2
$\begingroup$

$X\subset \mathbb{R}^d$, uncountable, not necessarily compact.

$(Y,\mathcal{F}_Y,\mu)$: measure space.

$f\colon\,X\times Y\to \mathbb{R}$ , continuous in $X$ for each $\mu$-a.e. $y$,, and $\mathcal{F}_Y/\mathcal{B}(\mathbb{R})$-measurable for each $x$.

Letting $g(y):=\sup_{x\in X}f(x,y)$, if $g(y)<\infty$ $\mu$-a.e., is this $\mathcal{F}_Y/\mathcal{B}(\mathbb{R})$-measurable?

My attempt:

Following, Measurability of the supremum of a Brownian motion I considered $$\sup_{x\in X\cap \mathbb{Q}^d}f(x,y),$$ noting that $\mathbb{Q}^d$ is dense in $(\mathbb{R},\|\cdot\|_2)$ (by $\|\cdot\|_2$ I mean Euclidean norm topology).

But I got confused by this question: Supremum over dense subset of banach space does $X$ need to be compact?


I have asked a similar question before on essential supremum: Is ess sup of product measurable function measurable?. For the supremum case I had the property of Caratheodory functions (18.19 in Infinite Dimensional Analysis: A Hitchhiker's Guide By Charalambos D. Aliprantis, Kim C. Border). But for that it seems $X$ needs to be compact.

I see many questions such as Supremum of a product measurable function..., but it doesn't seem they answer my question.

$\endgroup$
3
  • 1
    $\begingroup$ The compactness is probably to guarantee $g(y)$ is not infinite everywhere. $\endgroup$ – Jacky Chong Oct 1 '16 at 8:06
  • $\begingroup$ Thanks, I edited the question. $\endgroup$ – shall.i.am Oct 1 '16 at 8:14
  • $\begingroup$ I think I've solved your question. See my answer below. $\endgroup$ – Evan Aad Oct 1 '16 at 9:52
1
$\begingroup$

Yes, $g$ is measurable, whether or not $g$ is $\mu$-a.e. finite, whether or not $X$ is compact, and whatever $X$'s cardinality is. Moreover, we shall require $X$ to be merely a separable metric space, not necessarily $\mathbb{R}^d$.

We shall assume, w.l.g., that $x\mapsto f(x,y)$ is continuous for all $y \in Y$.

Let $D$ be a countable, dense subset of $X$ (every subspace of a separable metric space is separable too). Then, for every $y \in Y$, since $x \mapsto f(x,y)$ is a continuous function $X\rightarrow\mathbb{R}$, we have $\sup_{x\in X} f(x,y) = \sup_{x\in D} f(x,y)$. Therefore, $g = \sup_{x\in D} f(x,\cdot)$. This is the supremum of a countable number of $\mathcal{F}_Y/\mathcal{B}(\mathbb{R})$-measurable functions, and hence is $\mathcal{F}_Y/\mathcal{B}(\overline{\mathbb{R}})$-measurable, where $\overline{\mathbb{R}}$ is the extended real line. If, additionally, $g$ is finite, it is $\mathcal{F}_Y/\mathcal{B}(\mathbb{R})$-measurable.


As per OP's request, I will show that the result does not necessitate $X$ being compact.

Claim Let $(\Omega,\tau)$ be a topological space, and let $D$ be a $\tau$-dense subset of $\Omega$. Denoting the Euclidean topology on $\mathbb{R}$ by $\mathcal{E}$, let $f:\Omega\rightarrow \mathbb{R}$ be a $\tau/\mathcal{E}$-continuous function. Then $\sup_{x\in \Omega} f(x) = \sup_{x\in D}f(x)$.

Proof

Set $$ \begin{align} s_1 &:= \sup_{x\in \Omega} f(x) \\ s_2 &:= \sup_{x\in D}f(x). \end{align} $$ We wish to show that $s_1 = s_2$.

Since $D\subseteq \Omega$, $s_1 \geq s_2$. As for the other direction, it suffices to show that, for every $\varepsilon \in (0,\infty)$, $s_2 > s_1-\varepsilon$. Let then $\varepsilon \in (0,\infty)$. Choose $y \in \Omega$ such that $f(y) > s_1 - \varepsilon/2$. Since $f$ is $\tau/\mathcal{E}$-continuous at $y$, there is some $\tau$-neighborhood of $y$, $G$, such that, for every $z \in G$, $|f(y)-f(z)| < \varepsilon/2$. Let $z \in G\cap D$ ($D$ being dense in $\Omega$, $G\cap D \neq \emptyset$). Then $s_2 \geq f(z) > s_1-\varepsilon$, Q.E.D.

$\endgroup$
10
  • $\begingroup$ Ah, I was editing my question. Yeah I was thinking of the essentially the same thing, but found "Supremum over dense subset of banach space" (math.stackexchange.com/questions/535589/…) and got confused. $\endgroup$ – shall.i.am Oct 1 '16 at 9:59
  • $\begingroup$ So I guess the important thing to say sup taken over $X$=sup taken over $D$ is that the continuity of $\mathbb{R}$, not $X$, i.e., in the question above $\sup \overline{B}=\sup B$ for $B\subset\mathbb{R}$? $\endgroup$ – shall.i.am Oct 1 '16 at 10:10
  • $\begingroup$ I'm sorry, but I do not understand what you are trying to say. Is there any problem with my answer? If so, please indicate what the problem is, and I will do my best to rectify it. If not, please consider accepting it. Thanks. $\endgroup$ – Evan Aad Oct 1 '16 at 10:12
  • 1
    $\begingroup$ @shall.i.am: I see what you're saying now. I don't think this is a problem. I'll add an appendix to my answer proving this is not a concern, but it may take a couple hours, because I have to leave soon. $\endgroup$ – Evan Aad Oct 1 '16 at 10:28
  • 1
    $\begingroup$ It is so much clear now, thank you! Regarding the existence of a countable dense subset $D$, now 1. in my question I have $\Omega:=X\subset \mathbb{R}^d$ (a subset of a separable metric space $\mathbb{R}^d$), and 2. subsets of subsets of a separable metric space are separable (and thus by def. of separability there exists a countable dense subset)? $\endgroup$ – shall.i.am Oct 1 '16 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.