show/hide this revision's text 2 Some edits for readability, also pointing out error

Sorry for the necromancy. Even though Michael Greinecker demonstrated that there is not Here's an attempt at constructing a natural measure on $\sigma$-algebra using the space tensor product of measurable functions, there is a natural $\sigma$-algebra. Namely, \sigma$-algebras. This should likely not result in a measurable function space is equipped with Borel structure (i.e., a $\sigma$-algebra generated as the smallest Borel $\sigma$-algebra of a topological space), so that the evaluation map is measurable. I figured I'd answer the question to provide a quick reference for the futuredon't think it contradicts Aumann's work.

Let me introduce some notation If it's wrong, please correct it.

I figured I'd answer the question to make this preciseprovide a quick reference for the future.

That is

I think that $\Sigma_H$ should be well-defined, even though it's unlikely to be Borel in most interesting situations. There should always be some minimal solution, even if it's the whole power set $2^H$.

Here are some general thoughts on why it is important that the evaluation function is measurable, and why this is good enough for most interesting applications, e.g., applied analysis, physics or computation. This means that f $B \in \Sigma_Y$ is any measurable event in $Y$, then $$\operatorname{eval}^{-1}(B) = \big\{ (h,x) : h(x) \in B \big\} \in \Sigma_H \otimes \Sigma_X.$$This should be a good enough $\sigma$-algebra for most applications of measure theory, e.g., applied analysis, physics or computation.

show/hide this revision's text 1 [made Community Wiki]

Sorry for the necromancy. Even though Michael Greinecker demonstrated that there is not a natural measure on the space of measurable functions, there is a natural $\sigma$-algebra. Namely, a measurable function space is equipped with the smallest $\sigma$-algebra so that the evaluation map is measurable. I figured I'd answer the question to provide a quick reference for the future.

I made this answer community wiki, so feel free to edit it.

Let me introduce some notation to make this precise. Let $(X,\Sigma_X)$ and $(Y, \Sigma_Y)$ be two measurable spaces, and let $H = \operatorname{Hom}(X,Y)$ be the set of measurable functions from $X$ to $Y$. Define the evaluation map $\operatorname{eval} : H \times X \to Y$ by $$\operatorname{eval}(h,x) = h(x).$$

Now, simply define $\Sigma_{H}$ to be the minimal $\sigma$-algebra on $H$ so that the evaluation map $\operatorname{eval} : H \times X \to Y$ is measurable, where $H \times X$ is equipped with the tensor product $\sigma$-algebra $\Sigma_H \otimes \Sigma_X$.

That is, if $B \in \Sigma_Y$ is any measurable event in $Y$, then $$\operatorname{eval}^{-1}(B) = \big\{ (h,x) : h(x) \in B \big\} \in \Sigma_H \otimes \Sigma_X.$$ This should be a good enough $\sigma$-algebra for most applications of measure theory, e.g., applied analysis, physics or computation. For example, this always describes solution-sets to equations, since $$\{ h(x) = y \} = \operatorname{eval}^{-1}({y}).$$

When $Y$ is a measurable hierarchy (i.e., a pre-ordered measurable space), then this also includes inequalities, e.g., $$\{ h(x) \le y \} = \operatorname{eval}^{-1}(\downarrow{y}),$$ where $\downarrow{y} = \{ y' \le y \}$ denotes the down-set of $y \in Y$. Basically, $$\mbox{if you can write it down, it's probably measurable.}$$

This is very useful computationally, since the hom-set $\operatorname{Hom}(H \times X, Y)$ is adjoint to $\operatorname{Hom}(H,Y^X)$ via the process of currying. The adjoint to the evaluation map is called function application, and in computer science is known as Apply. Ultimately, this means that anything you work out computationally is measurable, which means no more appendices full of nasty measurability proofs by hand.

Note that $Y^X$ is a measurable space when equipped with the tensor-product $\sigma$-algebra, and in most cases of interest its $\sigma$-algebra is not generated by a topology (reference Jochen Wengenroth's answer to this question).

Furthermore, this should be useful in measure theory, and may lead toward an answer to Kenny Easwaran's question. If you can see a way to answer it, go ahead and edit this answer.