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Let $X,Y$ be non-empty sets. Define $$\Phi:\mathbb{R}^{X\times Y}\to ({\mathbb{R}^{Y}})^X, f\mapsto\left(x\mapsto\left(y\mapsto f(x,y)\right)\right).$$ Let $\mathcal{F}_X$ be a $\sigma$-algebra of subsets of $X$ and $\mathcal{F}_Y$ be a $\sigma$-algebra of subsets of $Y$. Let $\mathcal{M}$ be the set of measurable functions from $(X\times Y,\mathcal{F}_X\otimes\mathcal{F}_Y)$ into $\mathbb{R}$ and $\mathcal{N}$ be the set of measurable functions from $(Y,\mathcal{F}_Y)$ into $\mathbb{R}$. Then $$\forall f\in\mathcal{M}, \forall x\in X, \Phi(f)(x)\in\mathcal{N},$$ i.e. $$\forall f\in\mathcal{M}, \Phi(f):X\to\mathcal{N}.$$ Now, it's clear that for some choice of $(X,\mathcal{F}_X)$ (e.g. if $(X,\mathcal{F}_X)$ is $\mathbb{R}$ with its Borel $\sigma$-algebra) there exists $f\in \mathbb{R}^{X\times Y}$ for which $\Phi(f):X\to\mathcal{N}$ while $f\notin \mathcal{M}$. So, it seems that something is missing to characterize the measurability of $f$ from $(X\times Y,\mathcal{F}_X\otimes \mathcal{F}_Y)$ into $\mathbb{R}$ in terms of some property of $\Phi(f)$ (i.e. requiring that $\Phi(f): X\to\mathcal{N}$ is a necessary but in general not sufficient condition).

My guess is that maybe we can find a a $\sigma$-algebra $\mathcal{F}$ of subsets of $\mathcal{N}$ such that $f\in\mathcal{M}$ iff $\Phi(f)$ is measurable from $(X,\mathcal{F}_{\mathcal{X}})$ into $(\mathcal{N},\mathcal{F})$. So the question:

does there exist a $\sigma$-algebra of subsets of $\mathcal{N}$, say $\mathcal{F}$, such that $$\forall f\in\mathbb{R}^{X\times Y}, (f\in\mathcal{M})\iff\left(\Phi(f) \text{ is measurable from $(X,\mathcal{F}_X)$ into $(\mathcal{N},\mathcal{F})$}\right)?$$

If not, does there exist any results in this direction assuming more, e.g. that the functions are bounded or that the underlying spaces are (separable) metric spaces?

Any (also partial) result in this direction (or reference) is welcome.

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Are you aware of this question and the references therein ?

It tells that $\mathcal F$ cannot exist in general, if you suppose that it should not depend on $X$. Indeed, if $\mathcal F$ depends only on $Y$, then you could take $X = \mathcal N$ and $\mathcal F_X = \mathcal F$. Since the identity of $\mathcal N$ is measurable, the evalutation morphism $\mathcal N \times Y \rightarrow \mathbb R$ would be too, which means exactly that $\mathcal N$ is an exponential object between $Y$ and $\mathbb R$.

So it does not fully answer your question (a $\mathcal F$ depending on $X$ could still exist ; I do not have an answer for that), but you should definetely read it. In fact, since the answer to this weaker question is already far from being trivial, I would guess that your question is a difficult one.

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    $\begingroup$ "so it does not answer your question". why isn't this a comment? $\endgroup$ Feb 6, 2020 at 18:54
  • $\begingroup$ It seems that the post you linked is about showing that, in general, we can't find a $\sigma$-algebra $\mathcal{F}$ on $\mathcal{N}$ such that, defining $\Gamma : \mathcal{N}\times Y\to \mathbb{R}, (g,y)\mapsto g(y)$, it turns out that $\Gamma$ is measurable from $(\mathcal{N}\times Y, \mathcal{F}\otimes\mathcal{F}_Y)$ into $\mathbb{R}$. Actually, I'm a bit confused on how this result shows that the question in this post has a negative answer in general... can you please expand your answer showing the connection? $\endgroup$
    – Bob
    Feb 7, 2020 at 9:10
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    $\begingroup$ With some delay, I have just edited my answer to answer your comment. $\endgroup$
    – J. Darné
    Feb 14, 2020 at 12:50

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