, we use Theorem 5.3.2 and Fubini ' s Theorem to write � 2 (� ) = where g(� , u)
fu
g( � , u) du,
1� 1 < r,
- 1 ) [1 ( - oo ,O) (t - �wn (W(u)) ) - 1 ( - oo ,o) (t) ] f ( --J, (u, t) ) 8--J, (u, t) dt. ( - t: , t:
By elementary reasoning, one sees that, as � ---t 0,
V Changes of Variable
106
uniformly and boundedly for
.hm � 2 (x) e---. o �
=
u E U; and so (cf. (5.3. 18))
j f(u)wn ( W(u) ) 8W(u) .XR N- 1 (du) 1 =
u
aa
f(x) wn(x) Aaa (dx).
(5.4. 7) is more involved. To handle it, we introduce the D( y ) = q, - 1 ( Y ) N = (y - p( y ), n ( y ) ) JRN ' y E n. Observe that if p E f! n 8G, then ( cf. Exercise 5.3.20) . D(p + tv) - D (p) == 0 if v E Tp (8G) hm t---. o t 1 if v == n(p). The first part of notation
{
Hence,
( 5.4.8 )
V D(p) == n(p)
for p E
f!
n 8G.
Next, set
E( y , � ) == D (y - �wn ( Y ) ) - D( y - �w)
for
( y , �) E K ( - r, r). x
Clearly,
x - �w E G but x - �wn ( Y ) � G x - �w � G but x - �wn ( Y ) E G
===} ===}
0 < D (x - �wn (x) ) < E(x, � ) E(x, � ) < D(x - �wn (x)) < 0.
Thus,
� � � ( � ) � < ll f ll u AR N (r( � ) ) where r( � ) {X E K I D (x - Wn(x)) I < I E(x, � ) I }. =
In order to estimate
:
ARN ( r( � )) ' first observe that, for some cl < oo ,
Hence, since p ( x - �w0(x))
==
p(x),
which, together with ( 5.4.8 ) , leads to the existence of a c2 But, since w - w0(x) that
..l
Rm } , and use induction to define
rm, £ = � E E \
{
for 1 < f
L g
0
==
{ yk}1
{� E E :
}
Then, by the preceding,
F m dp
Moreover, it is easy to see that II I F m - F l I I L 1 (J.L ) 0 as m � oo . Thus, because C is closed, we now see that F dJL E C. At the same time, because g is continuous, g o F m � g o F in JL-measure as m � oo . Hence, by Fatou ' s Lemma, for each
m E
z+ .
JE
----7
We now need to develop a criterion for recognizing when a function is con cave. Such a criterion is contained in the next theorem. 6. 1 . 2 Lemma. Suppose that C C
�N is an open convex set, and let C be its Moreover, if g is continuous on C and g E C 2 (C) ,
closure. Then C is convex. then g is concave on C if and only if its Hessian matrix
is non-positive definite ( i.e., all of the eigenvalues of H9 (x) are non-positive) for each x E C. 0
VI Some Basic Inequalities
116
PROOF: The convexity of C is obvious. In order to prove that g is concave on C if H9 (x) is non-positive definite at every x E C, we will use the following simple result about functions on the interval [0 , 1] . Namely, suppose that u is continuous on [0 , 1] and that u has two continuous derivatives on ( 0, 1). Then u( t) > 0 for every t E [0, 1] if u(O) == u(1) == 0 and u"(t) < 0 for every t E (0, 1). To see this, let t > 0 be given and consider the function uf. - u - tt(t - 1). Clearly it is enough for us to show that uf. > 0 on [0 , 1] for every t > 0. Note that uf. (O) == uf. (1) == 0 and u� ( t) < 0 for every t E ( 0, 1). In particular, if uf. ( t) < 0 for some t E [0, 1], then there is an E (0, 1) at which uf. achieves its absolute minimum. But this is impossible, since then we would have that u� ( ) > 0. { The astute reader will undoubtedly see that this result could have been derived as a consequence of the strong minimum principle in Exercise 5.4.19 for N == 1.) Now assume that H9 ( x ) is non-positive definite for every x E C. Given x , y E C , define u(t) == g({1 - t)x + ty ) - (1 - t)g(x) - tg( y ) for t E [0, 1] . Then u(O) == u(1) == 0 and 0
s
s
0
u" (t) = (y - x , H9 ((1 - t)x + ty ) ( y - x ) ) R N < 0 for every t E {0, 1). Hence, by the preceding paragraph, u > 0 on [0, 1] ; and so g({1 - t)x + ty ) > ( 1 - t)g(x) + tg( y ) for all t E [0, 1] . In other words, g is concave on C, and, by continuity, it is therefore concave on C. To complete the proof, suppose that H9 ( x ) has a positive eigenvalue for some x E 6 . We can then find an w E s N - l and an t > 0 such that (w, H9 (x)w ) R N > 0 and x + tw E 6 for all t E ( - t , t ) . Set u(t) == g( x + tw) for t E ( - t , t ) . Then u"(O) == (w, H9 (x)w) R N > 0. On the other hand, 0
+ u( -t) - 2u{O) u(t) " (O) == 1. u ' t----.. 0 t2 liD
and, if g were concave,
( )
2u{O) == 2u t -2 t
(
)
2g 21 (x + tw) + 21 (x - tw) > g ( x + tw) + g( x - tw) == u(t) + u( - t),
==
from which would we would get the contradictory conclusion that
0. D
6 . 1 . 3 Lemma. Let A =
u"(O)
b2 . In particular, for each 2 0 -o: and ( x , y ) E [O, oo) 2 r-----+ a E {0, 1), the functions (x, y) E [O, oo) r-----+ x y 1 ( X 0 + yo: ) are continuous and concave. 1
a:
6. 1
117
Jensen, Minkowski, and Holder
PROOF: In view of Lemma 6. 1.2, it suffices for us to check the first assertion. To this end, let T == a + c be the trace and D == ac b2 the determinant of A. Also, let A and Jl denote the eigenvalues of A. Then, T == A + Jl and D == AJl. If A is non-positive and therefore A V Jl < 0, then it is obvious that T < 0 and that D > 0. If D > 0, then either both A and Jl are positive or both are negative. Hence if, in addition, T < 0, A and Jl are negative. Finally, if D == 0 and T < 0, then either A == 0 and Jl == T < 0 or Jl == 0 and A == T < 0. D -
6.1.4 Theorem (Minkowski's Inequality) . Let !I and /2 be non-negative,
measurable functions on the measure space [1 , oo ) ,
( E, B, JL ) .
Then, for every p E
PROOF: The case when p == 1 follows from (3.2.13). Also, without loss in generality, we assume that ff and ff are JL-integrable and that !I and !2 are [0, oo ) -valued. Let p E ( 1, oo ) be given. If we assume that JL ( E ) == 1 and we take a == ! , then, by Lemma 6. 1.3 and Jensen ' s inequality,
l (it + h ) p d/1 l [ (fit + (f�t] ! dJ1 ! < [ (l Ji dJ1 ) + (l g dJ1 ) ] [ (l fi dJ1) i + (l g dJ1) i r =
a
a
More generally, if JL ( E ) == 0 there is nothing to do, and if 0 < JL ( E ) < oo we can replace Jl by JL(E) and apply the preceding. Hence, all that remains is the case when JL ( E ) == oo . But if JL ( E ) == oo , take En == {!I V !2 > � } , note that JL ( En ) < nP J ff dJl + nP J ff dJl < oo , apply the preceding to JI , !2 , and Jl all restricted to En , and let n ---t oo . D 6. 1.5 Theorem (Holder's Inequality) . Given p E (1, oo ) define the Hol der conjugate p' of p by the equation 1p p1, == 1. Then, for every pair of
+
non-negative, measurable functions !I and !2 on the measure space
for every p E
( 1, oo ) .
(E, B, JL ) ,
VI Some Basic Inequalities
118
PROOF: First note that if either factor on the right hand side of the above inequality is 0, then !I f2 == 0 (a.e. , JL) , and so the left hand side is also 0. Thus we will assume that both factors on the right are strictly positive, in which case, we may and will assume in addition that both Jf and f� are JL-integrable and that !I and f2 are both [0, oo )-valued. Also, just as in the proof of Minkowski ' s inequality, we can reduce everything to the case when JL(E) == 1. But then we can use apply Jensen ' s Inequality and Lemma 6. 1.3 with a == � to see that I
l ft h dj1 l ( fi) o (f{ ) l - o dj1 < (l Jf dj1) o (l J{ dj1y -o (l Jf d11) (l J{ d11) ? D =
*
Exercises
6 . 1 . 6 Exercise: Here are a few easy applications of the preceding.
(i) Show that log is a continuous and concave on every interval [E, oo ) with E > 0. Use this together with Jensen ' s inequality to show that for any n E z + ' Jl i , . . . , Jln E (0, 1) satisfying L: � = I Jlm == 1, and ai , . . . , an E (0, oo ) , n
n
m= l
m= I
II a�m < L Jlm am .
In particular, when Jlm == � for every 1 < m < n this yields ( a I a ) :n < � L:� = l am , which is the statement that the arithmetic mean dominates the 1
·
geometric mean. (ii) Let n E z + ,
·
·
n
and suppose that !I , . . . , f are non-negative, measurable functions on the measure space ( E, B, Jl ) . Given PI , P E ( 1, oo ) satisfying L:� = I P� == 1, show that n
. . .
2,
,
n
Minkowski ' s and Holder ' s inequalities are in timately related and are both very simple to prove. Indeed, let !I and f2 be bounded, non-negative, measurable functions on the finite measure space (E, B, JL) . Given any a f= 0, observe that 6. 1. 7 Exercise: When p
==
6. 2
The Lebesgue Spaces
119
from which it follows that
2 r h 12 df-l < t r tt df-l + �t r Ji df-l JE JE JE
for every t > 0. If either integral on the right vanishes, show from the preceding that JE f1 f2 dJl < 0. On the other hand, if neither integral vanishes, choose t > 0 so that the preceding yields
(6. 1.8) Hence, in any case, (6.1.8) holds. Finally, argue that one can remove the restriction that f1 and f2 be bounded, and then remove the condition that JL (E ) < oo . In particular, even if they are not non-negative, so long as ft and fi are JL-integrable, conclude that f1 f2 must be JL-integrable and that (6. 1.8) continues to hold. Clearly (6.1.8) is the special case of Holder ' s inequality when p == 2. Because it is a particularly significant case, it is often referred to by a different name and is called Schwarz's inequality. Assuming that both ft and fi are JL integrable, show that the inequality in Schwarz ' s inequality is an equality if and only if there exist ( a , {3) E �2 \ {0} such that af1 + /3!2 == 0 ( a.e. , JL ) . Finally, use Schwarz ' s inequality to obtain Minkowski's inequality for the case when p == 2. Notice the similarity between the development here and that of the classical triangle inequality for the Euclidean metric on � N . 6. 1.9 Exercise: There is a "better" proof of Jensen ' s inequality which is
based on the following geometric fact. Namely, if g is a continuous, concave function on the closed, convex subset C in � N , then, for each p E C, there is a v E �N such that the line L { ( x, g (p) + (v, x - p) R N ) : x E �N } lies above 6 { (x, t) E C x � : t < g(x) } in �N + l . That is, g(x) < g (p ) + (v, x - p)R N for every x E C. Assuming this fact*, give another derivation of Jensen ' s Inequality. ( Hint: Take p == J F dJL. )
6.2 The Lebesgue Spaces.
I I £ 1 (JL)
and the space L 1 ( JL ) . We are now In Section 3.2 we introduced ready to embed L 1 ( JL ) into a one-parameter family of spaces. Given a measure space ( E, B, JL ) and a p E [1, oo ) , define ·
* When C is the closure of its interior and g is smooth, this fact is an easy consequence of Taylor 's Theorem. In general, it can be seen as an application of the Hahn-Banach Theorem for JR N .
120
VI Some Basic Inequalities
for measurable functions ( E, B) define
f on
(E, B) . Also, if f is a measurable function on
II J IIL= ( J.t ) = inf { M E [0 , oo] : I f I < M (a.e. , J-L) }. Obviously, as p varies 11 / II Lv(J.L ) provides different estimates on the size of f as it is "seen" by the measure Jl · Although information about f can be gleaned from a study of 1 1 / II L P(J.L ) as p changes (for example, spikes in f will be emphasized by taking p to be large) , all these quantities share the same flaw as 11 / IIL l (J.L) : they cannot detect properties of f which occur on sets having JL-measure 0. Thus, before we can hope to use any of them to get a metric on measurable functions, we must invoke the same subterfuge which we introduced at the end of Section 3.2 in connection with the space £ 1 (JL) . Namely, for p E [1, oo] , we denote by LP (Jl) == LP (E, B, JL) the collection of equivalence classes [/]'"'-� (cf. Remark 3.2.14) of �-valued, measurable functions f satisfying ll f ll £v (J.L ) < oo, and, once again, we will abuse notation by using f to denote its own equivalence class 1-L
[/]
1-L
I"V .
Note that, by
(3.2.13 ) and Minkowski ' s inequality,
(6.2. 1) for all p E [1, oo ) , /1 , /2 E LP (Jl) , and a , f3 E �- Moreover, it is a simple matter to check that (6.2. 1) continues to hold when p == oo. Thus, each of the spaces LP (Jl) is a vector space. In addition, because of our convention and Markov's inequality (Theorem 3.2.8) , 11 / II Lv (J.L ) == 0 if and only if f == 0 as an element of LP (JL) . Finally, (6.2. 1) allows us to check that l l /2 - /I IILP (J.L ) satisfies the triangle inequality and, together with the preceding, this shows that it determines a metric on LP (JL) . Thus, when {/n }1 U {/} C LP (JL) , we often write fn � f in LP (Jl) when we mean ll fn - f i ! Lv (J.L ) � 0. The following theorem simply summarizes obvious applications of the results in Sections 3.2 and 3.3 to the present context. The reader should check that he sees how each of the assertions here follows from the relevant result there. 6.2. 2 Theorem. Let
and
f, g E LP (Jl) ,
(E, B, JL)
be a measure space. Then, for any p E
[1, oo]
Next suppose that {In } ! C LP (Jl) for some p E [1 , oo] and that I is an � valued measurable function on (E, B) . (i) If p E [1, oo ) and fn � f in LP (JL) , then In � I in JL-measure. If fn � f in L CX) (JL) , then fn � f uniformly off of a set of JL-measure 0. (ii) If p E [ 1 , oo ] and In � f in JL-measure or ( a.e. , JL ) , then 11 / II L P(J.L) < lim n---+ CX) II fn II LP(J.L) . Moreover, if p E [1 , oo ) and, in addition, there is a g E LP (Jl) such that Ifn i < g ( a.e. , Jl) for each n E z+ ' then In � f in LP(JL) .
6.2
The Lebesgue Spaces
121
(iii) If p E [ 1 , oo] and limm � oo II In - lm iiLP(JL ) == 0, then there is an I E LP ( Jl ) such that In � f in LP (JL) . In other words, the space LP ( Jl ) is
SUPn > m
complete with respect to the metric determined by II · IIL P(JL ) · Finally, we have the following variants of Theorem 3.3. 13 and Corollary 3.3. 14. ( iv ) Assume that JL(E) < oo and that p, q E [1 , oo ) . Referring to Theorem 3.3. 13, define S as in that theorem. Then for each I E LP ( Jl ) n L q (Jl) there is a sequence { 'Pn }1 C S such that 'Pn � I both in LP ( Jl ) and in Lq (JL) . In particular, if is generated by a countable collection C, then each of the spaces LP (Jl), p E [1 , oo) , is separable. ( v ) Let ( E, p) be a metric space, and suppose that Jl is a measure on ( for which there exists a non-decreasing sequence of open sets En satisfying JL(En) < oo for each n > 1 . Then, for each pair p, q E [1, oo) and I E LP ( Jl ) n Lq (JL) , there is a sequence {cpn }1 of bounded p-uniformly continuous functions such that 'Pn 0 off of En and 'Pn � I both in LP (Jl) and in Lq (Jl) . The version of Lieb ' s variation on Fatou ' s Lemma for £P-spaces with p =I= 1 is not so easy as the assertions in Theorem 6.2.2. To prove it we will need the following lemma.
B
E, B ) E /E
6.2.3 Lemma. Let
p
E ( 1 , oo) , and suppose that {In } ! C LP (Jl) satisfies
supn > 1 ll ln iiLP(JL ) < oo and that In � 0 either in JL-measure Then, for every g E LP (JL) , l /n l p - 1 191 dJl == 0 == nlim l ln l l g l p - 1 dJL. nlim
J
or (a.e. , JL) .
J
� oo � oo PROOF: Without loss in generality, we assume that all of the In 's as well as g are non-negative. Given b > 0, we have that
J R:- 1 g dJ-L = }r{ Jn < 8g } ��- 1 g dJ-L + }r{ fn > 8g } ��- 1 g df-l < 8P - 1 II g ll iv (JL ) + r
}{ Jn >8 2 }
�� - 1 g dJ-L + r
}{ g 0. Thus, after another application of Lebesgue ' s Dominated Convergence Theorem, we get the first result upon letting b � 0.
VI Some Basic Inequalities
122
To treat the other case, apply the preceding with l!:,- 1 and gP- 1 replacing In and g, respectively, and with p' in place of p. D 6.2.4 Theorem (Lieb) . Let (E, B, JL) be a measure space, p
{ In } ! U { I } C LP (JL) . If SUPn> 1 ll ln ii L P ( JL ) < or (a.e., JL) , then
oo
E [1 , oo ) ,
and and In -----+ f in JL-measure
(6 . 2 . 5) and therefore ll ln - I II L P ( JL ) -----+ 0 if ll ln i i L P ( JL ) � II I I I LP ( JL ) · PROOF : The case when p == 1 is covered by Theorems 3.3.5 and 3.3. 12, and so we will assume that p E ( 1 , oo ) . Given such a p, we first check that there is a KP < oo such that (6.2.6) l l b i P - l a i P - l b - a l P I < Kp ( l b - a i P- 1 I a l + l a l p - 1 l b - a ! ) , a , b E �. Since (6.2.6) clearly holds for all a , b E � if it does for all a E � \ {0} and b E �' we can divide both sides of (6.2.6) by l a i P and thereby show that (6.2.6) is equivalent to
l l ci P - l - I e - l i P I < Kp ( l c - l l p - 1 + l c - 1 1 ) ,
cE
�.
Finally, the existence of a Kp < oo for which this inequality holds can be easily verified with elementary consideration of what happens when c is near 1 and when l ei is near infinity. Applying (6.2.6 ) with a == ln ( x ) and b == l ( x ) , we see that
pointwise. Thus, by Lemma 6.2.3 with In and g there replaced by In - f and I , respectively, our result follows. D We now turn to the application of Holder's inequality to the £P-spaces. In order to do so, we first complete the definition of the Holder conjugate p' which, thus far, has only been defined ( cf. Theorem 6. 1 .5) for p E ( 1 , oo ) . Thus, we define p' == oo or 1 according to whether p == 1 or oo . Notice that this is completely consistent with the equation p! + p1, == 1 used before. 6.2. 7 Theorem. Let (E, B, JL) be a measure space. (i) If I and g are measurable functions on (E, B) , then for every p E [1 , oo ]
( 6.2.8 ) In particular, if I
E £P (Jl)
and g
E LP (Jl) , '
then f g E
L 1 (JL) .
6. 2
(ii) If p E
123
The Lebesgue Spaces
[1, oo ) and f E £P (J-L ) , then
(6.2.9) In fact, if function
II!II L P(J.L )
>
0,
then the supremum in
(6.2.9)
is achieved by the
p - l sgn o f f l l . g= ll f ll i,tl' )
(iii) More generally, for any f which is measurable on ( E, B) ,
(6.2. 10) if p == 1 or if p E a-finite.
(1, oo )
and either J-L( I J I >
8)
0 or J-L
is
PROOF: Part (i) is an immediate consequence of Holder ' s inequality when p E ( 1, oo ) . At the same time, when p E { 1, oo } , the conclusion is clear without any further comment. Given (i) , (ii) is easy. When p == 1, (iii) is obvious; and, in view of (ii) , the proof of (iii) for p E (1, oo ) reduces to showing that, under either one of the stated conditions, IIJI ILP(J.L) == oo implies that the right hand side of (6.2.10) is infinite. To this end, first suppose that J-L( I f l > 8) < oo for every 8 > 0. Then, for each n > 1, the function Moreover, if
II 'l/Jn II LP' (J.L )
II J II Lv (J.L )
-----+
oo .
then, by the Monotone Convergence Theorem, ' Thus, Since II f'l/J n II L (J.L) == II 'l/Jn II pLP ' (J.L ) , we see that ==
oo , •
1
Finally, suppose that J-L is a-finite and that J-L( I J I > 8) == oo for some 8 > 0. Choose { En } ! C B so that En / E and J-L ( En ) < oo for every n > 1. Then it is easy to see that limn-+ II f 9n I I L (J.L) == oo when 00
Since
1
IIYn ii LP' (J.L) < 1, this completes the proof.
D
For reasons which will become clearer in the next section, it is sometimes useful to consider the following slight variation on the basic £P-spaces. Namely,
VI Some Basic Inequalities
124
(E I , 81 , J.1 1 ) PI , P2 E [ 1 , oo ) . let
and ( E2 , 82 , J.1 2 ) be a pair of a-finite measure spaces and let Given a measurable function f on (E1 x E2 , 8 1 x 82 ) , define
[l2 (l1 ! J(x 1 , x2 ) 1P1 JL 1 (dx 1 ) ) * tt2 (dx2 )]
l l f ii £(P t ·P2 l ( J.t 1 , J.1 2 )
pI2
,
and let £ (P I , p2 ) ( J.-t1 , J.-t 2 ) denote the mixed Lebesgue space of �-valued, 81 x 82 -measurable f ' s for which ll f ll £ c PI ·P2)(J.L I ,J.L 2 ) < oo . Obviously, when p 1 == p == P2 , II J II L c PI . P2 ) (J.L I ,J.L 2 ) == ll f ii £P(J.L I X J.L 2 ) and £ (P I , P2 ) (J.-t i , J.12 ) == LP(J.-t l x J.12 ) ·
6.2. 11
Lemma. For all f and g which are measurable on
and all a,
f3 E � '
(E1
x
E2 , 81
x
82 )
l l af + f3g ii £(PI ·P2) (J.L I ,J.L 2 ) < l a l l lf ll £cPI . P2) (J.L I ,J.L 2 ) + l /31 I IYI I £(PI . P2) (J.L I ,J.L 2 )
(6.2. 12)
Moreover, if {fn}! U { ! } C £ ( P I ,P2 ) (J.-t l , J.12 ), fn � f (a.e., J.-l l X J.12 ) , and I fn i < g (a.e. , J.-l l x J.12 ) for each n > 1 and some g E £ (P I ,P2 ) (J.-t i , J.12 ), then II fn - f II £(PI , p 2 ) (J.L I ,J.L 2 ) � 0. Finally, if J.-l l and J.-t 2 are finite and g denotes the class of all 'ljJ 's on E 1 x E2 having the form E� =l 1ri , m ( · I)'Pm ( · 2 ) for some n > 1 , { 'P m } f C L00 ( J.-t 2 ), and mutually disjoint r1 , 1 , . . . , rl, n E 8 1 , then, for every measurable f E £ (P I ,p2 ) ( J.-t1 , J.-t 2 ) and E > 0, there is a 'ljJ E g such that
II ! - '¢ 11 £(PI . P 2 ) (J.L I ,J.L 2 ) < E.
PROOF: Note that (6.2. 13)
Hence the assertions in (6.2.12) are consequences of repeated application of Minkowski's and Holder's inequalities, respectively. Moreover, to prove the sec ond statement, observe ( cf. Exercise 4. 1 . 1 1 ) that for J.-t 2 -almost every x 2 E E2 , fn ( · , x 2 ) � f ( · , x 2 ) ( a. e. , J.-t 1 ) , I fn ( · , x 2 ) I < g ( , x 2 ) ( a. e. , J.-t 1 ) , and g ( , x 2 ) E £P I (J.-t 1 ). Thus, by part (ii) of Theorem 6.2.2, ·
for J.-t 2 -almost every x 2
E
E2 .
for J.-t 2 -almost every x 2
E
E2 and, by (6.2. 13) with g replacing J ,
In addition,
·
6. 2
125
The Lebesgue Spaces
Hence the required result follows after a second application of ( ii ) in Theorem 6.2.2. We turn now to the final part of the lemma, in which the measures Jll and Jl 2 are assumed to be finite. In fact, without loss in generality, we will assume that they are probability measures. In addition, by the preceding, it is clear that, for each f E £ (P I ,p2 ) (JL I , Jl 2 ), Thus, we need only consider f ' s which are bounded. Finally, because Jl l x is also a probability measure, Jensen ' s inequality and (6.2. 13) imply that
JL 2
Hence, it suffices to show that, for every bounded measurable f on ( E1 x E2 , B1 x B2 ) and E > 0, there is a 'ljJ E g for which II! - 'l/J IIL q(J.L 1 xJ.L 2 ) < E. But, by part ( iv ) of Theorem 6. 2.2, the class of simple functions having the form n
L am lrl , m r2 ,m m=l with ri,m E Bi is dense in Lq(JL 1 JL 2 ). Thus, we will be done once we check X
'ljJ ==
x
that such a 'ljJ is an element of g. To this end, we use the same technique as we did in the final part of the proof of Lemma 3. 2.3. That is, set I == ( {0, 1 } ) and, for 'IJ E I, define r1,17 == n � =l ri � ) where r < o ) rC and r< 1 ) r. Then n
n
'¢ (x1 , x 2 ) == L am L 'TJmlr1 , 11 ( ) lr2 , m (x 2 ) == L lr1 ,11 (xi ) cp11 ( 2 ) , m=l 17 EI 17 E I x
xi
where
n
'P11 == Since the
L 'T/m amlr2 , m m= l
·
r1 , 17 ' s are mutually disjoint, this completes the proof. D
For our purposes, the most important fact that comes out of these consid erations is the following continuous version of Minkowski's inequality.
(Ei , Bi , Jli), i E { 1 , 2} , be a-finite measure spaces. Then, for any 1 < P I < P2 < oo and any measurable function f on ( E1 E2 , B l X 82 ), I I J II £(PI ·P2) (J.L I ,J.L2 ) < II J II £(P2.Pl ) (J.L 2 ,J.L l ) •
6.2. 14 Theorem. Let
x
PROOF: Since it is easy to reduce the general case to the one in which both Jll and Jl 2 are finite, we may take them to be finite. In fact, without loss in generality, we will assume, from the outset, that they are probability measures.
VI Some Basic Inequalities
126
'¢ rl ,m
Let g be the class described in the last part of Lemma 6.2. 11. Given == 's which is an element of Q , note that, since the L:� == 2::: � for any r E [0, oo ) and are mutually disjoint, 2::: � E � - Hence, by Minkowski ' s inequality for p == � ,
l r l , Tn ( . 1) 'Pm ( . 2 ) I am l r1 m l r l am l r l r1 , m , a1 , ... , an PI 1 P2 I 'I/J I L
1 be given and define
r
E [ 1 , oo ] by
( 6.2. 19 ) LP In particular, K maps LP (�-t 2 linearly into Lr (�-t 1 ). In fact, f Lr ( J-LI ) is the unique continuous mapping from LP (�-t2 finto £LrP (J-t(�-t21)) whoseKJrestriction to LP ( �-t 2 n Lq ( �-t 2 is given by the map K in ( 6. 2. 16 ) . PROOF: == == q', £ z+ . == (J-t ) fl f f n P 2 n, n ] [ I f(6.2.n 17L)P (J-t2 ) n L 00 (J-tI K2 )I ( I fn i , 6. 2. 20) ' fn LP (J-t2 ) n Lq (J-t2 ). < q' for
E
E
(�-t 2 ) . �
)
E
I
)
)
)
If r oo , and therefore p there is nothing to do. Thus, we will assume that r and therefore p are finite. be given, and set Let E for n E Because 0 E cf. Exercise p and E Hence, by applied to and I
j
, ) ( x1 x2 i l f( x2 ) l M 2 (dx2 ) K { x2 : I J (x2 )l < n}
r
I
In particular, by the Monotone Convergence Theorem, this proves both parts of ( 6.2. 19 ) . Furthermore, if E P ) and E � ' then
J, g £ (�-t2 a, {3 K (af+f3g) == a Kf+ {3 Kg AK(f) n AK(g). 0, )C )C AK( AK( g ! Lr ( �-t 1 K Kf == Kf on
Thus, since both a mapping into
and have �-t1-measure we now see that, as is linear. Finally, it is obvious that for ),
6. 2
129
The Lebesgue Spaces
J-L J-L ) ) Lq f E (J-L2 ) n Lq (J-L2 )· ( ( 2 2 ( L L (J-L ) (J-L ) r 2 {fn }1 ) I l K/ - K' / I £r (J.I t ) < }�--•� I l K/ - KJn I £r (J.I t ) I JC (f - fn) I Lr( 1 ) < M;!: M; - ; n---+ I J - fn i LP(J.I t ) 0. D I
I
£P n Hence, if /(' is any extension of /( r £P as a continuous, linear mapping from P to I , then with the same choice of as above
== limCX) n ---+
limCX)
J.L
=
Exercises
J-L ) f E Lq1 (J-L) nLq2 (J-L) ,
6.2.20 Exercise: Let ( E, B,
given. If
be a measure space and 1 show that for any t E (0, 1)
< QI < Q2 < oo be
1 t 1-t where - == - + Pt QI Q2
(6.2.21)
--
(
Note that 5. 2. 21) says that p �
- I f I LP (J..L) log
is a concave function of
�.
6.2.22 Exercise: The following exercises give some insight into the £P-spaces
in various situations. is a (i) If ( E, B, is a probability space, show that p E [1 , oo ] � non-decreasing function for any measurable on E, B ) . (ii) Let E == z + and define on B == P ( E ) by == 1 for all E z + . In this case show that p E (1 , oo ] � is non-increasing for every on E. (iii) Let ( E, B, be a measure space and : E � � is a B-measurable function. Show that limp/ CX) Further, assuming either that �-t E oo or that oo, show that == limp / CX) ( iv ) Let ( Ei , Bi ) , i E { 1 , 2}, be measurable spaces, and suppose that is a a-finite measure on E , B . Using part (iii) , show that, for every measurable is function on ( EI x E2 , BI x B , the function X I � measurable on ( EI , BI ) · In particular, we could have defined for all P I , P2 E [1 , oo] .
J-L)
J. .L f v ) i L ( l ( f J-L I J I Lv (J..L) J-L ( {n}) nf J-L) I J I Loo (J..L) < l ffi Lv (J..L) · ( )< I J I L l (J..L) < l f/-Li L2 P(J..L) · I J I Loo (J..L) ) ( 2 2 f 2) I J (xi,I J I ·£2c)v1l .LPoo2 ) ((J.J..L.L21 ),J..L2 ) J-L ) p E (1, oo ) f C E (0, oo)
6.2. 23 Exercise: Let ( E, B,
L (J-L)
of P for some for which there exists a
(6.2.24)
( i ) Set
v(r)
be a measure space, g a non-negative element , and is non-negative, B-measurable function such that
MU > t) < ct Jr{f> t } g dJ.L,
== fr g d�-t for
rE
t E (O, oo) .
B, note that (6.2.24) is equivalent to
J-L( f > t) < Tc v (f > t) ,
tE
(0, oo
)
,
VI Some Basic Inequalities
130
and use (5. 1 .4) ( cf. Exercise 5. 1 .8) to justify
I I I iP(J.I)
=p
r
J( o, oo)
tp - 1 M U > t) dt
Finally, note that conclude that
l f l iv �l (J.t)
l f l iv� l ( v)"
tP - 2 v(f > t) dt = C-p1 P J( o, oo) = J fP - 1 g dJ.-l, and apply HOlder ' s inequality to
< Cp
{
(6.2.25) (ii) Under the condition that
I J I Lv (JL)
E ) < oo for every E > 0. Hint: Given E > 0, consider M E == M r > E } ] , note that (6.2.24) with M implies itself with ME , and use (ii) to conclude that (6.2.26) holds with ME in place of M· Finally, let E � 0.
fn f
f
f
B[{f
6.2.27 Exercise: Recall the Hardy-Littlewood maximal function
Mf
defined in (3.4.2) , and observe that the definition extends, without change, to any measurable on � which is integrable on compacts. Show that
f
l i MJI I Lv (R) I J I Lv (R) , P E ) f E £P � (�) , f fE ECc£P(�) C Cc(�) , ( ) {fn}1 fn � f in LP � l i MJI I LP(R) < n oo l i Mfn i Lv (R) ·
(6.2.28)
Hint : When
handle general that
1/J+ ( t ) - 1/J+ ( s ) . Since � 1 1/J1 - � 1 1/J - == � 11/J 2 - �I'l/J+ for any interval I C J , we now see that 'lj;2 - 1/J + and 'lj; 1 - 1/J - are non-decreasing.
0 and 'lj; ( t ) == t sin ( ; ) , t E ( 0, 1] . Clearly 'ljJ is continuous on [0, 1] . On the other hand, if tn == 2n� 1 for n E N and Cn consists of the intervals [0, tn J , [tn , tn - 1 J , . . . , and [t 1 , to J , it is easy to check that S ('lj;; Cn) diverges as n ----t oo at least as fast as the harmonic series does. Thus, 'ljJ has unbounded variation on [0, 1] . 1 . 2.30: Define 1/J ( O )
==
Solution to Selected Problems
169
To handle the second part of the exercise, consider the function if t E (0, 1 ] \ { � } if t == � . Clearly var ('lj; ; [0, 1 ] ) == 2. On the other hand, for any cp E C ( [0, 1 ]) , one has that R ( cp l'l/J; C, �) == 0 as long as � is not an endpoint of any I E C. Thus, (R) f[o, I) cp (x) d'lj;(x) == 0 for every cp E C ( [0, 1 ]) .
2 . 1 . 20: Part (i) is obvious, since Ck is the union of 2 k mutually disjoint intervals of length 3- k . Furthermore, the first assertion in part (ii) is clear,
and the second one follows by induction on f. To prove the characterization of c�_, note that a is the left hand endpoint of one of the intervals I making up Ct if and only if ak ( a ) E {0, 2} for 0 < k < f, ak ( a ) == 0 for k > f, in which case the associated right hand endpoint is a + 3- l . Thus, X E c�_ if and only if ak ( x ) == ak ( a ) , 0 < k < f, for such an a and there is a k > f such that ak ( x ) f= 0. Once one has these facts, the remainder of (ii) is easy. Moreover, as a consequence of (ii) , we know that the subset n � 1 Ct of C is isomorphic to A which, in turn, differs from {0, 2}N by a countable set. Hence, the uncountability of C follows from that of { 0, 2 }N. 0
0
2. 2.4: Thinking of the vk ' s as column vectors, let [ VI . . . v N J denote the N x N-matrix whose ith column is vi . It is then clear that P ( v 1 . . . VN ) is the image of Q0 under [v 1 . . . v N ] ; and, therefore, Lebesgue would as sign P (VI , . . . , v N ) volume equal to the absolute value of the determinant of [vi . . . vN ] ·
In order to connect this with the classical theory, it is helpful to observe first that
where v[ denotes the row vector representation of vi and ( vi , vi ) R N is the inner (i.e. , "dot" ) product of vi and Vj · Next, we work by induction on N. Clearly there is no problem when N == 1. Next, assume that the two coincide for some N E z + and let VI , . . . ' V N + I E �N + I be given. Since there is no prob lem in the linearly dependent case, we will assume that vI , . . . , v N + I are lin early independent. Choose an orthogonal matrix 0 so that Tovi , . . . , TovN E H (ei , . . . , eN ) . Then, the Lebesgue volume of P ( VI , . . . , v N ) thought of as a
Solution to Selected Problems
170
subset of H ( VI , . . . , v N ) is the same as that of P (Tovi , . . . , Tov N ) thought of as a subset of H (ei , . . . , e N ) and is therefore equal to the square root of
Hence, all that remains to check is that
where u == V N + I w and w is the perpendicular projection of V N + I onto H ( VI , . . . , v N ) . But, since the determinant is a linear function of any one of its columns and because w E H ( VI , . . . , v N ) , we see that -
det
[ (( ==
( vi , vi ) R N
det
)) 1 0 there exist a closed F c r and an open G � r such that JL( G \ F) < E . To prove the opposite inclusion, we first check that It is clear that A C A is a a-algebra. Obviously, 0 E A, and r E A ===} rC E A. Moreover, if {r n } � c A, r == u � rn , and E > 0 is given, choose { Fn }� c � and {Gn } � c 0, n
Next, set
n
BE
{
Solution to Selected Problems
171
we can take find an n E z + such that JL ( r \ Fn ) = JL ( f ) - JL ( Fn ) < t ; and therefore we can take G = r and F == Fn . Finally, suppose that rl c r c r2 , where rl , r2 E BE and JL ( r2 \ rl ) = 0. To see that r E A, let t > 0 be given, and use the preceding to find F E � and G E L JL ( m =O
{
x
E
Qo :
Xi =
Qo and Xi = } ) = p Q
: }) > (
for every n E z + ; and therefore p = 0. Next, to see that JL ( [O, m,\] N ) = m N JL ( [O, ,X] N ) for note that
m
E
n
+ l)p
z+
and
,\ E (0, oo ) ,
and so, by the preceding and translation invariance,
From the preceding, we now know that JL ( [O, q] N ) = q N for any q E Q n ( O , oo ) ( Q is used to denote the rational numbers) . Thus, if r is any element of ( 0, oo ) , then by choosing {qn }1 C Q so that Qn � r, we see that Combining this with translation invariance, we conclude that JL ( Q) AJR N ( Q) for every cube Q in �N ; and, since every open set G in �N is the countable union of non-overlapping cubes, it follows immediately, from what we already know, that JL(G) AJRN (G) for all open G ' s in � N . Thus, since, for every R E (0, oo ) , the set of all open subsets of ( - R, R) N is 1r-system which generates BRN [( - R, R) N ] , we now know that JL coincides with AnN on BR N [( - R, R) N ] for every R E ( O , oo ) and therefore on the whole of BRN · ==
==
172 An C rn
Solution to Selected Problems
3. 1 . 12: (i) For m
for every
zz+ +, . Am nn> m rn ·
E
set Hence
E
n
r · U n n n> m z+ .
Then
==
m
(ii) Set B == Then B � lim n E Hence, if J;(B 1 ) < oo , then
Am / n---+ rn lim
n---+ rn rn c Bn and
CX)
CX)
and
for every
E � JL (rn) ( J-L (U� rn ) J-L (n---+ rn) n---+ J-L (Bn) m---+ n> m J-L (rn) 0. 2. Srnn r{1,n+ l ,. r 2 , n -J-L (rl rn+l ) ) ( F r - L - l ) card (F) JL (r F n ( r n r n +d ) ca d l F JL r ) ( ) L 0-#FCSn - 1 0-#FCSn - 1
(iii) If
0, g dJl - nlim fn dJl == lim hn " dJl > lim hn " dJl -----+> CX) n -----+- CX) n -----+- CX) == [g - n,�im-----+- CX) fn" ] dJl > g dJl - nlim ---+ CX) fn dJL. Given the preceding, the proof of part (ii) in the case of almost everywhere convergence is exactly the same as the derivation of Lebesgue ' s Dominated Convergence Theorem from Fatou ' s Lemma. Namely, one sets hn == l fn - J l and observes that hn < gn + g. The case of convergence in measure is then handled in precisely the same way as it was in the proof of Theorem 3.3. 1 1 . 3.3.20: (i) If /C is uniformly JL-absolutely continuous and M sup / E K ll f iiL l (J.L ) < oo , choose, for a given E > 0, 8 > 0 so that sup / E K fr I l l dJl < E whenever r E B satisfies JL(r) < 8 and choose R E (0, oo ) so that -% < 8. Then, by Markov ' s Inequality, sup / E K JL( I f l > R) < 8 and so sup / E K � / I > R I ll dJl < E. Next, suppose that /C is uniformly JL-integrable and define A(R) supK { J I dtt for R E (0, oo ) . I / E 1{ 1/ I > R } Clearly supK f I! I dtt < Rtt (r) + A(R) / E 1r for any r E B and R E (0, oo ) . Hence, if, for given E > 0, we choose R E (0, oo ) so that A(R) < � ' then fr I! I dJl < E for all f E /( and r E B with JL(r) < 2� ; and so /C is uniformly JL-absolutely continuous. In addition, when JL(E) < oo , by choosing R E (0, oo ) so that A(R) < 1, we see that II ! II L l (J.L ) < RJL(E) + 1 < oo for all f E /C. (ii) Note that, for any R E (0, oo ) , { J I dtt < R - o 1!1 1 + 6 dJL < R - 8 1!1 1 + 6 dJL. I 1{ 1/ I > R } 1/ I > R (iii) Suppose that fn � f in L 1 (JL). Clearly f E L 1 (JL) and supn EZ+ ll !n i i L l (J.L ) < In addition, for given E > 0, choose m E z + so that ll !n - ! I I L l (J.L ) < � for > m ; and, using Exercise 3.3. 15, choose 8 > 0 so that n < m 1fr I fn i dtt < � 1 <max
J
J
J
II
J
l
00 . n
J
II
J
J
J
1 75
Solution to Selected Problems
for all r E B with J.t ( r ) < b , and note that fr I l l d�-t < � whenever J.t ( r ) < Hence, for this choice of 8 > 0,
8.
for all r E B with J.t ( r ) < b. Conversely, if �-t(E) < oo , fn f in J.t-measure, and {fn }� is uniformly I-t-integrable, note that f is J.t-integrable, and ( cf. (i) above and Exercise 3.3. 15) choose, for a given t > 0, a b > 0 so that ------+
for all n E z + . Hence , if rn = J.t ( rn) < b for all n > m , then
{ l fn - f l > �-'tE) } and
m
is chosen so that
(iv) Tightness enables one to reduce each of these assertions to the finite
measure situation, in which case they have already been established. 3.3.22: Clearly the general case follows immediately from the case in which
�-t(E) V v(E) < oo; and therefore we will assume that J.t and v are both finite. Thus, by Exercise 3. 1 .8, all that we have to do is check that �-t(G) == v(G) for every open subset of E. Let G be an open set in E, and define
cp ( x ) =
(
dist ( x, GC ) 1 + dist ( x, GC )
)
:n 1
X
'
E
E and E z + ' n
where dist ( x, r ) inf { p ( x, r ) : y E r } is the p-distance from X to r. One then has that 'Pn is uniformly p-continuous for each n E z + and that 0 < 'Pn(x) / 1 a ( x ) as n oo for each x E E. Thus, by the Monotone Convergence Theorem, J.t ( G) == nlim 'Pn dj.t == nlim ---+ 'Pn dv == v( G). ------+
---+ CX)
J
CX)
J
4. 1 . 13: Without loss in generality, we will assume that a == 0 and that b == 1 . To prove the first assertion, define, for n > 2 ,
cpn ( t, X ) =
{�
( � X) '
if if
( t, x ) E ( t, x ) E
( kn 1 , � ) X E ( 1 - � , 1 ) X E.
Solution to Selected Problems
1 76
Clearly each VJn is measurable on ( (0, 1) x E, B( o , l ) x B) , and f is the point-wise limit of { VJn } �. Thus, f is itself measurable. Thrning to the second assertion, note that, for each t E ( 0, 1 ) , f' ( t , ) is the point-wise limit of functions which are measurable on (E, B). Thus, since f' ( , x) is continuous for each x E E, the measurability of f' follows from the first part. To prove the last assertion, let t E (0, 1) be given, and suppose that { tn } � C (0, 1 ) \ { t } is a sequence which tends to t. Set In == [t, tn] or [tn, t] according to whether tn > t or tn < t, and define ·
·
fn ( tn, x) - f(t, x) E z + and E E. ' tn - t f'(t, x) and l cpn( x) l I In l - l ( R) r f ' (s, x) ds < g (x)
VJn (x) Then
VJn(x)
�
=
n
=
x
}I n
x E E. Hence, by Lebesgue ' s Dominated Convergence Theorem, JE f ( tn, X) J.L ( dx) - JE f ( t, x ) J.L ( dx) r cpn(x) ( dx) r f' (t, x) J.L (dx); J.L tn - t JE JE from which we see that t E (0, 1) JE f(t, x) J,t( dx) is not only differen tiable at t but also that its derivative is equal to the J,t-integral of f ' ( t, ) . for every
=
�
�
·
Finally, given the preceding, the continuity of the derivative is simply another application of Lebesgue ' s Dominated Convergence Theorem.
5.2.4: (i) First, let r be a non-empty open subset of s N - l ' and set G == { X E B ( O, 1) \ {0} : �(x) E r} . Then, because � maps B ( O, 1) \ {0} continuously
s N G is a non-empty open subset of �N ; and therefore AsN ( r) NARN ( G) > 0. Next, let 0 be an orthogonal matrix and denote by R the corresponding rotation To on � N . Then � o R Ro � on �N \ {0} , l n ( O , l) o R == l n ( o,l ) , and R* ARN ARN ; and therefore the asserted invariance of As N follows directly from its definition. To prove the asserted orthogonality relations, define for each e E � N \ {0} the rotation Re : �N �N by 2( , x)RN Re x - X - e 2 e ' X E m N . lel onto
-I ,
==
-1
==
-1
�
�
Then, by rotation invariance,
==
Solution to Selected Problems
177
Similarly, for '11 ..l e ,
rsN- ( e , w)RN ( 7J , W )JRN AsN- l ( dw) = r (e , R� w) R N (7J , R� w ) RN AsN-l ( dw) 1 1 1sN- 1 = - rsN- 1 ( e , w)RN ( 7J , W )JRN AsN-l (dw) ; 1 and therefore, for any '11 E � N ,
Finally, if { e 1 , . . . , e N } is the standard orthonormal basis for � N , then, again by rotation invariance,
and so
to denote the rotation determined by the matrix Oo described in the hint, we see, by rotation invariance and Tonelli's Theorem, that (ii) Using
Ro
(
)
r f dv = 21 {g r f Ro(w) dO v( dw) 1S 1 1 1 1[0,2 1r] 7r
0
for any non-negative, Borel measurable f on 8 1 . Next, for fixed w E 81 , choose 'TJw E [0, 2 7r ) so that and observe that
{ f Ro(w) dO = { f (cos ( 'T7w + 0) , sin ( 77w + 0)) dO 1[0, 27r] 1[0 ,27r] f (cos 0, sin 0) dO + r f ( cos ( 21r + 0) , sin ( 21r + 0)) dO = 1[0, 7Jw ] [71w , 2 7r] = f f (cos O, sin O ) dO = f 1 f dJ.L . 1[0, 2 1r] 1S o
1
Solution to Selected Problems
178
Combined with the preceding, this now show that
f
be a non-negative, continuous function on �N with compact support. Then, by Tonelli ' s Theorem and Theorem 5.2.2, (iii) Let
N 2 p ) f (S (p,w) ) dp Ag N -1 (dw) dr J 1 - p2 ) � F( p) dp ( J [- 1 , 1) x SN - 1
( O,CX) )
(1 -
2 -1
-1
[- 1 , 1 )
where
F(p)
J
dy Hence,
2 p Jx S N - 1 ) � f(S(p, w) ) dp A8N - 1 (8w) dr (1 -
[- 1 , 1)
-1
(R) [-8( y ) , 8 ( y)]
where, for each
(0 , 1)
y E �N \ { 0 } ,
is chosen so that
1/;y (p) == (
PIYI 1 - p2 ) 2
,
p
E
[ -8( y ), 8( y )] , and 8( y ) E
p
== 0 for 8( y ) < < 1 .
Solution to Selected Problems
(
Since, by part i) of Exercise 5 . 1 . 6,
(R)
J f(y , 'l/Jy (p)) d'l/Jy (P)
=
[- 8 ( y ) , 8(y)]
=
y { J(o,oo)
for each
E � N \ {0} , we now see that
TN
(R)
179
J
['l/ly { -8( y )) , 'l/ly (8(y))]
f(y ,a)da
, ( J y JJR.l e7) de7 r
x s N -1 r 1 j ( z)dz r r j ( rw)>. sN (dw) ) dr ; ( J(o,oo) JsN f }JRN +
[- 1 , 1 )
TN
=
=
from which the desired result follows easily by taking ·
1J( I z l ) g ( 1 � 1 )
dr
to be of the form
5 .2.6: (iv) Define 1/J as in the hint and observe that 1
( )2
1/J s and therefore that
dd
==
1
-'lj;(s) 2 s
s+
( s 22 a a ) 2 + 4 ,B
1
sE�
------
==
1
2a
Thus, by Exercise 5. 1 .6, for any R E (0, oo ) :
e2;13 j
[ '1/J( - R ) � , 'ljJ ( R ) � ] 1
and so
1
1
==
(R)
=
_!_ (R)
C�
[ - R,R]
exp
2 [ -oh - � ] dt 2 2 /3 [- a '¢( s) - 1 ] d1j;(s) exp
2 e1 8 2a
�
'lj;( s) 2
�
2(s a ) 2 ds 1 2 2 ea /3 1 /3 t ] dt ! [-oh t a J [ - R, R]
exp
[ 'l/J ( - R ) � , '1/J ( R ) � ) 1
1+
s
+ 4 ,B
1
=
1
==
e - 8 2 ds ; e - 8 2 ds ,
[- R,R]
[ - R ' R]
and, after differentiating with respect to ,B and applying Exercise 4. 1. 13, one also has that
2 2 ,3 ] o: f3 e2 J t 2 [- a t - -t dt ,8 1 e - 8 2 ds . 1 1
['l/J( - R ) � , '1/J ( R ) � )
exp
==
3
Clearly the desired results follow when one lets R / oo .
[- R, R]
Solution to Selected Problems
180
5.3. 2 1 : Choose r > 0 and F : BJRN (p, r) � lR so that (5.3. 7) holds. Without
loss in generality, we will assume that r is taken so that F has bounded second order derivatives on BJRN (p, r) . In particular, since F vanishes on BJRN (p, r ) n M, there is a C < oo with the property that I F( y ) l < Cdist (y, M) for all y E BJRN (p, � ) . Hence, 1.
1m
e� o
dist (p + �v) "2
�
===}
Conversely, if v
E
1. F (p + �v) - F (p) < oo "2
< oo ===} 1m
e� o
( v, \7F (p)) JR N == 0
Tp (M) and 1 :
dist (p + �v, M)
0 } r n A == 0 JL(r) == ( JL1/J ({X} ) == JL1/J ( (X - E, X] ) == ( (X) - 'l/J (X - E) ) == 'l/J (X) - 'l/J (X - ) . == { 'l/J(x) - 'lj;(x-) > 0, JL }) ( x 'lj; (x) - 'l/J(x- ) JL'l/J JL1/J A A == 'l/J( x ) -'l/J( ) == ('l/J(t) - 'lj; (t-)) == 'l/Jd (x) . JL ( {t}) == x E r nS
then
:
===}
o.
is the set of atoms, and it is clear that
ii) Clearly,
lim
€� 0
lim 'ljJ
€� 0
Hence, x is an atom if and only if in which case . In particular, is non-atomic if and only if 'ljJ is continuous. On the other hand, if is purely atomic and is its set of atoms, then D( 'ljJ) and - oo
tE D ( 'lj; ) n ( - CX) , x )
t E D ( 'lj; ) n ( - CX) , x )
183
Solution to Selected Problems Conversely, if V; is purely discontinuous, then
JL� ( ( where
- oo ,
x] ) ==
tED( � ) n(
(V;(t) - V;(t - ) ) == v ( ( - oo , x] ) , x E JR, - oo , x]
v is the purely atomic measure given by v(r) ==
L (V; (x) - V;(x - ) ) ,
r E BR ·
xED( � )nr
Hence, since C == { ( - oo , x] : x E 1R} is a 1r-system which generates BR and JL� r c u {JR} == v r c u {JR} , it follows that JL� == v on BR . (iii) First suppose that
/L,p
JL�
