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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': 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

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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 $\bigcup F'\times X'$.

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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
cont. ...Kaboom! The measurable space generated by this topology would be discrete. As far as I know Aumann is concerned with defining measurable structures on the space of ALL measurable functions. That's actually my question, I want to know if I can do that, I don't know if I'm overlooking something or if there's something I should know. – Mario Carrasco Aug 10 '11 at 15:27

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