By Gobel R., Hill P., Liebert W. (eds.)

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Let (S, a) be a measurable space Y, be a separable metric space and 'lj; : S -+ 2 Y be a lower measurable correspondence. , Gf) E a 0,B(Y). Proof. Consider the function g : S X Y -+ [0,00 1 defined by g( s, y) = dist( y, 'lj;( s )). ) is lower measurable it follows that for each fixed y E Y, g(. , y) is measurable, for {s E S : dist(y, 'lj;(s)) < c} = {s E S: 'lj;(s) n B(y,c) ]1: 0} and the latter set belongs to a by the assumption of lower measurability. Obviously for each fixed s E S, g( s,·) is continuous.

Observe that {t E T : 'lj;(t) n U =J 0} T n U) n (I(t) + B(O,c)) =J 0} E O(t)} = proh(Gj n Gf)). 8. 3, proh(Gj n Gf)) E T. Therefore {t E T: 'lj;(t) n U =J 0} E T, and this completes the proof of the Fact. 9. Let (S, a) be a measurable space Y, be a separable metric space and 'lj; : S -+ 2 Y be a lower measurable correspondence. , Gf) E a 0,B(Y). Proof. Consider the function g : S X Y -+ [0,00 1 defined by g( s, y) = dist( y, 'lj;( s )). ) is lower measurable it follows that for each fixed y E Y, g(.

Such that: (i) for all k,fk is a measurable function from proh(Gx) into Y, and Nicholas C. Yannelis 43 (ii) for almost alit E proh(Gx),{fk(t): k = 1,2, ... } is a dense subset of X(t). Proof. For each n = 1,2, ... , let {Ei : i = 1,2, ... } be an open cover of Y such that diam(Ei) < 2~. For each n, i = 1,2, ... , define Tt = {t E T : X(t) n Ei =:J 0}. Since Tt = prohHt,y) E T X Y : y E X(t) n En and X(·) n Ei has a measurable graph in T X Y, Tt E r by virtue of the projection theorem. It can be easily checked that U~I Tt = proh(G x ) == S.