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0 such that OE(k)Xk ~ Vk. Put [ = min{E(l), ... , [(i)}. It remains to notice that
O"X ~ I1{O"Xk: k = 1, ... ,i} x I1{X k : k = i + 1,i + 2, ... } ~I1{Vk: k=l, ... ,i}xI1{Xk : k=i+1,i+2, ... }~U. In view of the arbitrariness of the point x E U the last fact means the openness of the set U with respect to the metric d. II. Take an arbitrary point x = {Xk: k = 1,2, ... } E X and an arbitrary number [ > O. Find a number i such that diam X k < [ for every k = i + 1, i + 2, .... For k = i + 1, i + 2, ... we have OcXk = X k. By virtue of our definition of the metric d on the product
O"X=I1{O"Xk: k=l, ... ,i}xI1{Xk : k=i+1,i+2, ... }, i.e., [-neighborhood of a point with respect to the metric d has the form of a Tychonoff neighborhood. So if a subset of the product is open with respect to the metric d, then it is open with respect to Tychonoff topology too. When we compare I and II we obtain that the Tychonoff topology on X coincides with the topology generated by the metric d. The theorem is proved. • In the next section we will show that the additional condition diam X k ---7 o for the existence of some metric on the product is inessential, but the above concrete description of the metric on the product will not appropriate. Every finite family of spaces may be extended to a countable family by taking as missing elements a one point space (a metric on it can be defined in a unique way). These additional elements change nothing in the consideration of the Tychonoff topology as well as in the introduction of the metric d. The condition diamX k ---7 0 is obviously fulfilled for the family completed in this way. Therefore the theorem proved may be automatically restated to a finite number of factors. In addition, we can propose to the reader as a simple exercise to revise the given reasoning, to reject conditions
Some properties of topological, metric and Euclidean spaces.
41
and remarks related with the infiniteness of the number of factors, and to give (an easier) direct proof of the existence of a metric on the product of a finite family of metric spaces. Example 1.1. The metric on the space JRn in Example 1.4.5 coincides with the metric introduced on this set as on the product of lines according to Theorem 1.2. Example 1.2. Let continuous real functions f and 9 be defined on a topological space X. By Theorem 1.1 the mapping : X --+ JR x JR, (x) = (f(x),g(x)), is continuous. Since the composition of continuous mappings is continuous, the remarks of Examples 1.7.2 and 1.1 imply the continuity of the function h( x) = f (x) + g( x). Example 1.3. The remarks of Example 1.7.4 and Theorem 1.7.7 imply the continuity of the function cp : JR --+ JR, cp( t) = e. Since
the remarks of the previous example imply the continuity of the mapping 'lj; : JR x JR --+ JR, 'lj;(x, y) = xy. The reasoning of the previous example now allows us to prove that under the same assumptions the function hI (x) = f(x)g(x) is continuous. Thus if a function can be constructed with help of the operations of addition and multiplication of continuous functions, then it is continuous. This remark and Lemma 1.1 imply the continuity of every mapping \]I : JRm --+ JRn with polynomials as coordinate functions (we consider the spaces with the metric of Example 1.4.5). Theorem 1.3. Let the hypotheses of Theorem 1.2 hold. Let the spaces Xb k = 1,2, ... , be compact. Then the product X is compact. Proof. Since by the previous theorem the space X is metric, to prove its compactness it is sufficient to check the fulfilment of condition (1.6.3). Let I be an arbitrary sequence of elements of the space X. We will choose sequentially subsequences of the sequence I' First we put 10 = I' Let the sequence Ik-I be constructed. Our next step is the construction of a sequence Ik. We do it as follows. Consider the sequence a of k-th coordinates of elements of Ik-I' Since the space X k is compact the sequence a contains a subsequence al converging to a point Yk E X k • We take as Ik the subsequence of the sequence Ik-I consisting of elements of Ik-I' the k-th coordinates of which belong to the sequence al' Now construct a subsequence 1* = {Xi; : j = 1,2, ... } of the sequence I' As Xit we take an arbitrary element of the sequence 11' Let elements Xi t , . . . ,Xi;_t be chosen. As Xi; we take an arbitrary element of the sequence Ij with the index > i j - I . Elements of 1* with indices > k belong to the sequence Ik' Therefore the k-th coordinates of the elements of 1* constitute
CHAPTER 2
42
a sequence converging to the point Yk. By Lemma 1.2 the sequence --y* converges to the point Y = {Yk: k = 1,2, ... }. The theorem is proved. • Theorem 1.4. Let X be a compactum and Y be an arbitrary topological space. Then the mapping F : Y -+ X x Y, F (y) = X x {y}, is upper semicontinuous. Proof. The proof use Theorem 1. 7.5. Let H be an arbitrary closed subset of the product X x Y and Y E Y\F- 1 (H). The last fact means that the intersection (X x {y} ) n H is empty: (X x {y }) n H = 0. Therefore for every x E X we can fix a Tychonoff neighborhood Ox x Vx of the point (x, y) in the product X x Y which does not meet the set H. The cover {Ox: x E X} of the compactum X contains a finite subcover {OX1,"" Oxd. Put Oy = n {Vx i l . . . , VXk } . Every point x E X belongs (at least) to one of the sets OXi' Therefore {x} x Oy 0, is compact. Proof. The inclusion 0,(0, r) ~ [-r, r]n (~ ]Rn) is obvious. The set [-r, r]n is compact by virtue of the compactness of the segment [-r, r] and Theorem 1.3. Lemma 3.3 and Theorem 1.6.2 imply what was required. The lemma is proved. •
Some properties of topological, metric and Euclidean spaces.
49
Theorem 4.1. A subset of an Euclidean space is compact if and only if it is closed and bounded at once. Proof. Sufficiency follows from remarks of the previous section, Lemma 4.1 and Theorem 1.6.2. Necessity follows from Theorem 1.6.1 and Lemma 3.7. The theorem is proved. • This theorem immediately implies that the closure of every bounded subset of an Euclidean space is compact, because this closure is a bounded set too. A topological space is called locally compact if every point possesses a neighborhood the closure of which (in the space under consideration) is compact. Evidently every compact space is locally compact. Every open subspace of a compactum is locally compact. The remark following Theorem 4.1 implies the local compactness of open subsets of the Euclidean space. So the following assertion is applicable directly to an arbitrary open subset U of the Euclidean space. Lemma 4.2. Let U be an arbitrary neighborhood of a nonempty compact subset B of a locally compact metric space X. Then there exists a number b > 0 such that the set 0f(B, b) (~O(B, b)) is compact and lies in U. Proof. Show first that there exists a neighborhood V of the set B such that the set [V] is compact and lies in U. According to the local compactness of the space X associate to every point b of the set B its neighborhood Wb such that the set [Wb ] is compact. According to the normalness of the space X (see Lemma 3.9) associate to each point b E X of the set Bits neighborhood Vb and a neighborhood Nb of the set Fb = X \ (Wb n U) such that Vb n Nb = 0. Then [Vb] n Fb = 0, [Vb] S;;; un W b. By virtue of the inclusion [Vb] S;;; Wb the set Vb is compact. The cover {Vb: b E B} of the compactum B has a finite subcover {Vb" ... , Vb.}. The set V = Vb, U·· ·UVb• lies in the compactum
that gives what was required. Our assertion now follows from Corollary of Lemma 3.8 (with V as the set U of the corollary) and from Lemma 3.3 .• Later we will use: Assertion 4.1. Let X be a locally compact metric space and M be its (at most) countable nonempty closed subset. Then set M has an isolated (in M) point. Proof. Take an arbitrary point p E M. By virtue of the local compactness of the space X for some c > 0 the set 0f(P,c) is compact. The set G = O(p, c) n M is open in the compactum K = O,(p, c) n M. It is (at most) countable. Enumerate its elements: G = {Xk: k = 1,2, ... }. Apply Baire's theorem 1.6.7 to the open subset G of the compact urn K. At least
CHAPTER 2
50
one of the sets {xd, k = 1,2, ... , has a nonempty interior. Let {xd be such a set. Since the set {xd consists of one point, the nonemptiness of its interior means that ({ xd) = {xd. Hence the set {xd is open (in K). Thus the point Xi is isolated (in G ~ K and hence it is isolated in M). The assertion is proved. • Corollary. Let U be an open subset of the space IRn. Let M be an (at most) countable nonempty closed (in U) subset of the set U. Then the set M has an isolated point (in M). • Let X and y be two arbitrary points of a vector space L. The set {tx + (1 - t)y: 0 ~ t ~ I} is called a segment (connecting the points x and y. The points x and yare called endpoints of the segment). A subset M of the space L is called convex, if for every two its points the set M contains the segment connecting the points. In order to investigate and to use properties of convex sets we will need the following notion. A function f : L ----t IR is called a linear functional, if f(u + v) = f(u) + f(v) and f(au) = af(u) for every elements u, v of the space L and for every number a. A very simple example of a linear functional on the space IR n is the function ai, i = 1, ... ,n, which associates to an element x = (Xl,' .. , x n ) of the space IRn its i-th coordinate Xi' This example may be generalized in the following way. Fix an arbitrary vector e = (al,"" an) E IRn. We define the linear functional
°
the set OEM is convex.
Proof. Let p and q be arbitrary points of the set OeM. By Lemma 3.4 there are points x and y of the set M such that Ilx - pil < f and Ily - qll < f. Take now an arbitrary point a of the segment connecting the points p and q. The point a may be represented in the form tp + (1 - t)q, t E [0,1]. By virtue of the convexity of the set M the point b = tx + (1 - t)y belongs to the set M. On the other hand,
Iia - bll
+ (1 - t)(q - y)11 xii + (1 - t)llq - yll
= Ilt(p - x) ~ tllp -
< tf + (1 =
t)f
f.
Hence a E OEM. The lemma is proved. • Lemma 4.4. The intersection of every family of convex sets is a convex set.
Proof. The proof is obvious: A segment connecting two arbitrary points of the intersection lies in every element of the family. Therefore it lies in the intersection. The lemma is proved. • Lemma 4.5. The closure of a convex subset of a normed space is a convex set.
Proof. Let M be a (nonempty) convex subset of the normed space under consideration. The representation [M] = : T x X - t Y be a continuous mapping of the product of topological spaces T and X into a topological space Y. Let 'Pt(x) = «I>(t, x) for t E T, x E X. Then the mapping F : T - t C(X, Y), F(t) = 'Pt, is continuous. Proof. Let to E T and F(t o) E O(K, U), where K is a closed compact subset of the space X and U is an open subset of the space Y. We have «I> ( {to} x K) = 'Pta (K) ~ U. By virtue of the continuity of the mapping «I> we can associate to every point x of the set K its neighborhood Ox and a neighborhood V(x) of the point to in a manner that «I>(V(x) x Ox) ~ U. The open cover {Ox: x E K} of the compact set K has a finite subcover {OXl, ... , Ox;}. Put V = V(xd n··· n V(Xi). Let t E V and x E K. For some j = 1, ... ,i the point x belongs to OXj. Therefore
In view of the arbitrariness of the point x E K this implies the inclusion 'Pt(K) ~ U. It means that F(t) = 'Pt E O(K, U). Referring to Theorem 1. 7.3 completes the proof. •
Spaces of mappings and spaces of compact subsets.
77
3. Vietoris topology Let X be a topological space. Denote by exp X the set of all nonempty closed subsets of the space X. For U I , . .. ,Uk s:;; X put O(UI , ... ,Uk) = {F: FE expX, F s:;; Uf=IUi , F nUl i- 0, ... ,F n Uk i- 0}. Notice two particular cases. For U s:;; X O(U) = {F: F E expX, F s:;; U} and O(U, X) = {F: F E expX, F n U i- 0}. The following equality is obvious: (3.1) On the set exp X introduce a so called Vietoris topology. It is the smallest topology, in which all set of the forms O(U) and O(U, X) are open (i.e., sets of the form mentioned constitute a sub-base of the Vietoris topology). Here U runs over the topology of the space X. Without any detailed discussion let us notice that the openness of sets of the form O(U) corresponds to the notion of convergence in the sense of condition (1. 7.1), and the openness of sets of the form O(U, X) corresponds to the notion of convergence in the sense of the lower limit. By virtue of (3.1) sets of the form O(UI , · · · , Uk), are open in the Vietoris topology. They are called Vietoris neighborhoods (of their points). In fact, Vietoris neighborhoods constitute a base of the Vietoris topology because elements of the above sub-base (the sets O(U) and O(U, X)) are Vietoris neighborhoods too, and, on the other hand, if FE O(UI , . . . , Uk) n O(VI , ... , Vi) then FE O(WI ,· .. , W m ) s:;; O(UI , .. ·, Uk) n O(VI , . . .
,
Vi),
where {WI,"" W m } is the renumbered family {Ui n Vi: i = 1, ... , k, j = 1, ... ,I, F n Ui n Vi i- 0}. The set exp X with the Vietoris topology is called the space of (nonempty) closed subsets of the space X. We will not give an account here of the large ammount of theory related to the notion introduced, see [Ku1], [FF], [En]. We will restrict ourselves to some technical remarks. Let 9 : X -> Y be a continuous mapping of a topological space X into a topological space Y. It is naturally to define the mapping g* of the space exp X as its domain by putting g* (F) = g(F). However in this way we do not always obtain a mapping of the space exp X into the space exp Y because the image of a closed set need not to be closed. In order to pass over this case consider the set E(g*) of all closed subsets of the space X, images under the mapping 9 of which are closed in Y. Consider the set E(g*) with the topology induced by the Vietoris topology. Although the space E(g*) in the general case need not coincide with the space expX, often it turns out to be quite large. So, if the spaces X and Yare Hausdorff, by virtue of
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78
Theorems 1.6.1 and 1.7.8 the space E(g*) contains all (nonempty) compact subsets of the space X. Theorem 3.1. Let 9 be a continuous mapping of a topological space X into a topological space Y. Then the mapping g* : E(g*) --+ exp Y is continuous.
Proof. The proof reduces to a direct verification. Let F E E(g*). Let O(UI , ... , Uk) be an arbitrary Vietoris neighborhood of the point g*(F). For VI = g-I(UI ), ... , V k = g-I(Uk ) we have
Therefore F E o (v;. , ... , Vk). This gives what was required. The theorem is proved. • Theorem 3.2. Let X be a regular space. Let generalized sequences 8 = {D,,: a E A} ~ expX and c = {E,,: a E A} ~ expX converge (in the Vietoris topology) with respect to a directed set cp to D and E, respectively. Let D" ~ E" for every a E A. Then D ~ E. Proof. Take an arbitrary t E X \ E. The regularity of the space X implies the existence of disjoint neighborhoods Ot of the point t and OE of the set E in the space X. By virtue of the convergence of the sequence c to E with respect to cp there exists F E cp such that E" E O(OE) for every a E F. Hence D" ~ E" ~ OE, D" nOt = 0 and D" rf. O(Ot, X). By virtue of the convergence of the sequence 8 to D with respect to cp the open set O(Ot, X) cannot be a neighborhood of the point D, i.e., D n Ot = 0 and t rf. D. This gives what was required. The theorem is proved. • Theorem 3.3. Let X be a normal space. Then the set expc X of all connected closed subsets of the space X is closed in the space exp X. Proof. By Theorem 1.3.3 it is sufficient to prove that the set H = exp X \ expc X of all disconnected closed subset of X is open in the space exp X. Let Fo E H. The set Fo may be represented as the union of two nonempty closed disjoint subsets FI and F 2 . By virtue of the normality of the space X there are disjoint neighborhoods UI and U2 of the sets FI and F 2, respectively. The set O(UI , U2) is a neighborhood of the point Fo in the space exp X. Every F E 0 (U I , U2 ) is equal to the union of two nonempty disjoint closed subsets F \ U2 = F nUl and F \ U I = F n U2· Therefore FE H, O(UI , U2 ) ~ H. The theorem is proved. • Corollary. Let FI ;;2 F2 ;;2 F3 ;;2 ... be a non-increasing sequence of nonempty connected compact subsets of a metric space X. Then the set F = n{Fk: k = 1,2, ... } is connected.
To obtain the corollary it is sufficient to restrict ourselves to the case FI = X. We have F = lim tOPk--+oo Fk (see Example 1.5.3). By Corollary of Theorem 1.6.3 the set F is nonempty. Let U be an open subset of the space X. If F E O(U, X), then the definition of the lower limit implies
Spaces of mappings and spaces of compact subsets.
79
that Fk E O(U, X), beginning with some k. If F E O(U), then by Theorem 1.7.2 Fk E O(U), beginning with some k. By Lemma 1.7.3 Fk --+ F in the Vietoris topology. This allows us to obtain the corollary. •
4. Hausdorff metric Let X be a Hausdorff space. By Theorem 1.6.1 every compact subset of the space X is closed, i.e., the set expc X of all nonempty compact subsets of the space X lies in exp X. The Vietoris topology was defined first on the set exp X. Now it induces a topology on the set expc X. The new topology is also called the Vietoris topology. Sets of the form Oc(U1, ... ,Uk ) = O(U1, ... ,Uk ) nexpcX, where U1"",Uk are open subsets of X (see the end of §1.3), constitute a base of the new topology. The set expc X with the mentioned topology is called the space of compact subsets of X. If X is a compact urn then by virtue of Theorems 1.6.1 and 1.6.2 expc X = exp X.
Figure 3.3
Figure 3.4
Let now (X, p) be a metric space. In this case consider one more possibility to introduce a topology on the set expc X. For F 1 , F2 E expc X put a(F1 ,F2) = inf{c: c > 0,F2 S;;; O[Fd, see Figures 3.3 and 3.4. Lemma 2.3.6 implies the nonemptiness of the set {c: c > 0, F2 S;;; O[Fd. The formula written defines a number a( Fl , F2)' The number is called the deflection of the set F2 from the set Fl' Notice properties of the deflection
a. For every (nonempty) compact subsets F 1 , F2 and F3 of the space X: 1) a(Fl,F2)~0; 2) a(F1 , F2) = if and only if F2 S;;; F 1 ; 3) a(Fl,F3):::; a(Fl,g) + a(F2,F3). Proof of these properties is simple. Notice only that in the case of the property 3) we need use Lemma 2.3.5. We have not mentioned the symmetry
°
80
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between properties of the deflection. The deflection does not possess this property and therefore it is not a metric. The Hausdorff distance between two compact subsets FI and F2 of the space X is defined as h(FI ,F2) = max{a(FI,F2),a(F2,Fd}. In contrast to the deflection the function h is symmetric (i.e.,
By virtue of the above properties of the deflection the function h is a metric on the set expc X (the Hausdorff metric). Theorem 4.1. The Vietoris topology on the space of compact subsets of a metric space is generated by the (corresponding) Hausdorff metric. Proof. Before this theorem is proved we must distinguish the space expc X of compact subsets of the metric space (X, p) with the Vietoris topology and the metric space H(X) with the same carrier set expc X and the Hausdorff metric h. Our aim is to prove that the identity mapping i : H(X) - t expc X is an homeomorphism (see §2.2). Since the bijectivity of this mapping is obvious we have to check the continuity of the mapping i and of its inverse mapping i-I. I. Let U be an open subset of the space X and Fo E Oc(U, X). Fix an arbitrary point x of the intersection Fo n U. Next, fix a number E > 0 such that O~x S;;; U (O~x denotes the E-neighborhood of the x point with respect to the metric p). If F E O~Fo (O~Fo denotes the E-neighborhood of the point Fo with respect to the metric h), then Fo S;;; O~F (the neighborhood with respect to the metric p). By Lemma 2.3.4 there exists a point t of the set F such that p(x, t) < E. SO t E O~x S;;; U. The intersection F n U contains the point t. Hence it is nonempty. Thus i(O~Fo) S;;; Oc(U,X). II. Let U be an open subset of the space X and Fo E 0 c(U). The condition Fo E Oc(U) may be rewritten as Fo S;;; U. By Corollary of Lemma 2.3.8 O,Fo S;;; U for some E > O. By the definition of the Hausdorff metric O~ Fo S;;; O(U), i.e., i(O~ Fo) S;;; O(U). III. The continuity of the mapping i follows from I, II and Theorem 1.7.3. IV. Let Fo E H(X) and E > o. The open cover {0,/4X: x E Fo} of the compactum Fo contains a finite sub cover , = {0,/4XI, ... , Oo/4xd. The following properties of the family, are essential for us:
and (4.2) if points s, t belong to an element of the family" then p( s, t)
o. For every k = 1,2, ... fix a family of intervals such that U, "2 Mk and k d('d ~ J.t*(Mk) + 2- E. Such a family exists by the definition of the outer measure. For the family, = U{ ,k: k = 1,2, ... } we have
'k
U {Mk: k = 1, 2, ... } ~ d(,)
= 2:)d('d:
k
u"
= 1,2, ... }
~L{J.t*(Md+2-kc: k=1,2, ... } 00
k=l
=
L{J.t*(Md: k
= 1,2, ... } +E.
Hence J.t*(U{Mk: k = 1,2, ... }) ~ {J.t*(Md: k = 1,2, ... } + c.
Since this is true for every number c > 0 the lemma is proved. • The outer measure of the empty set is equal to zero. In Lemma 1.2 we did not assume the nonemptiness of the sets M k • Therefore the analog of the inequality just proved for a finite family of sets is its particular case. Lemma 1.3. Let a subset of the real line be represented as a union of a finite number of pairwise disjoint segments. Then the outer measure of the subset is equal to the total length of these segments. Proof. Let M = U{[ak, bkJ: k = 1, ... , s} be the set and its representation from the statement of the lemma. Notice that by virtue of
97
Derivation and integration.
Theorem 2.6.7 these segments, and only they, are connected components of the set M and such a representation is unique. Let d = 2:~=1 Ib k - ak I. The inequality J.L*(M) ~ d follows from the consideration of the families = {( ak - E, bk + E): k = 1, ... , s}, E > O. The lemma will be proved when we show that d(,) ~ d for every (at most countable) family, of intervals covering the set M. By virtue of the compactness of the set M it is sufficient to prove this assertion under the assumption of the finiteness of the cover ,. It will be done with the help of an induction on the number of elements of, = {U1 , . . . , Ud. We will do it not for a fixed set M, but for all such sets at once. For t = 1 the assertion is obvious. Let the assertion be proved for t = i - I ~ 1. Prove it for t = i. We do not lose any generality when we assume in addition that the segments [ak, bd are enumerated according to their position in the line: al ~ b1 < a2 ~ b2 < ... < as ~ bs . We can also assume that just the interval Ut = (p, q) contains the point bs (in the opposite case we renumber the family,). If p < al then as in
,t:
V1
Vt
V3 ~
~
~
( ( ) () ) ........ ".... ( -+ -t-+---t----1f-+-t-at
() ( )
}
b1 a2 '-..--'
V2 Figure 4.2
......
the case t = 1, the assertion is obvious. Consider the case al ~ P « bs ) (Figure 4.2). Let Ml = M n [al,pl and M2 = M n [p, bsl. Each of these sets may be also represented as the union of a finite number of pairwise disjoint segments. If d1 and d2 are the total lengths of the components of Ml and M 2, respectively, then d = d 1 + d 2. The family,1 = {U1 , •.• , Ut-d covers the set MI. By the inductive hypothesis d(,I) ~ d 1 • The length of the interval Ut is not less than d 2. So d(,) = d( '1) + d( {Ut }) ~ d 1 + d 2 = d. The lemma is proved. • Lemma 1.4. Let an (at most) countable family, = {Mk : k = 1,2, ... } consist of pairwise disjoint segments, intervals and half intervals of the real line and E > O. Then: a) if d(,) < 00, then there are segments II"'" I j lying in different elements of the family, such that: J.L*(I1 U··· U I j ) ~ d(,) - c and J.L*«U,) \ (U{Il,"" Ij}) < E; b) if d(,) = 00, then there are segments I 1 , ••• ,Ij lying in different elements of the family, such that J.L*(I1 U ... U I j ) ~ c .
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Proof. Let ak ~ bk be endpoints of a segment, an interval or an half interval M k , k = 1,2, .... If ak = -00 or bk = 00 for some k, then we have b) and the validity of the assertion is obvious: the set Mk contains segments of arbitrarily large length. Below we assume that -00 < ak ~ bk < 00. For k = 1,2, ... denote by h the segment [ak + 2- k - 3 E, bk - 2- k - 3 E], if 2- k- 2E ~ bk - ak' If the last condition is not fulfiled we put h = 0. The length l(Id of the segment (or empty set) h is not less than the number max{bk - ak - 2- k- 2E,0}. For the calculus of d(,) we have the series L~=1 (b k - ad· For the calculus of the total length of the segments h, k = 1,2, ... we have the series L~=1 l(h). Let us estimate the difference of partial sums of these series: i
i
L
i
2)bk - ad l(h) k=1 k=1
~
L
Ibk - ak -l(h)1
k=1
~
i
L 2- k- 2
E
< ~.
k=1
Therefore the sums of these series, i.e., the limits of the sequences of its partial sums, differ at most by ! < ~, if db) < 00, and coincide if db) = 00. There exists a number j such that l(Id+· .. +l(Ij) > db) - ~ (> db) -E), for db) < 00 and l(Id + ... + l(Ij) > 10 for db) = 00. By Lemma 1.3 the last observation completes the proof in the case b). In the case a) the previous estimate proves the first part of the assertion. The second part of a) follows from the fact that difference (U,) \ (U{ II, ... ,Ij}) lies in the union of the sets [aI, al + 2- 4 10], [b 1- 2- 4 10, b1], ... , [aj, aj + 2- j - 3 E], [bj - 2- j - 3 E, bj ], [aj+l' bj+l], [aj+2' bj+2], .... Therefore Lemmas 1.1 and 1.3 and the inequality
imply the estimate
fL*((U,) \ (U{Il,'" ,Ij}) ~ 2- 3 10 + ... + 2- j - 2E + (bj+l - aj+l) + (bj+2 - aj+2)
there are a closed subset F and an open subset G (Figure 4.3) of the line such that F t;;;; M t;;;; G and J-L * (G \ F) < E.
°
--+----+ G: f-
-)
+------ + F:
)-
.
H: Figure 4.3
Lemmas 1.5 and 1.6 imply that open and closed sets are measurable. Evidently the complement to a measurable set is measurable (see the proof of Lemma 1.6). If a set M t;;;; IR is measurable its outer measure J-L*(M) is called the measure of the set M and is denoted by J-L(M). Lemma 1. 7. The intersection of two measurable subsets of the real line is measurable. Proof. Let M i , i = 1,2, be measurable subsets of the line and c > 0. Take a closed Fi and an open G i sets such that Fi t;;;; Mi t;;;; G i and J-L(G i \ F i ) < ~. Evidently Fl nF2
C;;;;
Ml nM2 t;;;; G l nG 2
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and By Lemma 1.2
/L*((GI n G 2 )
\
(FI n F2 ))
~ /L*(GI \ FI) + /L*(G 2 \ F2 ) < ~ + ~
=
E.
The lemma is proved. • Lemma 1.7 and the above remark about the measurability of a complement imply: Lemma 1.8. The difference of two measurable subsets of the real line is measurable. • In addition, MI U M2 = IR \ ((1R \ M I ) n (IR \ M 2)) and Lemma 1.7 and the same remark (or Lemma 1.8) imply the measurability of the union of two measurable set. An obvious induction on the number of elements gives: Lemma 1.9. The union of a finite family of measurable subsets of the real line is measurable. • Lemma 1.10. Let MI and M2 be disjoint closed subsets of the real line. Then ./L(MI U M 2 ) = /L(MI) + /L(M2 ).
Proof. The inequality /L(MI U M 2 ) ~ /L(MI ) + /L(M2 ) follows from Lemma 1.2. Our lemma will be proved when we establish the validity of the inverse inequality. Let, be an arbitrary cover of the set MI U M2 by intervals. By virtue of the normalness of the line there exist disjoint neighborhoods OMI and OM2 of the sets MI and M 2 , respectively. The sets Vi = (U,) n OMI and V2 = (U,) n OM2 are also disjoint neighborhoods of the sets MI and M 2 . The set of connected components of the set V = VI UV2 is split in two subset: components of the set Vi and components of the sets V2. By Lemmas 1.2-4 the measure of an open set is equal to the total length of its connected components. Therefore /L(V) = /L(Vd + /L(V2 ). By Lemma 1.2 /L(U,) ~ d(r). By Lemma 1.1 /L(V) ~ /L(U,) and
/L(Md
+ /-l(M2 )
~
/L(Vi)
+ /L(V2).
Thus
In view of the arbitrariness of the cover, we have
The lemma is proved.
•
101
Derivation and integration.
Theorem 1.1. Let measurable subsets M k , k = 1,2, ... , of the line be pairwise disjoint and L~l J.t(Md < 00. Then the set U{Mk : k = 1,2, ... } is measurable and 00
J.t(U{ M k : k
= 1,2, ... }) = L
J.t(Mk )·
k=l
Proof. Take an arbitrary c > O. According to the convergence of the series L~l J.t(Md fix an index j such that
Now Lemma 1.2 implies the existence of a family of intervals, such that
U{Mk: k = j
+ l,j + 2, ... } ~ U, and db) < ~.
For k = 1, ... , j fix a closed Fk and an open G k sets such that ~ Mk ~ G k and J.t(G k \ Fk ) < 2- k - 1 c. The sets F = FI U ... U Fj and G = G 1 U ... U Gj U (U,) satisfy the conditions
Fk
F ~ U{Mk: k = 1,2, ... } J.t(G \ F) ::::; J.t(G I \ Fd + ... + J.t(G j ::::;
c + ...
2- 2
+ 2-
In view of the arbitrariness of c
j - 1
c+
~ \
2- l
G, Fj ) c
+ J.t(U,)
< c.
> 0 this means the measurability of the
set
M = U{Mk: k = 1,2, ... }. By Lemmas 1.1 and 1.10 j
00
L J.t(Md k=l
c ::::;
L J.t(Mk) k=l
c
j
2 ::; L
J.t(Fk) = J.t(F)
k=l
::::; J.t(M) ::::; J.t(G) ::::; J.t(F) + J.t(G \ F) ::::; J.t(F)
k=l
k=l
In view of the arbitrariness c > 0 00
J.t(M) =
L
k=l
J.L(Mk).
+c
102
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The theorem is proved. • Lemma 1.11. A set M C ~ is measurable if and only if for every segment I ~ ~ the set M n I is measurable. Proof. Necessity follows from Lemma 1.7. Sufficiency. Take an arbitrary c > 0. For every k = 0, ±1, ±2, ... fix a closed Fk and an open G k subsets of the line such that
The set F = U{Fk: k = 0, ±1, ±2, ... } is closed. (Every point x E ~ lies in an interval (k, k + 2). If x f/. F, then x f/. Fk U Fk+I' The set Ox = (k, k+2)\(FkUFk+d is a neighborhood of the point x without points of the set F.) The set G = U{G k : k = 0, ±1, ±2, ... } is open and 00
F
~
M
~
J1.(G \ F) ~
G,
L k=-oo
00
2- lkl - 2 c
< L 2- k- I c = c. k=O
The lemma is proved. • Theorem 1.2. Let {Mk: k = 1,2, ... } be a family of measurable subsets of the real line. Then the set M = U{Mk: k = 1,2, ... } is measurable and
Proof. The proof use Lemma 1.11. Let [a, b] be an arbitrary segment of the line. We have the representation M n [a, b] = where
U"
, = {(Mk
n [a, b]) \
(U{MI , ... , Mk-d): k = 1,2, ... }.
The measurability of elements of the family , follows from Lemmas 1.8 and 1.9. By Theorem 1.1 and Lemmas 1.1 and 1.3 partial sums of the series l: {J1.(r) : r E ,} are bounded from above by the number b-a. Therefore the series converge. Theorem 1.1 implies the measurability of the set M n [a, b]. Referring to Lemma 1.11 completes the proof of the measurability of the set M. The validity of the inequality follows from Lemma 1.2. The theorem is proved. • Assertion 1.1. Under the hypotheses of Theorem 1.2 let MI ~ M2 ~ M3 ~ .... Then J1.(M) = .lim J1.(Mi)' '-+00 Proof. I. Lemma 1.1 implies that:
II. If limi-+oo J1.(Mi ) =
00,
then by I the fact in question is obvious.
103
Derivation and integration.
III. Let limi ..... oo J.t(Mi )
x t - x
Ar = limsup f(t) - f(x). t-+x, t>x t- x
t- x
t-+x, t<x
Example 3.1. Consider the functions
f(x) = {xosin
() { M
9 x =
o
~
for x =/: 0, for x = 0,
sin ~ x
=/: 0,
for x
for x = 0,
on the segment [-1,1]. For the function f at the point x = 0 we have Al = Ar = -1, Al = Ar = 1 (Figure 4.5). For the function 9 at the same point we have Al = Ar = -00, Al = Ar = +00 (Figure 4.6).
! .
.
' HI
, "
+- - ; .
Figure 4.5
(, 'I
lil[!:: ; 1[. .~,
~ iV" ': J I! ..
\/
Figure 4.6
For the existence of the (finite) derivative of a function f at a point x the fulfilment of the condition -00 < Al = Al = Ar = Ar < 00 is necessary and sufficient. Lemma 3.2. Let a non-decreasing real function f be defined and be continuous on the segment [a, b], a < b, of the real line. Then the function f has the derivative at almost every point of the segment [a, bJ. Proof. I. Show first that almost everywhere Ar < 00. Let A be the set of (all) points of the segment [a, bJ, at which Ar = 00. For every point x of the set A and every positive integer k there exists a )
Derivation and integration.
111
point t E (x, b] such that
f(t) - f(x) > k. t-x This means that f(t) - kt > f(x) - kx, i.e., the point x is a right increasing point for the function 9k(S) = f(s) - ks. By Lemma 3.1 the set Ak ofright increasing points of the function 9k is open. The point b cannot be right increasing point. We obtain from Corollary of Theorem 2.6.7 and Theorem 2.6.3 that the set Ak may be represented as the union of an (at most) countable set 'Yk of its (pairwise disjoint) components. The components are either intervals or half intervals with the closed endpoint a (no more than one such components). By Lemma 3.1 for every element of the set 'Yk with endpoints a < 13 we have 9k(a) ~ 9k(f3), 13 - a ~ Hf(f3) - f(a)). In view of the monotonicity of the function f this implies that d("(k) ~ tU(b) - f(a)). So J.L*(A) ~ J.L*(Ak) ~ Hf(b)- f(a)). Since this is true for every k = 1,2, ... , J.L*(A) = O. This was required. II. Show that almost everywhere Ar ~ A/. At the points a and b one of these numbers is not defined. Denote by B the set of all points of the interval (a, b), at which the inequality under consideration is not fulfilled, i.e., we have there the inequality A/ < Ar • Denote by P the set of all couples p = (q, r) of rational numbers, in which the first member is strongly less than the second one: q < r. The set P is countable (see Example 1.1.2 and Corollary of Theorem 1.1. 7). The set B may be represented as the union of (countable number of) sets B p , p = (q, r) E P, where Bp is the set of points, at which A/ < q < r < Ar . Our aim will be achieved when we show that the outer measure of every set B p , PEP, is equal to zero. Assume the opposite, i.e., that J.L*(Bp) = C > 0 for some p = (q, r) E P. There exists a covering "I of the set Bp by intervals such that d(,,() < ~c (with the notation of §1). Let V = (a,f3) E "I, By = Bp n V and x E By. At the point x we have A/ < q. Thus there exists a point t E (a, x) such that J(tt~(x) < q and hence f(t) - qt > f(x) - qx, i.e., the point x is a left increasing point of the function f(s) - qs on the segment Iv = [a, 13]. By Lemma 3.1 the set Mv of (all) such points is open in the segment Iv. The left endpoint of the segment Iv cannot be an left increasing point. Corollary of Theorem 2.6.7 and Theorem 2.6.3 imply that the set Mv may be represented as the union of an (at most) countable set "Iv of its (pairwise disjoint) components. The components are either intervals or an half interval with the closed endpoint sup Iv. By Lemma 3.1 for every element U of the set "Iv with endpoints a' < 13' we have the inequality f(a') - qa' ~ f(f3') - qf3' or (3.1)
f(f3') - f(a')
~
q(f3' - a').
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112
By virtue of the inequality Ar > r for every point x E UnBir there exists a point t E (x, {3') such that f(tt:~(x) > r, i.e., f(x) - rx < f(t) - rt. Thus the point x is a right increasing point for function f (s) - r s on the segment [a', {3']. By Lemma 3.1 the set M*(c/,{3') of (all) such points is open in the segment [a', {3']. By Corollary of Theorem 2.6.7 and by Theorem 2.6.3 the set M*(a', {3') may be represented as the union of an (at most) countable set ,*(a',{3') of its (pairwise disjoint) components. The components are either intervals or an half interval with the closed endpoint, which coincides with a'. By Lemma 3.1 for every element of the set (a', {3') with endpoints a" < {3" we have the inequality f(a") - ra" ~ f({3") - r{3", i.e.,
,*
{3" - a"
~ ~(f({3") r
The monotonicity of the function
d(r*(a', {3'))
f(a")).
f implies that
~ ~(f({3') r
f(a')).
By (3.1) the total length of elements of the set ,~= U{r*(a',{3'):
(a',{3') E 'v, ( (a',{3] E 'v)}
does not exceed ;J.L(V). The family ,* = U{rir: V E ,} covers the set Bp. By Lemmas 1.2 and 1.1 (and the measurability of open sets)
The contradiction obtained (J.L*(Bp) < J.L*(Bp)) shows that our assumption is false. Hence J.L*(Bp) = O. III. When we go from the function f (x) to the function (x) = - f ( - x ) the notions of left and right change places. From this observation we obtain immediately from the previous reasoning that at almost every point of the segment [a, b] we have Al ~ Ar . When we compare the above inequalities with the obvious estimates Al ~ Al and Ar ~ Ar we obtain that 0 ~ Ar ~ Al ~ Al ~ Ar ~ Ar < 00 at almost every point of the segment [a, b]. So 0 ~ Ar = Al = Al = Ar < 00. Therefore at almost every point of the segment [a, b] the derivative exists. The lemma is proved. • It is natural to ask in what measure the derivative describes the behavior of a function. There exist situations being a long way from the ones which we met when we considered continuously differentiable functions. For the construction of a corresponding example we will need
r
Derivation and integration.
113
Lemma 3.3. Let M be a closed subset of the segment [a, b], a < b. Let a real function f be defined and be non-decreasing on the segment [a, b], be continuous on the set M and on the closure of every component of the set [a, b] \ M. Then the function f is continuous. Proof. The continuity of the function f at points of the set [a, b] \ M follows from the openness of the set [a, b] \ M in the segment [a, b] and hypotheses of the lemma. Thus we need to prove the continuity of the function f at an arbitrary point x of the set M. Take an arbitrary number c > O. By virtue of the continuity of the function f on the set M there exists a number 8 > 0 such that If(t) - f(x)1 < c for every t E (x - 8, x + 8) n M. If (x - 8, x) n M "# 0 then denote by c an arbitrary point of the set (x - 8, x) n M. If (x - 8, x) n M = 0 then by virtue of the continuity of the function f on the closure of the component of the set [a, b] \ M, which contains the interval (x - 8, x), there exists a number c E (x - 8, x) such that If(c) - f(x)1 < c. Thus in this case too we have fixed a point c. If (x, x + 8) n M "# 0 then denote by d an arbitrary point of the set (x,x + 8) n M. If (x,x + 8) n M = 0, then by virtue of the continuity of the function f on the closure of the component of the set [a, b] \ M, which contains the interval (x, x + 8), there exists a number d E (x, x + 8) such that If(d) - f(x)1 < c. Thus in this case too we have fixed a point d. The interval (c, d) is a neighborhood of the point x. By virtue of the monotonicity of the function f we have
f(( c, d))
~
[f( c), f(d)]
~
(f(x) - c, f(x)
+ c).
The lemma is proved. • Example 3.2. We will construct a continuous mapping of the segment [0, 1] onto itself, a so called 'Cantor stairs'. Use the notation of Examples 1.6.4 and 2.2.8. In discussion of properties of this function the result of Example 1.2 will be also essential: the measure of the Cantor perfect set is equal to zero. In Example 2.2.8 we have pointed an homeomorphism f : K --t DN. Define first the function h ('Cantor stairs') on the set K and next extend it to the entire segment [0, 1]. For a point x = {Xl, X2, ... } E DN put Sets VX1, ... ,Xk
=
{y
= {YI, Y2,'" }: Y E D N , YI = Xl,··· ,Yk =
= {Yl, Y2,' .. } E VX1, ... ,Xk' then h ( )1 ~ IXk+l - Yk+d + IXk+2 - Yk+21 + 0 Y '" 2k+1 2 + 2 ' ..
constitute a base at the point x. If Y
Ih o( x ) -
xd
k
~
111 + 2k+2 + ... = 2k '
2k+l
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114
This implies the continuity of the mapping ho and hence the continuity of the mapping h = hof. Let [e, d] be a component of the set Sk([O, 1]). Then (with the notation of Example 2.2.8) for every i = 1, ... , k the set fi([e, d]) consists of one point. Denote it by ai. The values of the mapping f at the points e and d may be written as
f(e) = {al, ... ,ak,O,O, ... }, Let el
f(d) = {al, ... , ab1, I, ... }.
= e + ~(d - e), d l = e + Hd - e). We have:
f(el) = {al, ... ,ak,O,I,I, ... }, Therefore
and (3.2)
h() = h(d ) = al el
1
2
+
...
ak
_1_ = h(e)
+ 2k + 2k+l
+ h(d) . 2
Thus values of the function h at the endpoints of a component (el, dd of the complement to the Cantor perfect set coincide. Extend the function h to the segment [0,1] and put h(x) = h(ed for every x E (el,dd. The equality (3.2) helps to imagine the graph of this function (Figure 4.7). The function h is monotone. Lemma 3.3 implies its continuity. The continuity, the equalities h(O) = and h(l) = 1, Corollary 1 of TheoFigure 4.7 rem 2.6.4, and Theorem 2.6.3 imply the coincidence h([O, 1]) = [0,1]. The complement of the Cantor perfect set in the segment [0,1] has the full measure. At every its point the derivative of the function h exists and is equal to zero. We have constructed the continuous non-decreasing nonconstant function, the derivative of which almost everywhere is equal to zero. Recall for comparison that if the derivative of a function exists and is equal to zero everywhere (but not 'almost everywhere') then the function
°
Derivation and integration.
115
is constant. It seems that Lemma 3.2 gives a possibility to introduce the notions of the integral and of the primitive for a new rather large class of functions. Example 3.2 shows that here we go away from relations between functions and derivatives to which we are accustomed in dealing with continuously differentiable functions. However in Lemma 3.2 we made a step in the direction needed. Our nearest aim will be to select a class of functions which have derivatives almost everywhere and which are uniquely determined by their derivatives up to constant terms. We will pass to it in the next section, but now Theorem 3.1. A convergent series of continuous non-decreasing functions f(x) = fl (x) + f2(X) + h(x) + ... on a segment [a, b], a < b, admits almost everywhere the termwise derivation:
f'(x)
= f~(x) + f~(x) + f~(x) + ....
Before to prove the theorem let us make some introductional remarks. As usual, the sum of the series means the limit of the sequence of the partial sums
as k -+ 00. In the statement of the theorem we did not define exactly the type of convergence of the series. Assume now the convergence of the series at points a and b only. By virtue of the monotonicity of the functions fk' k = 1,2, ... for every point t of the segment [a, b] and for every index i < j we have In view of the arbitrariness of the point t E [a, b] this implies that
where in the left hand side the norm of the uniform convergence appears. This inequality, the convergence of the sequences {s d a): k = 1, 2, ... } and {s k (b): k = 1, 2, ... }, and the fact that every convergent sequence is fundamental (see §1.6) imply that the sequence {Sk: k = 1,2, ... } is fundamental with respect to the norm of the uniform convergence. From Corollary of Theorem 3.1.2 we obtain the convergence of the sequence {Sk : k = 1,2, ... } to a continuous function f. This guarantees the summability of Our series with respect to the norm of the uniform convergence. The sum f of our series is non-decreasing. By Lemma 3.2 terms of the series and the sum have derivatives almost everywhere on the segment [a, b]. Proof of Theorem 3.1. Consider a subset M of full measure of the segment [a, bj, at points of which derivatives of all terms and of the sum of
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116
the series exist. For every point x E M and for every point t E [a, b] \ {x} we have =
t-x
t- x
Signs of terms of the numerator in the left hand side coincide with the sign of the denumerator. Thus for every number N
When we pass to the limit as t
--+
x we obtain
N
L f~(x) ~ J'(x). k=1
Since this is true for every N = 1,2, ... the nonnegativity of the terms implies the summability of the series of the derivatives and the estimate (3.3)
f{(x) +
f~(x)
+ f~(x) + ...
~
J'(x).
Now for every k = 1,2, ... fix a number i(k) = 1,2, ... such that SiCk) II ~ 2- k. The above remarks may be applied to the series
Ilf -
g(x) = gl (x) + g2(X) + g3(X) + ... , where gk(X) = f(x) - Si(k)(X) = fi(k)+l(X) + fi(k)+2(X) + .... Thus at almost every point x of the segment [a, b] we have
g'(x)
~ g~ (x)
+ g~(x) + g;(x) + ...
The convergence of a numerical series implies that their terms tend to zero. In our case this means that S~(k)(X) --+ f'(x) as k --+ 00. Now (3.3) and the nonnegativity of terms of the series in the left hand side of the inequality (3.3) imply that Si(X) --+ f'(x) as i --+ 00 for almost all x E [a, b]. The theorem is proved. • 4. Derivative with respect to a set Let the domain of a function f with values in the Euclidean space ~n (for n = 1 in the real line ~) contains a set M ~ R Let x be a non-isolated point of the set M. If the limit of the quotient I(tt:~("') as t E M, t --+ x exists, it is called a derivative of the function f with respect to the set M at the point x and is denoted by f~(x).
117
Derivation and integration.
Following the standard method we prove that if two functions have the derivatives with respect to a set M ~ IR at a point x E M then their sum, difference, and product of each of these functions by a number also have such a derivative and its value is equal, respectively, to the sum, the difference and the product by the number of the values of the derivatives of initial functions. A mapping f (t) = {fl (x), ... , f n (t)} of the set M into the space IR n has the derivative at a point x E M with respect to the set M if and only if the coordinate functions fl(X), ... , fn(x) have the corresponding derivatives (necessity follows from Lemmas 2.1.1 and 1.7.2, sufficiency follows from Lemma 2.1.2). In this case fk(x) = {(fl)~(X), ... , (fn)~(x)). Lemma 4.1. Let the domain of a function f with values in the Euclidean space IR n contain sets A ~ B (~ 1R). Let x be a non-isolated point of the set A (hence x is a non-isolated point of the set B too). Let the derivative f~(x) exist. Then the derivative f~(x) exists and f~(x) = f~(x). Proof. The proof is obvious. • Lemma 4.2. Let the domain of a real function f contain a (nonempty) compact set M ~ R Let the function f be continuous and non-decreasing on the set M. Then the function f has a derivative with respect to the set M at almost every point of M. Proof. It is sufficient to prove the assertion under the additional assumption 7r(f) = M. Let a = inf M and b = supM. The set L = [a, bj \ M is open in the segment [a, bj and does not contain the points a and b. Theorems 2.6.7 and 2.6.3 imply the representation of the set L as the union of a set 'Y of pairwise disjoint intervals. Endpoints of these intervals belong to M. Extend the function f to the entire segment [a, bj by putting f(t) = f(a)({3 - t) + f(f3)(t - a) (3-a
for t E [a,bj\M, where t E (a,{3) E 'Y, see Figure 4.8. By Lemma 3.3 the extended function f is continuous. Evidently it is non-decreasing. By Lemma 3.2 there exists a set Mo ~ [a, bj of measure zero such that for every point x of the set [a, bj \ Mo the derivative f'(x) exists. Lemma 4.1 implies the existence of the derivative fk(x) for every point x E M \ Mo. It remains to notice, that by Theorem 1.1 J.t(M \ Mo) = J.t(M). The lemma is proved. •
-M a
M
fJ Figure 4.8
M
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118
Lemma 4.3. Let the domain of a function f with values in the Euclidean space Rn contain a measurable set M ~ R Let the (finite) derivative f~ (x) exist at almost every point x of the set M. Then the function f~ is measurable. Proof. Let Mo be the set of all points x EM, at which the derivative f~(x) exists. If x E M o, then the quotient f(tt:~(x) tends to a finite limit, f(t) - f(x) = f(t) - f(x) (t - x) t-x
~
0
as t ~ x in the set M. Therefore the function f IM is continuous at x. Fix a point tik E (2- i k, 2- i (k+1))nM for i = 1,2, ... , k = 0, ±1, ±2, ... satisfying the condition (2- i k, 2- i (k + 1)) n M =I 0. The set Ml of all so fixed points is (at most) countable. Therefore IL(Md = O. For i = 1,2, ... , k = 0, ±1, ±2, ... and x E (2- i k, 2-i(k + 1)) n (Mo \ (Ml U Q)) put gi(X) = f(x) - f(t ik ). x - tik
The function gi is continuous on every set
i = 1,2, .... Therefore it is continuous (see Theorem 1.7.7). Thus the func-
tion is measurable (on Mo \ (Ml U Q)). For a fixed x E Mo \ (Ml U Q) by virtue of the existence of the derivative f~(x) the sequence gi(X) converges to f~(x). Theorem 2.4 implies the measurability of the function f~IMo\(MluQr Lemma 2.1 implies the measurability of the function f~· The lemma is proved. • 5. Absolutely continuous functions Let the domain of a function f with values in a metric space (X, p) contain a set M ~ R The function f is called absolutely continuous on the set M if for every number c > 0 there exists a number > 0 such that if
°
(*,0, M) a finite family,
= {(ai, (3i):
i
= 1, ... ,k} consists of pairwise
disjoint intervals with endpoints from the set M and with d(,) < 0, then E{p(f(ai),f({3;)): i
= 1, ... ,k}
~
c.
If the set M coincides with the domain of the function f then the function f is called absolutely continuous (without the mentioning of M). Lemma 5.1. Let a function f with values in a metric space (X, pd be absolutely continuous on a set M ~ R Let a mapp'ing 9 of the space X into
Derivation and integration.
119
a metric space (Y, P2) satisfy the Lipschitz condition. Then the function gf is absolutely continuous on M. Proof. Take an arbitrary number c > O. Let L (~ 0) be a Lipschitz constant for the mapping g. Select a number 0 > 0 such that as soon as a family'Y = {(ai, .Bi): i = 1, ... ,k} satisfies condition (*,0, M), then
For every such a family 'Y 'L3p2(gf(ai), gf(.Bi)): i = 1, ... ,k}
~ L 'L3P1 (f(ai), f(.Bi)): i = 1, ... , k}
~ L
Lc
+ 1 < c.
The lemma is proved. • Lemma 5.2. The diagonal product f of functions fj, j = 1, ... ,n, with values in metric spaces (Xj, pj) is absolutely continuous on a set M ~ IR (with respect to the metric on the product from Theorem 2.1.2) if and only if every function Ii is absolutely continuous on M. Proof. Necessity follows immediately from Lemma 5.1. Sufficiency. Let P be the metric on the product. Take an arbitrary number c > O. For every j = 1, ... ,n fix a number OJ > 0 in a manner that
for every family'Y = {(ai,.Bd: i = 1, ... , n} satisfying condition (*, OJ, M). If 0 = min{ 01,'" ,On} and a family 'Y = {(ai, .Bi): i = 1, ... ,k} satisfies condition (*,0, M), then L{p(f(ai),f(.Bi)): i = 1, ... ,k}
~L
{t
pj(Jj(ai), Ii (.Bi)) : i
= 1, ...
'k}
3=1
n
= LL{pj(Jj(ai),fj(.Bi)):
i
= 1, ... ,k}
i=1 ~
c n· n
The lemma is proved.
= c.
•
120
CHAPTER 4
Lemma 5.3. A vector function f = {fl, ... , fn} with values in the Euclidean space IRn is absolutely continuous on a set M ~ IR if and only if every coordinate function fl, ... , fn is absolutely continuous on M. Proof. To be precise, in the statement we speak about the Euclidean metric of the space IRn. For the metric of Example 1.4.5 an analogous assertion follows immediately from Lemma 5.2. Because of the inequality (2.2.1) relating these two metrics the assertion in question follows from Lemma • 5.1. The lemma is proved. Lemma 5.4. Let the domains of functions f and 9 contain a set M ~ R Let the functions take values in the Euclidean space IRn and be absolutely continuous on the set M. Let>. be a real number. Then the functions f + 9 and >.f are absolutely continuous on the set M. Proof. Lemma 5.3 reduces the proof to the scalar case. Here we obtain what was required from Lemma 5.2, remarks of Examples 1.7.2 and 1.7.3 and Lemma 5.1. The lemma is proved. • Example 5.1. The identity mapping of a set M ~ IR into itself is absolutely continuous. This follows immediately from definitions. So Lemma 5.1 implies that if a function satisfies the Lipschitz condition then it is absolutely continuous. Assume that a function f is defined, has the continuous derivative on the segment [a, bj, and m = sup f'([a, b]). For every point s < t of the segment [a, bj there exists a point c E [s, tj such that f(t) - f(s) = f'(c)(t - s) (Lagrange formula). So If(t) - f(s)1 ~ mit - 81, i.e., the function f satisfies the Lipschitz condition. Hence it is absolutely continuous. Lemmas 5.4 and 5.3 extend immediately our list of examples of absolutely continuous functions. Every absolutely continuous function is uniformly continuous. We obtain it, when we take in (*,0, M) a family 'Y consisting of one element. Lemma 5.5. Let a function f with values in a metric space (X, p) be absolutely continuous on a set M ~ R Then for every number EO > 0 there exists a number > 0 such that if
°
(**,o,M) an (at most) countable family 'Y = {(ai,,Bi): i = 1,2, ... } consists of pairwise disjoint intervals with endpoints from the set M and with d(,) < 8, then 'L-{p(f(ai), f(,Bi)): i = 1,2, ... } ~ EO. Proof. For an arbitrary number EO > 0 find a number 8 > 0 according to the definition of the absolute continuity. If a family 'Y = {( ai, ,Bi) : i = 1,2, ... } is infinite and satisfies condition (**,8, M) then for every k = 1,2, .. .
Derivation and integration. When we pass to the limit as k
-t
00,
121
we obtain
(For a finite family, we obtain what was required directly from the definition.) The lemma is proved. • Theorem 5.1. Let a real function j be defined and be absolutely continuous on a segment [a, b]' a < b, of the real line. Then for every number c > 0, there exists a number 8 > 0 such that if M ~ [a, b] and J.L*(M) < 8, then J.L*(f(M)) :::;; c. Proof. I. For arbitrary c > 0 fix 8 according to Lemma 5.5. Let M ~ [a, b] and J.L*(M) < 8. II. Consider the particular case, where the set M is open in the segment [a, b]. By Corollary of Theorem 2.6.7 the set M may be represented as the union of an at most countable set, of its connected components. By Theorem 2.6.3 every element J of the set, is either an interval or a half interval or a segment. Let a(J) and (3(J) be its initial point and the end, s(J) = inf j([a(J), (3(J)]) and t(J) = sup j([a(J), (3(J)]) , see Figure 4.9. Remark 2.3.4 By s(J) = j(a'(J)) and t(J) t( J) = j ({3' (J)) for some cl(J), (3' (J) E [a( J), (3(J)] 1(J) (Figure 4.9). This means that under the action of j the image of the segs(J) ment I(J) with the endW(J) ~ a.'(J) . points a' (J) and (3' (J) co---------------a.(J)-;---~(J) incides with the image of the segment [a(J),{3(J)]. Figure 4.9 By the choice of 8 and by results of §1
J.L(f(M)) :::;; J.L(U{f([a(J),{3(J)]): J E ,}) = J.L(U{[s(J), t(J)]: J E ,}) :::;; 2)If(a'(J)) - j({3'(J))I: J E ,} :::;; c, because I(J) ~ [a(J), (3(J)] and
I)Ia'(J) - {3'(J)I: J E ,} :::;; 2)la(J) - {3(J)I: J E ,}
= J.L(M) < 8.
III. Let us go on to the general case. The definition of the outer measure implies the existence of an at most countable family of intervals, covering the set M with db) < o. By Theorem 1.2 the measure of the set G 1 =
U,
CHAPTER 4
122
is less than o. The measure of the (see Lemmas 1.7 and 1.1). The set definition of the induced topology. theorem is proved. Corollary. Let a real function f
set G 2 = G 1 n [a, bj is less than 0 too G 2 is open in the segment [a, bj by the By II JL(f(M)) ~ JL(f(G 2 )) ~ E. The •
be defined and be absolutely continuous on a segment [a, b], a < b, of the real line and M be a subset of the measure zero of the segment [a, bj. Then JL(f(M)) = o. • Lemma 5.6. Let a function f with values in a metric space (X,p) be defined and be continuous on a subset M of the real line R Let it be absolutely continuous on a set Mo ~ M. Then function f is absolutely continuous on the set Ml = [Moj n M. Proof. Take an arbitrary number E > O. According to the definition of the absolute continuity find 0 > O. Let a family 'Y = {(ai, ,8i) : i = 1, ... , k} satisfy condition (*,0, Md and s = db) « 0). For every n = 1,2, ... and i = 1, ... ,k fix points ar E O{ai n Mo and ,8': E 0e,8i n M o, where ~ = 2-nk(0 - s), such that intervals of the family 'Yn = {( ar,,8,:): i = 1, ... , k} are pairwise disjoint. We have
The family 'Yn satisfies condition (*,0, M). Thus z)p(f(a~),f(,8~)): i = 1, ... ,k} ~
E.
Since
- p(f(ai), f(,8i)) I ~Ip(f(a~), f(,8~)) - p(f(a~), f(,8i)) I + Ip(f(an, f(,8i)) - p(f(ai), f(,8d) I ~p(f(,8~), f(,8i)) + p(f(a~), f(ai))
Ip(f(a~), f(,8~))
(the last passage is made by Lemma 2.3.1), z)lp(f(a~), f(,8~)) - p(f(ai), f(,8i)) I : i = 1, ... , k} ~ L {p(f(a~), f(ai)): i = 1, ... ,k}
+ L{p(f(,8~),f(,8d): Terms in the right hand side tend to zero as n
L
-+ 00.
Therefore
= 1, ... ,k} {p(f(ai), f(,8i)): i = 1, ... ,k}.
L)p(f(ar), f(,8':)): i -+
i = 1, ... ,k}.
123
Derivation and integration.
When we pass here to the limit, we obtain what was required:
z)p(f(ad,f(,8i)): i = 1, ... ,k}:S; c. The lemma is proved. • Lemma 5.7. Let a function f with values in a metric space X be absolutely continuous on a set M ~ JR. and MI ~ M. Then the function f is absolutely continuous on the set MI' Proof. The proof is obvious. •
6. Derivation of absolutely continuous functions Lemma 6.1. Let a real function f be defined and be absolutely continuous on a (nonempty) compact subset M of the real line JR.. Then there are absolutely continuous and non-decreasing functions g and h defined on the set M such that f = g - h. Proof. 1. For points a < ,8 of the set M denote by V(a,,8) the upper bound of sums
where a = Xo :S; ... :S; Xk+l = ,8 and {Xl,'" , xd ~ M. For now we do not assert that this upper bound is finite. II. Let a :S; ,8 :S; , be points of the set M. Consider an arbitrary sum
where a = xo:S; ... :S; Xk+l =, and {xI, ... ,xd such that X j :S; ,8 :S; X j+l' Then
s:S; (l)lf(xi+l) - f(xi)l: i = O, ... ,j - I}
+ (If(xj+d - f(,8)1 + 2: {If(XHI) :S; V(a,,B) + V(,8,,).
~
M. Fixj
+ If(,8)
= O, ... ,k
- f(xj)l)
f(xi)l: i = j
+ 1, ... , k})
When we pass to the upper bound in the left hand side of this inequality we obtain (6.1)
V(a,,) :S; V(a,,B)
On the other hand, for every c >
V(,8,,) we can fix points a
= Xo
+ V(,8, ,).
°
by the definition of V(a,,8) and
~ ... :S; Xj+l
= ,8
~ Xj+2 ~ ... ~ Xk+l
of the set M such that
L {If(xHd L {If(xHd -
f(xi)l: i = 0, ... ,j} ~ V(a,,8) - c,
!(xi)l: i = j + 1, ... , k} ~ V(,8,,) - c.
=,
CHAPTER 4
124 Therefore
By the definition of V(a,,)
V(a,,)
~
V(a,.8)
+ V(.8,,)
- 2£.
In view of the arbitrariness in the choice of £ > 0 (6.2)
V(a,,)
~
V(a,.B)
+ V(.8, ,).
When we compare (6.1) and (6.2), we obtain
V(a,,) = V(a,.8)
+ V(.8, ,).
III. For an arbitrary number £ > 0 take a number Ii > 0 according to the condition which occurs in the definition of the absolute continuity. Show that as soon as a family, = {( ai, .8i): i = 1, ... , k} satisfies the condition (*, Ii, M) of §5, then
Assume the opposite, i.e., assume the existence of a family , = {( ai,.8d: i = 1, ... ,k} satisfying condition (*, Ii, M) such that the number 'fJ = L {V(ai' .8i) : i = 1, ... , k} - c is positive. By the definition of V(ai' .8i) for every i = 1, ... ,k we can choose points
of the set M in a manner that
The family U{{(Xi,/,Xi,l+d: (*, Ii, M) and the estimate
1 = O, ... ,j(i)}:
~ {If(Xi,I+l) - f(Xi,/)I: 1 = 0, ... ,j(i), i
i = 1, ... ,k} satisfies
=
1, ... , k} > c.
This contradicts the choice of Ii. Thus our assumption is false. This gives what was required. IV. Let a = inf M. For t E M put g(t) = V(a, t). It follows from II and from the nonnegativity of values of V(a, .8), that the function 9 is non-decreasing. Show that for every t E M the value g(t) is finite. The set )
Derivation and integration.
125
Mo = {t: t E M, g(t) < oo} is nonempty because it contains the point a: g(a) = O. Let to = sup Mo.
By virtue of the monotonicity of the function g we have Mn[a, to) ~ Mo. For c = 1 find a number 6 > 0 according to III. Next, fix an arbitrary point s E (to-~, to+~)nMo. For every point t E (s, to+~)nM (and, in particular, for t = to) g(t) = g(s) + V(s, t) ~ g(s) + 1 < 00. Therefore t E Mo. The set Mo is open in the subspace M. If to =1= sup M, then the set M \ Mo is nonempty. In this case by virtue of the openness of the set Mo the point tl = inf(M \ Mo) is different from the point to and (to, td n M = 0. The last equality implies that V(to, td = If(td - f(to)l. So
Hence tl E Mo. This contradicts the definition of the point t 1 . Thus our assumption is false and to = sup M. This means the finiteness of values of the function g. V. The absolute continuity of the function g follows from II and III. VI. The absolute continuity of the functions f and g implies the absolute continuity of the function h = g - f. If points a < (3 belong to the set M, then the definition V( a, (3) implies immediately that If((3) - f(a)1 ~ V(a, (3).
Therefore h((3) - h(a) = (g((3) - g(a)) - (f((3) - f(a)) ~
V(a, (3) - If((3) - f(a)1 ~ 0,
i.e., the function h is non-decreasing. The lemma is proved. • Theorem 6.1. Let a function f with values in the space IR n be defined and be absolutely continuous on a compact set M ~ R Then almost everywhere on the set M the function f has the derivative.
Proof. Lemma 5.2 reduces the proof to the scalar case. Here the assertion follows from Lemmas 6.1 and 4.2. The theorem is proved. • 7. Lebesgue integral Lemma 7.1. Let a real function f be defined and be continuous on a segment [a, b], a < b, of the real line. Let it be absolutely continuous on a closed set M ~ [a, bj and at almost every point of the set M its derivative with respect to the set M be positive. Let the function f be increasing on every component of the set [a, bj \ M. Then the function f is non-decreasing. Proof. If a point x E M \ {b} is not the left endpoint of a component of the complement to the set M (but almost every point of the set M satisfies
126
CHAPTER 4
this condition), the derivative Jk(x) exists and is positive, then there exists a point t E M n (x, b) such that
J(t) - J(x) > t-x
:........:...-'-----=--..:........:...
o.
Hence J(t) - J(x) > 0, i.e., the point x is a right increasing point of the function J. Every point of the complement to the set M (except the point b, if it belongs to the complement) is also a right increasing point. This follows immediately from the hypotheses of the lemma. Thus the set 8 of right increasing points of the function J has the full measure on the segment [a, bj. By Lemma 3.1 the set 8 is open. For endpoints Q < f3 of every its component we have J(Q) ~ J(f3). Enumerate elements of the set 'Y of connected components of the set 8 and for k = 1,2, ... denote by 'Yk the set of components with numbers 1, ... ,k. Elements of the set 'Yk are pairwise disjoint. Therefore their positions in the segment [a, bj are ordered: if we take two its different elements then one of them lies entirely on the left of the another element. Enumerate anew elements of the set 'Yk in order to guarantee that every element with smaller number be placed in the left of every element with larger number: 'Yk = {II, ... ,ld. Introduce the numbers
+ ... + (J(f3~) - J(Q~)), J(f3f)) + ... + (J(Q~) - J(f3LI))'
Pk = (J(f3f) - J(Q~)) qk = (J(Q~) where Q~
< f3; are endpoints of Ii. Since J.t(U'Yk) ~ f3~ - Q~ ~ b - a b- a
f3t -
Q~
---t
= 1'(8) = J.t(U'Y) = "{J.t(l): L....J
and
I E 'Y}
= k--+oo lim J.t(U'Yk),
b - a. The inequalities
a ~ ... ~ Q~ ~ Q~ ~ f3i ~ f3~ ~ ... ~ b, imply that Q~ ---t a and f3t ---t b. Since Pk + qk = J(f3t) - J(Q~), we obtain the convergence Pk + qk ---t J(b) - J(a) as k ---t 00. The endpoints of the intervals (f3f, Q~), ... ,(f3LI' Q~) belong to the set M. Their total length (i.e., by virtue of Theorem 1.1, the measure of their union) is ~ (b - a) - J.t(u'Yd. Since the set 8 has the full measure on the segment [a, b] the total length tends to zero as k ---t 00. Now the absolute continuity of the function J on the set M implies that
Iqkl ~ If(Q~) - f(f3f)1
+ ... + If(Q~)
as k ---t 00. On the other hand, Pk f(b) - f(a) ~ 0, i.e., f(b) ~ f(a).
~
o.
- f(f3:-I)1
---t
0
When we compare, we obtain
Derivation and integration.
127
If we take two arbitrary points 8 < t of the segment [a, b]' then we see that the hypotheses of the lemma hold for the segment [8, t], the function fl[s,t] and the set M n [8, tj. By the just proved assertion f(t) ~ f(8). The lemma is proved. • We can pass immediately to a more precise result. Lemma 7.2. Let a real function f be defined and be continuous on the segment [a, b], a < b, of the real line, be absolutely continuous on the closed set M ~ [a, bj. Let at almost every point of the set M its derivative with respect to the set M be non-negative. Let the function f be non-decreasing on every component of the set [a, b] \ M. Then the function f is non-decreasing. Proof. For every e > 0 the function f(x) + eX satisfies the hypotheses of the previous lemma. Apply it. When we pass to the limit as c ~ 0 we obtain what was required. The lemma is proved. • Corollary. Let a real function be defined and be absolutely continuous on the segment [a, bj. Let at almost every point of the segment [a, bj its derivative be non-negative. Then the function f is non-decreasing. • The following version of Lemma 7.2 will be helpful too. Proposition 7.1. Let a real function f be defined and be absolutely continuous on a compact subset M ~ [a, bj of the real line. Let at almost every point of the set M its derivative with respect to the set M be non-negative. Let f(a) ~ f(fJ) for every component (a, f3) of the set [inf M, sup Mj \ M. Then the function f is non-decreasing on the set M. Proof. Extend the function f on the segment [inf M, sup Mj and define it at a point t of a component (a, fJ) of the set [inf M, sup Mj \ M by the formula
f(t)
=
t-a fJ-t fJ _ a f (f3) - fJ _ af(a).
The so extended function f satisfies the hypotheses of Lemma 7.2 (see • Lemma 3.3). This gives what was required. Lemma 7.3. Let a function f with values in the Euclidean space ~n be defined and be absolutely continuous on a segment [a, b], a < b, of the real line. Let at almost every point of this segment its derivative be equal to zero. Then the function f is constant . Proof. In the scalar case Corollary of Lemma 7.2 implies that both functions f and - f are non-decreasing. This may be only when the function f is constant. In the vector case the hypotheses of the lemma imply that almost everywhere derivatives of the coordinate functions are equal to zero. The previous remark implies that these functions are constant. Thus the function f is constant too. The lemma is proved. •
128
CHAPTER 4
Example 3.2 ('Cantor stairs') shows that in Lemma 7.3 we cannot omit the requirement of the absolute continuity of the function f without its compensation by any other restrictions. Let a function f be defined almost everywhere on the segment [a, b]' a < b, of the real line and take values in the space IRn. The function f is called Lebesgue integrable if there exists an absolutely continuous function F : [a, b] - t IRn such that F'(t) = f(t) for almost every t E [a, b]. The function F is called a primitive (or an indefinite integraQ of the function f. As in the case of the Riemmann integral (Le., of the integral of a continuous function whose theory is usually accounted at first steps of the course of Mathematical Analysis) the primitive is defined up to an additive constant: if F and G are two primitives of a function f on [a, b], then for almost all x E [a, b] we have (F(x) - G(x))' = f(x) - f(x) = O. By Lemmas 7.3 the difference F - G is constant. On the other hand, if F is a primitive of f on [a,b] and c E IRn , then (F(x) + c)' = F'(x) at every x E [a,b]' at which the derivative F'(x) exists. Thus F + c is also a primitive to f. By virtue of the last remark for every points s, t of the segment [a, b] the difference F(t) - F(8) does not depend on the choice of the primitive F. This vector (number in the scalar case) is called a (definite) integral of the function f from 8 to t (or over the segment [8, t], when 8 ~ t). It is denoted by t
J
f(x)dx.
8
Evidently t
J
J
8
t
f(x)dx = -
8
f(x)dx.
The derivation is made in each coordinate independently. Thus the finding of a primitive, which is a vector function, and calculation of a definite integral, which is a vector, may be also made in each coordinate independently. If a function f is integrable on a segment [a, b], then it is integrable on every smaller segment I, i.e., the function fll is integrable. If F be a primitive of the function f then the function FII is a primitive of the function fl/' On the other hand, if 8 E [a, bJ and a function f is integrable on the segments [a,8J and [8, bJ then it is integrable on the segment [a, bJ. Proof of this assertion may be proposed to the reader as an easy exercise. J
129
Derivation and integration.
f
If s, t and u are three arbitrary points of the segment [a, b], a function is defined and Lebesgue integrable on [a, bj, then u
Jf(x)dx = F(u) - F(s) 8
= (F(t) - F(s)) t
=
+ (F(u)
- F(t»
u
J
f(x)dx
+
J
f(x)dx,
t
8
where F is an arbitrary primitive of the function f. Let F be a primitive of a function f on a segment [a, bj and s E [a, bj. For t E [a, bj put t
G(t) =
J
f(x)dx = F(t) - F(s).
s
The function G differs from the primitive F by a constant -F(s). Because of above remarks it is a primitive for the function f on the segment [a, bj (moreover G(s) = 0), i.e., a definite integral when we consider it as a function on its upper limit of integration (i.e., of the number t in the above formula), is a primitive. Let a function f be defined on a interval or on a half interval J. The function f a is called locally integrable if it is integrable on every segment lying in J. For every s E J the formula t
G(t) =
Jf(x)dx s
defines a primitive of the function f on J (i.e., for every segment I ~ J the function Gil is a primitive of the function fII). For a function f being locally integrable on a half interval [a, (0) the limit s
8li.~
Jf(x)dx a
(under the natural assumption that it exists) is denoted by
J 00
f(x)dx.
a
130
CHAPTER 4
Analogously we define the value of b
J
f(x) dx
-00
Put also
a
J
J
00
f(x)dx =
-00
J 00
f(x)dx
+
f(x)dx,
a
-00
where a E (-00,00) (evidently the sum in the right hand side does not depend on the choice of a). In these formulae we can use every other suitable symbol instead of x to denote of the independent variable. Let M be a subset of the set X. The function
X(X)={~
if x E M, if x E X \ M.
is called a characteristic function of the set M (in X). Let M be a (measurable) subset of the real line. We regard the symbol
J
f(x)dx
M
as equipotent to the symbol 00
J
f(x)X(x)dx,
-00
where X is a characteristic function of the set M (in 1R). Example 7.1. Every continuous function is Lebesgue integrable (see Example 5.1). Its Lebesgue integral coincides with its Riemann integral. Example 7.2. Let -00 < a < (3 < 00. The characteristic function of a segment, of an interval or of a half interval with endpoints a, (3 is Lebesgue integrable: the function
F(x) = {x ~ a (3-a
if x < a, if a ~ x ~ (3, if x > (3
is its primitive (the absolute continuity of the function F may be easy checked directly).
131
Derivation and integration.
Theorem 7.1. Let functions f and g with values in the space ~n be defined almost everywhere on a segment [a, b], a < b, of the real line. Let them be Lebesgue integrable. Let>. be an arbitrary real number. Then the functions f + g and >.g (defined almost everywhere on the segment [a, bj) are Lebesgue integrable and b
b
+ g(x))dx
jU(x)
=
a
b
j f(x)dx a
+j
g(x)dx,
a
b
b
j >.f(x)dx = >. j f(x)dx. a
a
Proof. The integrability follows from Lemma 5.4. The equality may be • easily checked directly. The theorem is proved. Lemma 7.4. Let a non-negative function f be defined (almost everywhere) and be integrable on a segment [a, b], a < b, of the real line. Then b
j f(x)dx
~ O.
a
Proof. By Corollary of Lemma 7.2 (each) primitive of the function f is a non-decreasing function, that immediately gives what was required. The lemma is proved. • Theorem 7.2. Let non-negative real functions
