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On the Representation of Sets in Finite Measure Spaces
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.
200, 506]510 Ž1996.
0219
NOTE
On the Representation of Sets in Finite Measure Spaces Kali P. Rath Department of Economics, Uni¨ ersity of Notre Dame, Notre Dame, Indiana 46556 Submitted by Dorothy Maharam Stone Received June 26, 1995
This note provides a simple proof of the Bollobas]Varopoulos theorem on ´ the representation of measurable sets for finite measure spaces. Q 1996 Academic Press, Inc.
1. INTRODUCTION Let Ž X, X , m . be an atomless measure space, A i 4 , i g I, a family of sets in X having finite measure and a i 4 , i g I, a collection of nonnegative real numbers, I an index set. A i 4 is said to be a-representable if there is a family of measurable sets Ei 4 such that Ei : A i , m Ž Ei . s a i for all i g I and m Ž Ei l Ej . s 0 for all i and j in I, i / j. A classic result on representation of measurable sets is the following theorem in w1x. THEOREM 1 ŽBollobas ´ and Varopoulos.. A i 4 is a-representable iff
m
žDA / G Ý a i
igF
i
Ž 1.
igF
for all finite subsets F of I. The generality of this result is underscored by the facts that the measure of the underlying set X can be infinity and the index set I is arbitrary, in particular it can be uncountable. The proof requires the scalar version of 506 0022-247Xr96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
NOTE
507
Lyapunov’s convexity theorem, the Krein]Milman theorem, and is rather involved. It was recently exhibited in w5, 6x that Theorem 1 has important implications for the theory of large games and for games with private information. However, since these authors work with finite measure spaces, the full power of the theorem is really not necessary. More generally, in many applications, it is quite likely that the set X is a finite or a probability measure space. The purpose of this note is to provide a simple proof for this case, using only the scalar version of Lyapunov’s convexity theorem and some elementary measure theoretic facts. A simple and elegant proof of the scalar version of the convexity theorem, when the underlying measure space has certain additional topological structures, is given in w4x. Thus, the proof in this paper will be useful in the context of topological or metrizable spaces with finite measure. If m Ž X . - ` then one can assume without loss of generality that I is countable. In what follows it will be assumed that m Ž X . - ` and I will denote the set of positive integers.
2. A PROOF OF THE THEOREM In the proof of Theorem 1, the following two lemmas will be used. Lemma 1 is a finite version of the theorem and has been used in a slightly different form in the proof of Lemma 2 in w1x. It can be found in w2, p. 171; 3, p. 74x. Lemma 2, whose proof requires Lemma 1, furnishes the essential ingredient of the proof of Theorem 1. It asserts that given any pair of sets, it is possible to make them disjoint in such a way that the relevant inequality still remains valid with the new collection of sets. LEMMA 1. Let Ž X, X , m . be an atomless measure space, m Ž X . - `. Let A1 , . . . , A m 4 be a finite family of measurable sets of X and a 1 , . . . , a m 4 a finite family of nonnegati¨ e real numbers. Suppose that m ŽD i g F A i . G Ý i g F a i for any F : 1, 2, . . . , m4 . Then there exists E1 , . . . , Em 4 such that Ei : A i , m Ž Ei . s a i and m Ž Ei l Ej . s 0 for i / j. LEMMA 2. Let Ž X, X , m . be an atomless measure space, m Ž X . - `, and I the set of positi¨ e integers. Let A i 4 be a collection of measurable subsets of X and a i 4 a collection of nonnegati¨ e numbers, i g I. Suppose that A i 4 and a i 4 satisfy Ž1.. Then for any j and k in I, j - k, there exist sets Bj : A j and Bk : A k such that: Ž a. Ž Bj j Bk . s Ž A j j A k ., Ž b . m Ž Bj l Bk . s 0, and Ž c . the collections A1 , . . . , A jy1 , Bj , A jq1 , . . . , A ky1 , Bk , A kq1 , . . . 4 and a 1 , . . . , a jy1 , a j , a jq1 , . . . , a ky1 , a k , a kq1 , . . . 4 satisfy Ž1..
508
NOTE
Proof of Theorem 1. Clearly, if the sets are a-representable then Ž1. must be satisfied. Therefore, to prove the theorem, it suffices to show that Ž1. implies that the sets are a-representable. Towards this end, consider the following triangular array of sets. T11 T12 T13 T14 ?
T22 T23 T24 ?
T33 T34 ?
T44 ?
?
The properties of these sets are as follows. Ža. T11 is A1. Žb. For any fixed n n n, Ti n : A i if i F n and D is1 Ti n s D is1 A i . m ŽTi n l Tjn . s 0 for any Ž . i F n, j F n, and i / j. c For any fixed n, the collections T1 n , . . . , Tn n , A nq1 , A nq2 , . . . 4 and a 1 , . . . , a n , a nq1 , a nq2 , . . . 4 satisfy Ž1.. Žd. for any i G 1 and n G i, Ti n = Ti, nq1. The actual construction follows from Lemma 2. T11 is simply A1. Lemma 2 yields T12 and T22 by taking j s 1 and k s 2. Now suppose that T1 n , T2 n , . . . , Tn n have been constructed for some n. Application of Lemma 2 to the pair ŽT1 n , A nq1 . will yield the sets T1, nq1 , T2 n , . . . , Tn n , and say C1, nq1. By applying Lemma 2 to the pair ŽT2 n , C1, nq1 ., one obtains T1, nq1 , T2, nq1 , T3 n , . . . , Tn n , and say C2, nq1. Thus successive applications to each of the sets T1 n , T2 n , . . . , Tn n leads in a finite number of steps to the desired sets T1, nq1 , T2, nq1 , . . . , Tn, nq1 , Tnq1, nq1. For any i, let Pi s F nG i Ti n . m Ž Pi l Pj . s 0 for i / j because m ŽTi n l Tjn . s 0 for n G max i, j4 . For any i, m ŽTi n . G a i for every n G i. So, m Ž Pi . G a i . Lyapunov’s theorem ensures that there exists Ei : Pi with m Ž Ei . s a i . This collection of sets Ei 4 has the required properties. This completes the proof.
3. PROOFS OF THE LEMMAS Now all that remains is the proofs of the two lemmas. Lemma 1 can be proved by induction; see w2, 3x. There, it has been proved under the m additional hypothesis that m ŽD is1 A i . s Ým is1 a i . The two versions with or without this equality are equivalent since the measure space is finite. If strict inequality Žgreater than. holds, equality can be restored by defining a m m new set and a new number by A 0 s D is1 A i and a 0 s m ŽD is1 Ai . y m Ý is1 a i . The proof Lemma 2 is actually constructive. Naturally, it is difficult to manufacture two disjoint sets with the required properties in one stroke.
NOTE
509
Therefore, a sequential approach will be taken. It will be exhibited that there are sets with arbitrarily small intersections which satisfy the required conditions. A limit argument will then complete the proof. Proof of Lemma 2. Without loss of generality let j s 1 and k s 2. Suppose that there are two decreasing sequences of sets Pn4 and Q n4 such that: ŽI. A1 = P1 = ??? = Pn = Pnq1 . . . , ŽII. A 2 = Q1 = ??? = Q n = Q nq 1 . . . , ŽIII. Ž Pn j Q n . s Ž A1 j A 2 ., m Ž Pn l Q n . F Ž1rn. for each n, and ŽIV. Inequality Ž1. holds for any n when Pn and Q n are substituted for A1 and A 2 , respectively. Then by letting B1 s F `ns1 Pn and B2 s F `ns1 Q n it is easily seen that Ž B1 j B2 . s Ž A1 j A 2 ., m Ž B1 l B2 . s 0 and the collections B1 , B2 , A 3 , . . . 4 , a 1 , a 2 , a 3 , . . . 4 satisfy Ž1.. In order to prove the existence of such sets Pn4 and Q n4 it suffices to show that given any e ) 0, there are sets P1 and Q1 such that P1 : A1 , Q1 : A 2 , Ž P1 j Q1 . s Ž A1 j A 2 ., and m Ž P1 l Q1 . F e . Successive applications to Pn and Q n will then generate Pnq1 and Q nq1 for any n. Let e ) 0 be given. Denote by F1 the set of nonempty finite subsets F of I such that 1 g F and 2 f F and by F2 the set of nonempty finite subsets F of I such that 2 g F and 1 f F. Suppose that either Ži. for every F in F1 , m Žw A1 _ A 2 x j wD i g F _14 A i x. G Ý i g F a i , or, Žii. for every F in F2 , m Žw A 2 _ A1 x j wD i g F _24 A i x. G Ý i g F a i . In the first case, let P1 s Ž A1 _ A 2 ., and Q1 s A 2 . In the second case, let P1 s A1 and Q1 s Ž A 2 _ A1 .. In either case, Ž P1 j Q1 . s Ž A1 j A 2 . and Ž P1 l Q1 . s B. It is easy to verify that the collections P1 , Q1 , A 3 , A 4 . . . 4 and a 1 , a 2 , a 3 , a 4 . . . 4 satisfy Ž1.. Therefore, posit that there are F1 g F1 , F2 g F2 and 0 - u F e such that: ŽA. m Žw A1 _ A 2 x j wD i g F1 _14 A i x. - Ý i g F1 a i y u and ŽB. m Žw A 2 _ A1 x j wD i g F 2 _24 A i x. - Ý i g F 2 a i y u . Since m Ž X . - ` and F1 and F2 are finite, there exists m such that m ) sup F1 , m ) sup F2 , and Ý`ismq1 a i - Ž ur2.. The finite collections A1 , A 2 , . . . , A m 4 and a 1 , a 2 , . . . , a m 4 satisfy the hypothesis of Lemma 1. So, there exists S1 , S2 , . . . , Sm 4 such that Si : A i , m Ž Si . s a i , and m Ž Si l S j . s 0 if i / j. S1 : A1 and S2 : A 2 in conjunction with ŽA. and ŽB. imply that m Ž S1 l Ž A1 l A 2 .. G u and m Ž S2 l Ž A1 l A 2 .. G u . Let V1 : Ž S1 l Ž A1 l A 2 .. and V2 : Ž S2 l Ž A1 l A 2 .. be such that m Ž V1 . s m Ž V2 . s Ž ur2.. Lyapunov’s theorem ensures the existence of such sets. Define P1 s Ž A1 _ S2 . j V2 and Q1 s Ž A 2 _ A1 . j S2 j V1. Clearly, P1 : A1 , Q1 : A 2 , Ž P1 j Q1 . s Ž A1 j A 2 ., and m Ž P1 l Q1 . s m Ž V1 j V2 . s u F e . It remains to be shown that for any nonempty finite subset F of I, the collections P1 , Q1 , A 3 , A 4 , . . . 4 and a 1 , a 2 , a 3 , a 4 . . . 4 satisfy Ž1.. Clearly, it suffices to establish this Ži. for any F in F1 and Žii. for any F in F2 .
510
NOTE
Consider any F in F2 and let F0 s F _ m q 1, m q 2, . . . 4 . F0 / B since 2 g F0 . m Ž Q1 j wD i g F _24 A i x. G m Ž Q1 j wD i g F 0 _24 A i x. and, from the construction of these sets, m Ž Q1 j wD i g F 0 _24 A i x. G m Ž S2 . q m Ž V1 . q Ý i g F 0 _24 m Ž Si . s ŽÝ i g F 0 a i . q Ž ur2.. Moreover, ŽÝ i g F 0 a i . q Ž ur2. G Ý i g F a i since Ý i g F _ F 0 a i - Ž ur2.. Therefore, Ž1. holds for any F in F2 . Analogous arguments can be used to show that Ž1. holds for any F in F1. Therefore, the desired inequality Ž1. holds for every nonempty finite subset F of I. This completes the proof.
ACKNOWLEDGMENT This work was inspired by conversations with Thomas E. Armstrong, M. Ali Khan, and Yeneng Sun. I am grateful to them for helpful comments and suggestions.
REFERENCES 1. B. Bollobas ´ and N. Th. Varopoulos, Representation of systems of measurable sets, Math. Proc. Cambridge Philos. Soc. 78 Ž1974., 323]325. 2. S. Hart and E. Kohlberg, Equally distributed correspondences, J. Math. Econom. 1 Ž1974., 167]174. 3. W. Hildenbrand, ‘‘Core and Equilibria of a Large Economy,’’ Princeton Univ. Press, Princeton, NJ, 1974. 4. R. E. Jamison, A quick proof for a one-dimensional version of Liapounoff’s theorem, Amer. Math. Monthly 81 Ž1974., 507]508. 5. M. Ali Khan and Y. Sun, The Marriage Lemma and large anonymous games with countable actions, Math. Proc. Cambridge Philos. Soc. 117 Ž1995., 385]387. 6. M. Ali Khan and Y. Sun, Pure strategies in games with private information, J. Math. Econom., 24 Ž1995., 633]653.
200, 506]510 Ž1996.
0219
NOTE
On the Representation of Sets in Finite Measure Spaces Kali P. Rath Department of Economics, Uni¨ ersity of Notre Dame, Notre Dame, Indiana 46556 Submitted by Dorothy Maharam Stone Received June 26, 1995
This note provides a simple proof of the Bollobas]Varopoulos theorem on ´ the representation of measurable sets for finite measure spaces. Q 1996 Academic Press, Inc.
1. INTRODUCTION Let Ž X, X , m . be an atomless measure space, A i 4 , i g I, a family of sets in X having finite measure and a i 4 , i g I, a collection of nonnegative real numbers, I an index set. A i 4 is said to be a-representable if there is a family of measurable sets Ei 4 such that Ei : A i , m Ž Ei . s a i for all i g I and m Ž Ei l Ej . s 0 for all i and j in I, i / j. A classic result on representation of measurable sets is the following theorem in w1x. THEOREM 1 ŽBollobas ´ and Varopoulos.. A i 4 is a-representable iff
m
žDA / G Ý a i
igF
i
Ž 1.
igF
for all finite subsets F of I. The generality of this result is underscored by the facts that the measure of the underlying set X can be infinity and the index set I is arbitrary, in particular it can be uncountable. The proof requires the scalar version of 506 0022-247Xr96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
NOTE
507
Lyapunov’s convexity theorem, the Krein]Milman theorem, and is rather involved. It was recently exhibited in w5, 6x that Theorem 1 has important implications for the theory of large games and for games with private information. However, since these authors work with finite measure spaces, the full power of the theorem is really not necessary. More generally, in many applications, it is quite likely that the set X is a finite or a probability measure space. The purpose of this note is to provide a simple proof for this case, using only the scalar version of Lyapunov’s convexity theorem and some elementary measure theoretic facts. A simple and elegant proof of the scalar version of the convexity theorem, when the underlying measure space has certain additional topological structures, is given in w4x. Thus, the proof in this paper will be useful in the context of topological or metrizable spaces with finite measure. If m Ž X . - ` then one can assume without loss of generality that I is countable. In what follows it will be assumed that m Ž X . - ` and I will denote the set of positive integers.
2. A PROOF OF THE THEOREM In the proof of Theorem 1, the following two lemmas will be used. Lemma 1 is a finite version of the theorem and has been used in a slightly different form in the proof of Lemma 2 in w1x. It can be found in w2, p. 171; 3, p. 74x. Lemma 2, whose proof requires Lemma 1, furnishes the essential ingredient of the proof of Theorem 1. It asserts that given any pair of sets, it is possible to make them disjoint in such a way that the relevant inequality still remains valid with the new collection of sets. LEMMA 1. Let Ž X, X , m . be an atomless measure space, m Ž X . - `. Let A1 , . . . , A m 4 be a finite family of measurable sets of X and a 1 , . . . , a m 4 a finite family of nonnegati¨ e real numbers. Suppose that m ŽD i g F A i . G Ý i g F a i for any F : 1, 2, . . . , m4 . Then there exists E1 , . . . , Em 4 such that Ei : A i , m Ž Ei . s a i and m Ž Ei l Ej . s 0 for i / j. LEMMA 2. Let Ž X, X , m . be an atomless measure space, m Ž X . - `, and I the set of positi¨ e integers. Let A i 4 be a collection of measurable subsets of X and a i 4 a collection of nonnegati¨ e numbers, i g I. Suppose that A i 4 and a i 4 satisfy Ž1.. Then for any j and k in I, j - k, there exist sets Bj : A j and Bk : A k such that: Ž a. Ž Bj j Bk . s Ž A j j A k ., Ž b . m Ž Bj l Bk . s 0, and Ž c . the collections A1 , . . . , A jy1 , Bj , A jq1 , . . . , A ky1 , Bk , A kq1 , . . . 4 and a 1 , . . . , a jy1 , a j , a jq1 , . . . , a ky1 , a k , a kq1 , . . . 4 satisfy Ž1..
508
NOTE
Proof of Theorem 1. Clearly, if the sets are a-representable then Ž1. must be satisfied. Therefore, to prove the theorem, it suffices to show that Ž1. implies that the sets are a-representable. Towards this end, consider the following triangular array of sets. T11 T12 T13 T14 ?
T22 T23 T24 ?
T33 T34 ?
T44 ?
?
The properties of these sets are as follows. Ža. T11 is A1. Žb. For any fixed n n n, Ti n : A i if i F n and D is1 Ti n s D is1 A i . m ŽTi n l Tjn . s 0 for any Ž . i F n, j F n, and i / j. c For any fixed n, the collections T1 n , . . . , Tn n , A nq1 , A nq2 , . . . 4 and a 1 , . . . , a n , a nq1 , a nq2 , . . . 4 satisfy Ž1.. Žd. for any i G 1 and n G i, Ti n = Ti, nq1. The actual construction follows from Lemma 2. T11 is simply A1. Lemma 2 yields T12 and T22 by taking j s 1 and k s 2. Now suppose that T1 n , T2 n , . . . , Tn n have been constructed for some n. Application of Lemma 2 to the pair ŽT1 n , A nq1 . will yield the sets T1, nq1 , T2 n , . . . , Tn n , and say C1, nq1. By applying Lemma 2 to the pair ŽT2 n , C1, nq1 ., one obtains T1, nq1 , T2, nq1 , T3 n , . . . , Tn n , and say C2, nq1. Thus successive applications to each of the sets T1 n , T2 n , . . . , Tn n leads in a finite number of steps to the desired sets T1, nq1 , T2, nq1 , . . . , Tn, nq1 , Tnq1, nq1. For any i, let Pi s F nG i Ti n . m Ž Pi l Pj . s 0 for i / j because m ŽTi n l Tjn . s 0 for n G max i, j4 . For any i, m ŽTi n . G a i for every n G i. So, m Ž Pi . G a i . Lyapunov’s theorem ensures that there exists Ei : Pi with m Ž Ei . s a i . This collection of sets Ei 4 has the required properties. This completes the proof.
3. PROOFS OF THE LEMMAS Now all that remains is the proofs of the two lemmas. Lemma 1 can be proved by induction; see w2, 3x. There, it has been proved under the m additional hypothesis that m ŽD is1 A i . s Ým is1 a i . The two versions with or without this equality are equivalent since the measure space is finite. If strict inequality Žgreater than. holds, equality can be restored by defining a m m new set and a new number by A 0 s D is1 A i and a 0 s m ŽD is1 Ai . y m Ý is1 a i . The proof Lemma 2 is actually constructive. Naturally, it is difficult to manufacture two disjoint sets with the required properties in one stroke.
NOTE
509
Therefore, a sequential approach will be taken. It will be exhibited that there are sets with arbitrarily small intersections which satisfy the required conditions. A limit argument will then complete the proof. Proof of Lemma 2. Without loss of generality let j s 1 and k s 2. Suppose that there are two decreasing sequences of sets Pn4 and Q n4 such that: ŽI. A1 = P1 = ??? = Pn = Pnq1 . . . , ŽII. A 2 = Q1 = ??? = Q n = Q nq 1 . . . , ŽIII. Ž Pn j Q n . s Ž A1 j A 2 ., m Ž Pn l Q n . F Ž1rn. for each n, and ŽIV. Inequality Ž1. holds for any n when Pn and Q n are substituted for A1 and A 2 , respectively. Then by letting B1 s F `ns1 Pn and B2 s F `ns1 Q n it is easily seen that Ž B1 j B2 . s Ž A1 j A 2 ., m Ž B1 l B2 . s 0 and the collections B1 , B2 , A 3 , . . . 4 , a 1 , a 2 , a 3 , . . . 4 satisfy Ž1.. In order to prove the existence of such sets Pn4 and Q n4 it suffices to show that given any e ) 0, there are sets P1 and Q1 such that P1 : A1 , Q1 : A 2 , Ž P1 j Q1 . s Ž A1 j A 2 ., and m Ž P1 l Q1 . F e . Successive applications to Pn and Q n will then generate Pnq1 and Q nq1 for any n. Let e ) 0 be given. Denote by F1 the set of nonempty finite subsets F of I such that 1 g F and 2 f F and by F2 the set of nonempty finite subsets F of I such that 2 g F and 1 f F. Suppose that either Ži. for every F in F1 , m Žw A1 _ A 2 x j wD i g F _14 A i x. G Ý i g F a i , or, Žii. for every F in F2 , m Žw A 2 _ A1 x j wD i g F _24 A i x. G Ý i g F a i . In the first case, let P1 s Ž A1 _ A 2 ., and Q1 s A 2 . In the second case, let P1 s A1 and Q1 s Ž A 2 _ A1 .. In either case, Ž P1 j Q1 . s Ž A1 j A 2 . and Ž P1 l Q1 . s B. It is easy to verify that the collections P1 , Q1 , A 3 , A 4 . . . 4 and a 1 , a 2 , a 3 , a 4 . . . 4 satisfy Ž1.. Therefore, posit that there are F1 g F1 , F2 g F2 and 0 - u F e such that: ŽA. m Žw A1 _ A 2 x j wD i g F1 _14 A i x. - Ý i g F1 a i y u and ŽB. m Žw A 2 _ A1 x j wD i g F 2 _24 A i x. - Ý i g F 2 a i y u . Since m Ž X . - ` and F1 and F2 are finite, there exists m such that m ) sup F1 , m ) sup F2 , and Ý`ismq1 a i - Ž ur2.. The finite collections A1 , A 2 , . . . , A m 4 and a 1 , a 2 , . . . , a m 4 satisfy the hypothesis of Lemma 1. So, there exists S1 , S2 , . . . , Sm 4 such that Si : A i , m Ž Si . s a i , and m Ž Si l S j . s 0 if i / j. S1 : A1 and S2 : A 2 in conjunction with ŽA. and ŽB. imply that m Ž S1 l Ž A1 l A 2 .. G u and m Ž S2 l Ž A1 l A 2 .. G u . Let V1 : Ž S1 l Ž A1 l A 2 .. and V2 : Ž S2 l Ž A1 l A 2 .. be such that m Ž V1 . s m Ž V2 . s Ž ur2.. Lyapunov’s theorem ensures the existence of such sets. Define P1 s Ž A1 _ S2 . j V2 and Q1 s Ž A 2 _ A1 . j S2 j V1. Clearly, P1 : A1 , Q1 : A 2 , Ž P1 j Q1 . s Ž A1 j A 2 ., and m Ž P1 l Q1 . s m Ž V1 j V2 . s u F e . It remains to be shown that for any nonempty finite subset F of I, the collections P1 , Q1 , A 3 , A 4 , . . . 4 and a 1 , a 2 , a 3 , a 4 . . . 4 satisfy Ž1.. Clearly, it suffices to establish this Ži. for any F in F1 and Žii. for any F in F2 .
510
NOTE
Consider any F in F2 and let F0 s F _ m q 1, m q 2, . . . 4 . F0 / B since 2 g F0 . m Ž Q1 j wD i g F _24 A i x. G m Ž Q1 j wD i g F 0 _24 A i x. and, from the construction of these sets, m Ž Q1 j wD i g F 0 _24 A i x. G m Ž S2 . q m Ž V1 . q Ý i g F 0 _24 m Ž Si . s ŽÝ i g F 0 a i . q Ž ur2.. Moreover, ŽÝ i g F 0 a i . q Ž ur2. G Ý i g F a i since Ý i g F _ F 0 a i - Ž ur2.. Therefore, Ž1. holds for any F in F2 . Analogous arguments can be used to show that Ž1. holds for any F in F1. Therefore, the desired inequality Ž1. holds for every nonempty finite subset F of I. This completes the proof.
ACKNOWLEDGMENT This work was inspired by conversations with Thomas E. Armstrong, M. Ali Khan, and Yeneng Sun. I am grateful to them for helpful comments and suggestions.
REFERENCES 1. B. Bollobas ´ and N. Th. Varopoulos, Representation of systems of measurable sets, Math. Proc. Cambridge Philos. Soc. 78 Ž1974., 323]325. 2. S. Hart and E. Kohlberg, Equally distributed correspondences, J. Math. Econom. 1 Ž1974., 167]174. 3. W. Hildenbrand, ‘‘Core and Equilibria of a Large Economy,’’ Princeton Univ. Press, Princeton, NJ, 1974. 4. R. E. Jamison, A quick proof for a one-dimensional version of Liapounoff’s theorem, Amer. Math. Monthly 81 Ž1974., 507]508. 5. M. Ali Khan and Y. Sun, The Marriage Lemma and large anonymous games with countable actions, Math. Proc. Cambridge Philos. Soc. 117 Ž1995., 385]387. 6. M. Ali Khan and Y. Sun, Pure strategies in games with private information, J. Math. Econom., 24 Ž1995., 633]653.