 .//ce:title  On a theorem of DieudonnÃ©  .//dc:description   .//xocs:srctitle  Advances in Mathematics  .//dc:title  On a theorem of DieudonnÃ©  BODY  ADVANCES iN MATHEMATIcs 36, 165168 (1980) On a Theorem of DieudonM JAMES K. BROOKS Department of Mathematics, University of Florida, Gainesville, Florida 32601 A beautiful theorem due to Dieudonn6 [2] states that if a sequence of regular measures, defined on the Borel subsets of a compact metric space, converges on all open sets, then it converges on every Borel set. In this paper we present an extension of this theorem and a direct proof which uses an interesting lemma recently established (independently) by R. V. Chacon [1] and H. P. Rosenthal [4]. The lemma and its proof are given here for completeness. The proof is due to Chacon. In the sequel, S will be a Hausdorff topological space. The Borel field is the afield generated by the open subsets of S. Regularity of a measure is defined in terms of open and closed sets as in [3, 111.5.11]. A sequence of measures (scalar or vector) (p,,) is uniformly bounded if supâ€ž,E I pn(E)I < 00. The total variation measure of a measure p is denoted by I p 1. LEMMA. Let (pn) be a uniformly bounded sequence of measures defined on the measurable space (T, E), each absolutely continuous with respect to A, where A is a finite positive measure. Then there exists a subsequence (pâ€ž{{{{) of (pn) such that for every e > 0 there is a set AE e E such that A(A,) < e and (pn) is uniformly absolutely continuous with respect to A (u.a.c. A) on T  AE . Proof. If we deny the conclusion, then by a diagonal argument, one can show that there exists an e > 0 and a subsequence (pâ€ž{{{{) such that whenever A e E and A(A) < e, then no further subsequence of (pn{{{{) is u.a.c. A on T  A. Consider the set A1 of all c > 0 such that for any rl > 0 there exists a set Eâ€ž such that A(Eâ€ž) <,q and I pn,(E,)I > c for infinitely many i. Note A1 is bounded and nonempty. Let c,. = 1/2 sup A,. Choose a set Cl with measure less than e/22 such that I pl,i I(Ci) > c1 L, where (p,,i) is a subsequence of (pn{{{{). Let A2 be the set of all c > 0 such that for any 77 > 0 there exists a set Eâ€ž such that Eâ€ž C T  Cl , A(Eâ€ž) < 71 and I pl,i I(E,,) > c for infinitely many i. Choose a set C2 C T  Cl , A(C2) < e/2g, a subsequence (p's.i) of (fi) such that I p2,i I(C2) > c2 for all i, where c2 = 1/2 sup A,. Continue in this fashion and obtain (pk.i), Ck , ck such that the Ck are disjoint, A(Ck) < E/2kt1, I pk.i I (Ck) > ck, i = l, 2,... . If the ck were bounded away from zero, this would contradict the boundedness assumption on the (pn). On the other hand, if ck > 0, the diagonal sequence (pk.k) would be u.a.c. A on T  Uk Ck , a contradiction since A(Uk Ck) < e. 165 00018708/80/05016504$05.00/0 Copyright Â® 1980 by Academic Press, rnc. All rights of reproduction in any form reserved. 166 JAMES K. BROOKS THEOREM. Let (p,,) be a sequence of regular measures defined on the Borel subsets of a topological space S. Assume either (a) S is compact or (b) S is normal and (/â€ž,,) is uniformly bounded. If lim pn(U) exists for every open set U, then lim p,,(B) exists for every Borel set B. The above conclusions also hold when the pn are Banach valued regular measures. Proof. We shall treat the scalar case first. Assume S is compact and (p,,(U)) is a bounded sequence for every open set U. We assert (pn) is uniformly bounded. Since S is compact, it suffices to prove that every point in S belongs to an open set on which (p,,) is uniformly bounded. If we deny this, there exists a y e S such that sup,, I pn I (U) = oo, whenever U is an open set containing y. Let M = sup,, I p,,(y)I; note M is finite. Let U be any open set containing y; fix an integer N > 1. There exists a Borel set A C U and an integer nl arbitrarily large so that lp,,,(A) I > N }}}} 2M + 1. By regularity there is a closed set Co C A such that I pnl(Co) I > N }}}} 2M }}}} 1. Let Cl = Co V (y). Then I 1_,.l(C1)I > N }}}} M {{{{ 1. Again by regularity there is an open set W such that U D W D Cl and I p,,1(W)I > N }}}} M }}}} 1. Choose an open set Q such that W D Q D (y) and J p,,2 I (Q  (y)) < 1/2. Let V be an open set such that y e V C V C Q (17 is the closure of V). Let Dl = W  V. Note that Dr C U, Dl n V = 0, and I p,nl(D1)I > N. Since y E V and (p,,) is unbounded on V, by the above method obtain for any integer N', an integer n2 > nl , open sets D21 Vl contained in V such that D2 n Vl = o1, y e Vl and I p..(D2) I > N'. Note that Dl n D2 = o. Thus we may inductively define a sequence of disjoint open sets Dx and a sequence (nk) such that for every k, I pnk(D,)I > k. Define xki = klp,,k(Di). Let A be a subset of the positive integers. Since I Zicn xki I kla, , where as = supra I pn(Ui.a DJ I < oo, it follows that limkYi,a xki = 0. By the Schur theorem [3, IV. 8.14], "MI L I xki I = 0, a contradiction since I xxk I > 1 for every k. Assume now that S is normal, (g,,) is uniformly bounded, and lim p,,(U) exists for open sets U. If limp,,(B) does not exist for a Borel set B, there exists an e > 0 and (ni) such that I pn;+1(B)  pn,(B)I > e for each i. Let of  p,,.+l  p,,, . The sequence (Qi) of uniformly bounded regular measures converges to zero on open sets and for each i, I ai(B)I > e. Define the regular measure A = 2i Z I of 1. Since ai < A, by the above Lemma there exists a (a,,;) such that for every 77 > 0 there is a set Aâ€ž such that A(Aâ€ž) < 71 and (a,,,) are u.a.c. A on S  Aâ€ž . By the regularity of A we may assume that Aâ€ž is open. For notational convenience, assume ni = i. Let 77,,77,,% > 0 be given. There exists a 8 > 0 such that A(A) < 8 implies I al I(A) < 77,. Choose Al open so that A(A,) < 8 and (ai) are u.a.c. A on S  A,. Assert (*) There exists an nz such that I a,,(B n A,) I > e  37)2 for n > nl. To see this, obtain a 81 > 0 such that for each n, I a,, I(Q) < rte whenever A(Q) < 81 and Q C S  A,. Since A is regular, obtain an open set U D B ON A THEOREM OF DIEUDONNA 167 such that A(U  B) < 81. Note that I uâ€ž(B  Al) I < I aâ€ž(U  Al) + I oâ€ž I ((U  B)  Al) < I aâ€ž(U)  aâ€ž(U n Al) I }}}} % < 3"12 when n > nl for some nl since an(U) and un(U n A,) tend toward zero. Now let C be any closed subset of A,. Note al(B  A,) I > e 77, ; hence I al(B  C)I > I al(B  A1)I  I al I(A,) > e  2711. Let Ul be an open set containing B such that I al J (U1  B) < 771. Then I al(Ul  C) I > e  3771. Hence (* *) There exists an open set U, such that for every closed set C C A,, Ial(ULC)I >e3711. We now assert (* * *) There exists a closed set C C A1 and an integer k2 > ni such that I an(B n C) I > e  371,  2"la for all n > k,. If we deny the above assertion, then in view of (*), I aâ€ž(B n (Al  e)) I > 27]3 for infinitely many n. Pick a closed subset Cl of A1 and an integer tl (as large as we please) so that I a,,(B n (Al  Cl)) I > 277,. By regularity, find a closed set Ci such that Cl C C1 C A1 and I aq I (Al  Cl*) < 77,. By normality there exists an open set Vl such that Ci C Vl C Vl C A,. Note that I atl(B n (Vl  C1)) I > I at,(B n (Al  Ci)) I  I at, I(B n (Al  V1)) > 77.. Let L,. be an open subset of Vl containing B n (V1  Cl) such that if Dl =LI n (V1  Cl), then I at,(DI) I >71,. Thus Dl is open and Dl C Vl. Let CZ = V,. Choose t2 > tl such that I at,(B n (Al  C2)) J > 2713. As before, obtain a closed set C$ and open set V2 such that CZ C CZ C V2 C V2 C A1 satisfying I at, I(Al  Cz) < 77s and I %(B n (V2  C2))I > 77,. There exists an open subset of V2, say L, such that L2 D B n (V2  C2 ), and if D2 =L2 n (VI  C2), then I t9(D2)1 >'73. Dl and D2 are disjoint. In general, one can obtain tl < t2 < ... < t,, disjoint open sets Dl ,..., D,, and open sets V1,..., Vk such that D= C Vi C Vi C A,, i = 1,..., k, and I at,(Dj)I >713 . Let Ck+1 = Vk ; choose tk+1 > tk and construct Dk+1 = Lk+1 n (Vk+l  Ck+1) for suitable Lk+1 I Vk+1, where I at k+l(Dk+1) I > CJs . Thus we can obtain a sequence of disjoint open sets Dk and a sequence (tk) such that I atk(Dk) I > "la . Let xki = atk(D,). Since (atk) converges to zero on open sets, we have limk Eica xk= = 0, for any d, which is a contradiction by the Schur theorem since I xkk I > 773 for all k. This establishes the assertion (***). Thus there is a closed set Cl contained in A1 and an integer k2> nl such that n > k2 implies I aâ€ž(B n Cl) I > E3'92 2%. Choose an open set Wl such that Cl C Wl C Wl C A1 . By (* *), if Zl = Ul  Wl , then I al(Zl) I > .E 377, ; hence Z, is open and Z, C S  Wl . Note that 7h , 712 , '7s are independent; let 377, = e/4; 3772+ 277, = e/4; set kl = 1. Repeat this construction in the space Wl with measures (an)n>k, , where I an(Bl)I > c E/4 for n > k2 and Bl = B n Cl . Given X111&1& > 0, find an open set Q such that A(Q) < 8, where 8 > 0 is a number for which I ak, I (A) < "h whenever A(A) < 8, and (aâ€ž) is u.a.c. A on S  Q. Let A2 =W1 n Q. Find an integer n2 > k2 such that, I a,,(B, n A2)I > e  e/4  3"12 for n > n2*. Obtain an 168 JAMES K. BROOKS open set U2 C Wl such that U2 D B,, and I aks(U2 11 (WI  C)) I > e  E/4  3~1 whenever C is a closed subset of A,. Find an integer k3 > n2 and a closed set C2 C A2 such that I un(B, n C2) I > E  E/4  31,  2~3 for all n > k3 . Choose an open set W2 such that C2 C W2 C W2 C A,. Let Z2 = U2 n (Wl  V2). Set 3~1 = 3~2 + 2~s = E/42. Thus I a,,(Z2)I > E  E/4  E/42, Zl n Z2 = 0 and I an(B2)I > e  e/4  e/42 for all n > k,, where B2 = Bl n C2. The new space to work in is W2. By induction obtain a sequence of disjoint open sets Zi and (ak,) such that I ak,(Zi)I > E  e/4    e/4i. Let x, = ak,(Zi). The convergence of (an) to zero on open sets implies limt Pica xti = 0, which is a contradiction in view of the Schur theorem, since I xii I > e/2 for each i. This proves the scalar case. Now assume the 1L. are Xvalued regular measures, where X is a Banach space with dual space X*. Note that if S is compact, then by the above, (x*1'n) is uniformly bounded for each x* e X*, which implies, by the BanachSteinhaus theorem, that ([kJ is uniformly bounded. Suppose S is normal, (p.â€ž) is uniformly bounded and converges on open sets. If lim p,â€ž(B) does not exist for some Borel set, then there exists an e > 0, (ni), xi E X*, I xi I < 1 such that the measures ai = xi (ant+1  ani) satisfy I ai(B)I > e for each i. (ai) is a uniformly bounded sequence of regular scalar measures converging to zero on open sets; hence by the above, ai(A) > O(A) for every Borel set A, where 0 is a measure by the Nikodym theorem. Since Oi < A (A defined as above) for each i, the VitaliHahnSaks theorem implies ai << A uniformly in i; thus ~ < A and the regularity of 0 follows from that of A. Since 0 vanishes on open sets,  0, which contradicts I ai(B)I > E, for each i. Remarks. (1) If regularity is defined in terms of open and compact sets, then (a) can be replaced by "S is locally compact" by using the one point compactification and assigning zero mass to the new point. (2) It follows from the above proof that if (pn) is a sequence of regular measures on a compact space such that (pn(U)) is a bounded sequence for each open set U (the bound depending, in general, on U), then (1"n) is uniformly bounded. Note added in proof. The author and R. V. Chacon have recently extended the Lemma presented in this paper in several directions (to appear in this journal). By using a finitely additive version of the Lemma, it can be shown that a sequence of finitely additive regular measures, which converges on open sets converges on every Borel set. The proof is essentially the one presented here. REFERENCES 1. R. V. CHACON, unpublished notes, 1977. 2. J. DIEUDONNt, Sur la convergence des suites de mesures de Radon, An. Acad. Brasil. Ci. 23 (1951), 2138, 277282. 3. N. DUNFORD AND J. T. SCHWARTZ, "Linear Operators. 1: General Theory," Interscience, New York, 1958. 4. H. P. ROSENTHAL, Subsequences of Li with the strong Schur property, unpublished. t k+l(Dk+1) I > CJs . Thus we can obtain a sequence of disjoint open sets Dk and a sequence (tk) such that I atk(Dk) I > "la . Let xki = atk(D,). Since (atk) converges to zero on open sets, we have limk Eica xk= = 0, for any d, which is a contradiction by the Schur theorem since I xkk I > 773 for all k. This establishes the assertion (***). Thus there is a closed set Cl contained in
