1. ) is measurable in s as well. Proof of theorem 2.1: By Lemma 2.2, B is isotone. Moreover, by Proposition 2.1, ? T d is a chain complete poset. Hence, by Markovsky's Theorem (see Appendix, Theorem A.2), B has a chain complete poset of fixed points, with the greatest and the least element. Denote the greatest element of the set by ? * . Then, by definition, ? * constitutes distributional BayesianNash equilibrium of ?. Next, we will prove that ? * is the greatest equilibrium of the game. Take any other equilibrium of the game ? . Fix s ? S, such that ? ({(?, a)|a ? m(?, s, a)} |s) = 1. Clearly {(?, a)|a ? m(?, s, a)} is a set of full measure;Lemma 9, it is also measurable on L, hence v is Carath�odory. By Assumption 2.1(ii),(iii), as well as Corollaries A.3, A.6 and Proposition 2.1, v has single crossing differences in (a, f ), with respect to �-a.e. pointwise order?T order? order?T,2006
2. Proof of corollary 2.2: By Assumptions 2.1-2.3, for any ? ? ? T d , B(? ) (respectively B(? )) is increasing in ? and monotonically inf-preserving (respectively sup-preserving) on?Ton? on?T d . Therefore, by Theorem A.8, ? * (�) and ? * (�) are increasing on ?. Proof of lemma 3.1: Since A ? R n , any compact subset of A is closed. Hence, by Assumption 3.1(i), ? A has non-empty, closed values. Moreover, it maps measurable space into a complete metric space (hence, a Polish space). Therefore, by Kuratowski-Ryll-Nardzewski Selection Theorem (see Aliprantis and Border;? ? T B Therefore;Since T is isotone, by Markovsky's Theorem ? * ? T ? . We prove existence of the least equilibrium analogously, using operator B. Proof of corollary 2.1: In order to prove the result, we shall use Tarski-Kantorovich Theorem (see Appendix, Theorem A.3),1994
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