THE UNIVERSITY OF MI CHI GAN COLLEGE OF LITERATURE, SCIENCE, AND THE ARTS Department of Mathematics Technical Report No. 22 CONVEXITY OF THE RANGE OF CERTAIN INTEGRALS Lamberto Cesari -z ORA Project 024160 submitted for: UNITED STATES AIR FORCE AIR FORCE OFFICE OF SCIENTIFIC RESEARCH GRANT NO. AFOSR-69-1662 ARLINGTON, VIRGINIA administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR May 1971

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ADDENDUM III. CONVEXITY OF THE RANGE OF CERTAIN INTEGRALS Lamberto Cesari In this appendix we consider any vector function f(t) = (fl,...,fn) whose components are L-integrable in [a, b], and prove that the set function K(E) = fEf(E)dt has for range a convex closed set when E describes all measurable subsets of [a, b]. This result, which is proved here rather elementarily through a set of lemmas, is actually a particular case of an analogous one concerning nonatomic vector valued measure functions and due to A. Lyapunov [76]. III 1. SOME PRELIMINARY LEMMAS If [a, b] is any given interval of length X = b - a, and a, 0 < a 1, any number, then the point t = a + a(b - a) = a + al divides [a, b] into two parts of measures ca and (1 - a)O. If we divide [a, b] into two equal parts, and we divide each part as above, the corresponding set 2 D = [e < t < a + a~/2] U [a + ~/2 < t < a + ~/2 + a~/2] has still measure 9a, and is the union of two disjoint intervals. In general, if we divide [a, b] into 2 equal parts, and in each part we take corresponding subintervals, then the set Da= U=1 + 2 (i - 1)1 a + 2k (i - 1 + a) (III. 1.1) U i=l L. has measure, and is the union of 2 disoint intervals. Also, for has measure ai, and is the union of 2 disjoint intervals. Also, for

k k k ki 0 < Ca < 1 and the same k we have D c D and meas D - D ( - - - a nd m e~Ca at' a a (III 1.i) Given any vector function f(t) = (fl..,f ), a <t < b, whose components are L-integrable in [a, b], and any E > 0, there is an integer K such that for all k > K and C O a <a < 1, we have f f(t)dt - afb f(t)dt < E. D a a In other words, if a = (al...,a) denotes the integral of f(t) on [a, b], then the integral on Dk thought of as a function of a, 0 < a < 1, is uniformly approximated by the linear function a a, a 0 < a 1. Proof. It is not restrictive to assume a = O, b = 1. We know that there is a continuous vector function g(t), 0 < t < 1, such that f |f(t) - g(t)Idt < E/4. O Then g(t) is uniformly continuous in [0, 1], and hence there is that t, t' E [0, 1], It - t'I < 5 implies Ig(t) - g(t')| < ~/4. smallest integer with 1/2 < 6. For any k > K let g (t), 0 < t step function defined by gk(t) = g(t l) for all t i t t where ti = i/2. Then Ig(t) - gk(t)l < /4 for all 0 < t < 1. 6 > 0 such Let K be the < 1, be the i T 1h,.., Thus 1 1 1 A = of(t) - gk(t)lt < olI - gldt + /og - gkldt< /4+ /4 = E/2, and 2

f( - 1 f( t)dt f (t) dt - k dt k + Ct k kk Dk D D cc a a + f k gkdt - f kdt + af gkdt - af f(t)dt = s1 + s + s3 D Here s2 = igk(t )(a/2k) a- a ig(ti)(1/2) = 0, Il I f J -f gkldt < f f - glut < ~/2 D a Is3 I f |g - fldt < ~/2 and finally A < E/2 + 0 + e/2 = E. Statement (III 1.i) is thereby proved. Statement (III l.i) has a stronger form which of course is less easy to prove. (III l.ii) Given any vector function f(t) = (f,...,f ), a < t < b, whose components are L-integrable in [a, b], then for every ac, 0 < a 1, there is a measurable subset E ofEa, b] such that f f(t)dt = ab f(t)dt, 0 < ac 1. (III. 1.2) Fa a In other words, if a = (a,...,a ) is the integral of f on [a, b], the integral at the first number of (III 1.3) thought of as a function of a, is a linear function of oa, say a ac, 0 < a < 1. This statement is a particular case of the following one which we shall prove below. 3

(IlI l.iii) Given any vector ['unction f(t) = (fl,...,f ), a < t < b, whose components are L-integrable functions in [a,b], and any measurable subset A of [a, b], then for every, 0 < a < 1, there is a measurable subset B B cA [a, b] with J f(t)dt = a f f(t)dt, 0 <a < 1. (III. 1.3) B A Proof. The proof of (III l.iii) is made up of parts. (a). Let us prove (iii) for n = 1 and f a nonnegative scalar function. If cp(t) denotes the characteristic function of A, say cp = 1 on A and cp = 0 otherwise, then t f(t) cp (t) is L-integrable in [a, b], and hence F(t) = t f(T) cp (T)dT a a < t < b, is a continuous function taking all values from F(a) = 0 to F(b). Thus, there is some c, a < c < b, with F(c) = aF(b), and, if B = [a,c] n A, also o a f dt = f p dt = F(c) =aF(b) = a f cp dt = a f dt. Ba A Thus, (III l.iii) is proved for n = 1 and f scalar nonnegative. (b) Let us assume that we know how to determine B/2 for every A and a given vector function f = (fl,...,f ) whose components are nonnegative Lintegrable, and let us prove that we can determine all sets B, 0 <a < 1, a' and that we can determine them in such a way that a < a' implies B c B For the sake of simplicity we shall use the notation (EE) = f dt, Hi(E) = fidt i = 1,.i.,n. (III. 1.4) = j ( I I 1.4) First, for B/2 = A - B1/2 we have

~I(B/2) = p(A) - (B /2) = i(A) - (1/2)i(A) = (l/2)4(A). Then let us determine sets B1/4 c B,/4 B/ such that p^B ) (1/2) (B ^, = ( 1/2)(B2 B./4) = (I/2( B1/2), B( B/4) = ( 1/2) 3(B1/2) and then, for B,/4 = B/2 U /4 also (1/4) = (l/4)i(A), ( B3/4) = (3/4)t(A), and if B =, B = A, we have B c B c B c B c B ~~~o 1~ o ~~1/4 /2 ( /4 f / By repeating this process we obtain sets B/2r i = O 1,..,2r = 1,2,... so that [t(Bi/2r) = i/2r, and for i < j, X = i/2r, = j/2r also B B ' Now let Q be any number 0 < a < 1, and let [\ ], [\'] be sequences of numbers s s = i/2, = j/2, such that < X < << ' < a < s a,\ ' a. For s s s s+1 s+l s s s B = U B% % = n BX s s we have B c B' and X / f dt =. f dt < I f dt < / f dt < S dt = \' f f dt, A B B B, A s s where < means that such a relation holds for each component. As s + oo we obtain a = f dt = / f dt, 0 <a < 1. Ba B' This proves (b). 5

(c) Assume that (iii) has been proved for some vector function f = (fl'...,f ) whose components are nonnegative L-integrable. Let E, F be any two measurable subsets of A c [a, b]. Then for every o, 0 < Ca < 1, there is some subset C(a) of E U B with C(O) = E, C(1) = F, such that f f dt = (1 - a) f f dt + a f f dt, 0 <a < 1, C(a) E F I f dt - dt -a (; f dt + f dt c(a0) c(al' ) E-F F-E / 0 < a<, a' < 1, i = 1,...,n. (III.1.5) Indeed, let us apply (iii) to the sets E-F and F-E, and the number cx. Let B c E-F, B' c F-E be the corresponding sets and take ) ( F) U (-F-B) U B C(a) = (E n F) U (B-F-B ) U B'. U a Then 11(c(CO),(E n F) + ((E - F) - |i(B ) + ~(B') at a = ((E n F) + I(E - F) - a ((E - F) + a 4 (F - E) = i(E n F) + (1 - a)((E - F) + a 4 (F - E) = (1 - a)J.[(E n F) U (E - F)] + a 4 [(E n F) U (F - E)] = (1 - c)4(E) +a i(F). In addition, for each component fi and 0 <.X, a' < 1, we have 6

|Ii(C(a)) - i(c(a')) = I(- a+a' )(E - F) + (a - a')>(F - E) i i < la - a'I (E(E - F) + jL(F - E)), i = l,...,n. Thus (c) is proved. (d) Statement (iii) has been proved for n = 1 and f Assume that (iii) has been proved for n - 1 and vectors f components, and let us prove it for n. Let f be the (n - f (f.f n- f ) with f =f + f and let A 1 n-2- n-i n-i n-1 n functions defined by (III.1.4) with f replaced by f, fi' fi. a = 1/2 applied to f there is a subdivision of A into two E n F =, E U F = A, and (E) = (F) (/2(A) scalar nonnegative. with nonnegative 1) -vector i' i i be the set First by(iii) with parts E, F, with (III 1.6) Also, by force of (c), there are sets C(a) c E U F = A, < a < 1, with C(O) = E, C(1) = F, and (c(a)) = (1 - a)(E) + a i(F) = (1 - a)(1/2)(A) + a(l/2)4(A) = (1/2)H(A), 0 <a <1. (III 1.7) Let us prove that jn1 (C(a)) is a continuous function of a in [0, 1]. Indeed, n (C(a)) is a scalar, namely the integral of f > 0 over C(a), an- - and 7

pnl (C(a)) - n- (C(a'))l = Ipn [C(a) - C(a')] - p. [C(a') - C(a)]I n-I n-i n-i n-i < Ipnl [c(a) - c(W')] + n1i [c(') - c(a)]I < lIn-l [c(a) - c(a )] + n-1 [C(a') - C(a)] n- - n- n < - '] (n (E - F) ( E)) < 21a - a' I f(fr fn)dt. This proves that. (C(a)) is a continuous function of a for 0 <a < 1. On n-1 - the other hand, n-i (C(O)) = np (E), np. (C(l)) = n (F). Since E and F are complementary in A then 1(E) (1/2)L p (A) according as n-1 7 n-1 p _(F) > (1/2)p l(A). Thus, as a describes [O, 11, p.(C(a)) describes an n-i < n-i interval which contains (1/2)p (A). We conclude that there is some a, n-1 o <a < 1, such that p. (C(a)) = (1/2)p (A). For this particular value of n-i n-1 a, we have from (III 1.7) f f.dt = (1/2) f f.dt, i =,...n-2 C(a) 1 A f (f + f )dt. (1/2) (f + f )dt, C(a) n-1 n A n-l n f f dt =(1/2) J f dt, C(a) n- A and hence, by difference, also f f dt = (1/2) f f dt, C(a) n A n or f f dt = (1/2) f f dt. (III 1.8) C(a) A 8

We have proved that for the n-vector f = (fl,...,f ) we can determine a subset B1/2 = C(a) c A satisying (III 1.6), where A is any measurable subset of [a, b]. Thus, by (b), we can determine analogous sets B for all a, 0 <a < 1, and (III l.iii) is proved for vector valued functions with nonnegative components. (e) We have now to prove (III l.iii) for vector functions f = (fl'...,f ) with L-integrable components of arbitrary signs. For every i = l,...,n, and j = 1,2, we consider the sets Ail where f. > 0 and Ai2 where f < 0. We divide [a, b] into 2 disjoint measurable sets A = A. n A2 n...n Anj, where r denotes any one of the 2n systems r j1 2 n (jlj2',...jn) of indices 1 and 2. On each set A the components f. have constant signs, and there are, therefore, sets B c A with f f dt = ra r B Qa f dt, 0 < a < 1. The sets B = U B then satisfy the requirements of A - rCr r (III l.iii). Statement (III l.iii) is thereby proved. (III l.iv) Given any vector function f(t) = (fl,...,fn), a < t < b, whose components are L-integrable in [a, b], and any two fixed measurable sets E, F c [a, b], then for every C, 0 < a< 1, there is some set C(a) c E U F, with C(O) = E, C(l) = F, and f f dt = (1 - a) f f dt + I f dt. C(a) E F This statement is a consequence of parts (b) and (c) of the proof of (III 1.iii). 9

III 2. THE MAIN STATEMENTS (III 2.i) Given any vector function f(t) = (fl,....,f), a < t < b, whose comn ponents are L-integrable, and any measurable subset A of [a,b], then [(E) = f f(t)dt (III 2.1) E describes a convex set H as E describes all possible measurable subsets E of A (in other words, the range of >(E) is convex). Proof. If [1' p2 e H, then there are measurable sets E1, E2 in A such that Hi = i(E ) = f f dt, i = 1,2. Among all measurable subsets of i i E. A there certainly are the sets C(a), 0 <a < 1, defined in (III l.iv). Then ^i(C(a)) = (1 - a) fEf dt +a fFf dt = (1 - a)41 + 12 that is, all points of the segment (1 - a)> + a2' 0 < a < 1, belong to H, and H is proved to be convex. (III 2.ii) Given any two vector functions f(t) = (fl,...,f ), g(t) = n (gly...,g ), a < t < b, whose components are L-integrable, let E denote any measurable subset of [a, b] and hE(t), a < t < b, the function hE(t) = f(t) for t E E, hE(t) = g(t) for t e F = [a, b] -E. Then (E) (t)dt (III 2.2) 1(E) = fa hE(t)dt (III 2.2) a E describes a convex subset H of the space E as E describes all measurable n subsets of [a, b]. 10

Proof. For every E as above and F = [a,b] -E, we have b b 1(E) = fa hEdt Ef dt + Fg dt = E(f - g)dt + fa g dt If p, is the fixed value of the last integral, and we apply (III 2.i) to the function f-g, we see that the set H of (III 2.ii) is simply a translation of the convex set H of (III 2.i) relative to f-g..(III 2.iii) The set H of statement (III 2.i) is closed. Proof of (III 2. iii) for n = 1. We have [L(E) = f f dt where f is a E scalar. If f > 0 the statement is trivial since the values taken by B(E) fill the closed segment [0, I(A)]. Otherwise, let A, A be the subsets of all + + - t e A where f > O, or f < 0, and thenA n A =, A U A =A. For every + - _ + - set E cA, let E = E A, E = E A, and then i(E) = >(E ) + (E ). Then + + ((E) takes on its maximum value L = i(A ) > 0 for E = A, and its minimum value [i = (A ) < 0 for E = A, and the values taken by [((E) fill the segment [IE' t ]. We shall prove (III 2.iii) for n > 1 below. (III 2.iv) If H is the convex set of (III 2.i) and i: p * x - c = 0 any supporting plane for H with n n cl H # 0, then I n H # 0, that is, there is some c H with p ~ - c = 0, and some measurable set E c Awith f = E f dt. o E Proof. We may assume p ~ x - c > 0 for all x c H, and hence also for all x e cl H. Thus, (cl H) n i # 0 implies c = Inf (p * x) where Inf is taken for all x e H, that is, c = Inf p * fEf dt = Inf /E(p ~ f)dt, where Inf is taken for all measurable subsets E of A. On the other hand v(E) = fE(p f f) dt is our usual function i relative to the scalar function 11

g(t) = p * f(t), t e A. By (III 2.iii), v(E) takes on its maximum and minimum values. Thus, there is some measurable set E c A with C = v(E ) = JE (P f)dt = p fE f dt, that is, p - c = 0 for = E f dt. Proof of (III 2.iii) for n > 1. We have proved (III 2.iii) for n = 1. Let us assume that (III 2.iii) has been proved for 1,2,...,n- l, and let us prove it for n. Let f(t) =(fl,...,f), t A, A measurable, and let Hbe the range of the function p(E) = J f dt as E describes all measurable subset E of A. By E (III 2.i) we know that His convex, andwe have to prove that H is closed. Suppose that this is not true, so that cl H - H, and let 5 be apoint ~ e cl H - H. Then 6 6 bd H, and by (Vol. I, App. C2) there is a supporting hyperplane IT: p x - c = O through,, thus p x - c > 0 for all x c H, and p ~ = c. By (III 2.iv) there is a point ~' e H on I, that is, p ~* ' - c = 0, and since E' e H, there is a measurable subset E of A with f' = E f dt. Thus 0 O c = p ~ ' = p E f dt= E (p f)dt. Then for every measurable set E cA o o we have v(E) = p * t(E) - c = p * fEf dt - p E (p f)dt o = -E P f dt + E E(-P. f)dt. 0 0 If g(t), t e A denotes the scalar g(t) = p * f(t) for t E A - E, g(t) = - p ~ f(t) for t e E, then v(E) = f g(t)dt > 0 (E-Eo) + (Eo-E) for every measurable subset E cA. This implies that g(t) > almost everyfor every measurable subset E c A. This implies that g(t) > 0 almost every 12

where in A. Let A [A, 6 > O], be the set of all t e A with g(t) < 0 [g(t) < 5]. Then all sets A, A5 are measurable, A c AA c A, A - A + 0, meas 6 & o 6 o o (A - A ) O0 as 6 0 + 0. Let.', pi" be the functions i'(E) = 4(E - A ) = 6& 0 0 f f dt, "(E) = [(E n A ) = J f dt, both defined for all measurable subsets E-Ao EFlA E of A, and let H', H" be the ranges of i' and t"t. Since p - J"(E) p ~ i(E n Ao) = J (p ~ f)dt = 0, we see that the range H" of Ot" is contained AnAo in the hyperplane H: p * x = 0. By a change of coordinates we could, therefore, represent H" by means of an (n - 1)-vector function, that is, as the range of the values f f(h)dt, h = (h...hn- ), for an L-integrable (n - 1)EnAo -vector function g. By the induction argument, H" is therefore a convex closed subset. Let us prove that (*)v(E ) + 0 implies p'(E ) + 0 for any sequence [E ] of measurable subsets of A. Let E' = E - A, and let us assume that this s s o statement is not true. Then there is some m > 0 and a sequence, say still [E ], with Ij'(E )I > m > 0, and hence also I '(E')I = I|(E- A )l = s s s s O I '(E ) > m > 0. Since fis L-integrable inA, there is some a > 0 such that.fE fldt < m/2 on every measurable subset E ofA of measure < a. This implies that meas El > a for every s, since otherwise I g(E's)I = I|E f dt| < JE Ifldt < s '-E 'E Ifd K' s. s m/2, a contradiction. Finally, if we take 5 > 0 so that meas (A - A ) < 0/2, we see that the set E" = E' - Ab = E' - (A - A ) has measure > a/2, and hence s s s 6 o -0 p * f > 6 everywhere in E", and v(E ) = v(E') > v(E") > 5 a/2, while v(E ) + O0 a contradiction. We have proved (*). Since g e bd H, there is a sequence [s ] with s e H, -> +, and hence 13

a sequence of sets B c A with p * i(B ) - = p * - c - p P'(B ) - 0 as s oo, or -(B s s u(B n B ) = j(B ) -,u(B - A ) s o s s o Since H" is closed we have e c e = P"i(E ) =,(E1 n A ). This 1 1 o E E1 n A C A C A with i(E) closed, and thus, by induction li(B ) = S, s = 1,2,.... Then v(B ) = * % - c = 0. By force of (*) we have then A ) + 0. On the other hand i"(B ) = o s -+ as s - oo. This proves that e e cl H". H", that is, there is some set E c A with proves that there is some measurable subset = i, that is, e H. We have proved that H is argument, (III 2.iii) is proved for every n. Bibliograhical notes. Beside the original paper by A. Lyapunov [76] on the convexity of the range of vector valued nonatomic measure functions, we mention here the work of D. Blackwell [12], of J.F.C. Kingman and A. P. Robertson [64], and W. de Wilde [119]. The paper by P. Halmos [Bull. Am. Math. Soc. 54, 1948, 416-421] contains an error which was corrected in [Bull. Am. Math. Soc. 1949].

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