# A question about measurable structures on function spaces

Hey, I was just wondering, I'm using some of Robert Aumann's ideas about measurable structures on function spaces (From his paper 'Borel structures for Function spaces': http://projecteuclid.org/euclid.ijm/1255631584) and I had a question. Let $(X,\Sigma_{X})$, $(Y,\Sigma_{Y})$ be measurable spaces, let $Y^{X}$ be the space of measurable functions from $X$ to $Y$, I want to define a measurable structure on a subset $F$ of $Y^{X}$, using his construction I managed to define a measurable structure on $F$, now the thing here is that I need that structure to be discrete for some stuff I'm working on. I was thinking and couldn't find a way around it so I remembered that $F$ is actually the set of continuous functions from $X$ to $Y$ (which are of course measurable), so I was thinking, what if I simply use the function space $Y_{C}^{X}$ (the space of continuous functions from $X$ to $Y$) and equip it with the discrete topology? I was thinking that since it is a topological space I can define a measurable space generated by the open sets (which are all the subsets) on $Y_{C}^{X}$, denoted by $(Y_{C}^{X},\Sigma_{Y_{C}^{X}})$, which would have a discrete measurable structure, but I'm not sure about doing this, so my question is:

If I take the set of continuous functions $Y_{C}^{X}$ between 2 topological spaces $X$ and $Y$ and define a topology on it (in this case the discrete topology), can I then define a Borel structure on it generated by the open sets? Because then I'd have the discrete measurable space I'm looking for, this seems like the logical thing to do but I don't know if I'll run into some conceptual problems if I do this, I don't know if this has been done as Aumann doesn't mention it in his paper

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, $X=[0,1]$, $Y=\mathbb{R}$ with the usual topology. Then $C(X,Y)$ is a complete, separable metric space when endowed with the uniform metric and the evaluation function is continuous, hence measurable when $C(X,Y)$ 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 $\bigcup F'\times X'$.
• Thanks Michael. Now how about this: Let $(X,T_X)$ and $(Y,T_X)$ be topological spaces, let $Y^{X}_C$ be the set of all continuous functions from $X$ to $Y$, let $Y^{X}_M$ be the set of all measurable functions from $X$ to $Y$, we can define a topology on $Y^{X}_C$. Can we define a measurable space (Borel space) on $Y^{X}_C$ generated by the open sets in this topology? This would work for me because the set I'm trying to define a measurable structure on is $F = Y^{X}_C \subset Y^{X}_M$, a subset of the set of ALL measurable functions. Now if I put a discrete topology on $Y^{X}_C$... cont. – Mario Carrasco Aug 10 '11 at 15:27