It depends. What Aumann is concerned with in the paper are admissible structures, that is, he looks for a sigma-algebras on sets of measurable functions $F\subseteq Y^X$ such that th evaluation $e:Y^X\times X\to Y$ given by $e(f,x)=f(x)$ is measurable with respect to the product-sigma-algebra.
A well behaved case is when, say, $X=Y=\mathbb{R}$ X=[0,1]$,$Y=\mathbb{R}$with the usual topology. Then$C(A,B)$C(X,Y)$ is a complete, separable metric space when endowed with the uniform metric and the evaluation function is continuous, hence measurable when $C(A,B)$ C(X,Y)$is endowed with the Borel sigma-algebra. The discrete sigma-algebra is even larger, so the evaluation is certainly measurable in this case. Now let$X=Y$be discrete topological spaces with cardinality larger than the continuum. Let$y\in Y$. We show that$e^{-1}(\{y\})$is not in the product-sigma-algebra. Suppose it is. By the first two lemmata here,$e^{-1}(\{y\})$has to be the union of continuum many product sets of the form$F'\times X'$,$F\subseteq Y^X$and$X'\subseteq X$. For such a product set, we have that every$f\in F'$is constant and equal to$y$on$X'$. W.l.o.g. we can assume that none of the$X'$is empty or equal to$X$. So we can construct an$X''\subseteq X$such that$X''$intersects every$X'$and every$(X')^C$. Let$F''=\{f\in Y^X:f(x)=y\text{ iff }x\in X''\}$. Then$F''\times X''\subseteq e^{-1}(\{y\})$but$F''\times X''$is not a subset of$\bigcup F'\times X'$. 1 It depends. What Aumann is concerned with in the paper are admissible structures, that is, he looks for a sigma-algebras on sets of measurable functions$F\subseteq Y^X$such that th evaluation$e:Y^X\times X\to Y$given by$e(f,x)=f(x)$is measurable with respect to the product-sigma-algebra. A well behaved case is when, say,$X=Y=\mathbb{R}$with the usual topology. Then$C(A,B)$is a complete, separable metric space when endowed with the uniform metric and the evaluation function is continuous, hence measurable when$C(A,B)$is endowed with the Borel sigma-algebra. The discrete sigma-algebra is even larger, so the evaluation is certainly measurable in this case. Now let$X=Y$be discrete topological spaces with cardinality larger than the continuum. Let$y\in Y$. We show that$e^{-1}(\{y\})$is not in the product-sigma-algebra. Suppose it is. By the first two lemmata here,$e^{-1}(\{y\})$has to be the union of continuum many product sets of the form$F'\times X'$,$F\subseteq Y^X$and$X'\subseteq X$. For such a product set, we have that every$f\in F'$is constant and equal to$y$on$X'$. W.l.o.g. we can assume that none of the$X'$is empty or equal to$X$. So we can construct an$X''\subseteq X$such that$X''$intersects every$X'$and every$(X')^C$. Let$F''=\{f\in Y^X:f(x)=y\text{ iff }x\in X''\}$. Then$F''\times X''\subseteq e^{-1}(\{y\})$but$F''\times X''$is not a subset of$\bigcup F'\times X'\$.