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Let $(\mathbb{X}, \mathcal{X})$ and $(\mathbb{R}, \mathcal{R})$ denote a metric space and the real line, both endowed with their Borel $\sigma$-algebras. Let $\mathcal{X}\otimes\mathcal{R}$ denote the usual product $\sigma$-algebra and $f: \mathbb{X} \times\mathbb{R}\mapsto [-\infty, \infty]$ be a $\mathcal{X}\otimes\mathcal{R}$-measurable function.

Q.1 Is it true that the map $x \mapsto \sup_{y \in \mathbb{R}}|f(x,y)|$ is $\mathcal{X}$-measurable?

The question might be silly, though I'm not managing to get a final answer. In particular, we know that, under the previous hypothesis, the map $x \mapsto f(x,y)$ is $\mathcal{X}$-measurable for every $y \in \mathbb{R}$. Thus, for any $\epsilon>0$, we have $B_{\epsilon,y}:=\{x \in \mathbb{X}: |f(x,y)|< \epsilon\} \in \mathcal{X}$. Though, we can not guarantee that the uncountable intersection $B_\epsilon:=\cap_{y \in \mathbb{R}}B_{\epsilon,y}$ is still contained in $\mathcal{X}$. Whence I would conclude that the answer to Q.1 is: no. Is this correct?

If so:

Q.2 are there additional conditions we can assume on $\mathbb{X}$ in order to have a positive answer? Can we obtain such a positive answer without requiring the continuity of the maps $y \mapsto f(x,y)$, $\forall x \in \mathbb{X}$?

Jack London
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1 Answers1

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Q1: In general, no, not even when $\mathbb{X} = \mathbb{R}$. There exists a Borel subset $A \subset \mathbb{R}^2$ whose projection $\pi(A)$ onto the first coordinate is not a Borel set. So taking $f$ to be the indicator function of $A$, the map $x \mapsto \sup_y |f(x,y)|$ is exactly the indicator of $\pi(A)$, and thus is not a Borel function.

Q2: Since $\mathbb{R}$ is about the nicest nontrivial space in measure theory, I don't think you will find any useful conditions on the space $\mathbb{X}$ to make this true. I am not sure offhand as to conditions on $f$ that would help.

Nate Eldredge
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  • I was thinking to the following: let's replace $\mathbb{R}$ with $(0,1)$ and consider the sets $B_{\epsilon, n}=\cap_{i=1}^{n-1}{x \in \mathbb{X}: |f(x, i/n)| < \epsilon}$. Clearly, $B_{\epsilon, n} \in \mathcal{X}$. We also have $\cap_{n \geq 1} B_{\epsilon,n} \in \mathcal{X}$. Could this pave the way to prove that $x \mapsto \sup_{y \in (0,1)}|f(x,y)|$ is $\mathcal{X}$-measurable? – Jack London Mar 16 '20 at 08:22
  • @JackLondon: That will only get you as far as showing that $x \mapsto \sup_{y \in (0,1) \cap \mathbb{Q}} |f(x,y)|$ is measurable, which in general tells you very little about the sup over all of (0,1). Unless of course $f$ is continuous in $y$. – Nate Eldredge Mar 16 '20 at 12:30
  • @JackLondon: A positive result, however, is that if $\mathbb{X}$ is Polish, then your function will be universally measurable; i.e. for every $\sigma$-finite Borel measure $\mu$ on $\mathbb{X}$, your function is $\mu$-a.e. equal to a Borel function (where the null set may of course depend on $\mu$). This is because the sets ${x : \sup_y |f(x,y)| > a}$, while not necessarily Borel, are analytic. – Nate Eldredge Mar 16 '20 at 12:36