Suppose that $(X,d)$ is a Polish metric space and $A$ is a set of continuous bounded functions $f:X\to \mathbb{R}$. Suppose that $\mu:X\to[0,1]$ is a Borel probability measure. Define

$$\sup A:X\to (-\infty,+\infty],\quad \sup A(x):=\sup\{f(x)\colon f\in A\}, $$ On the other hand, we have the essential suppremum, that is, a Borel measurable function $$\text{ess.sup}_\mu A:X\to (-\infty,+\infty], $$ which is unique up to $\mu$-null sets an such that $$ f\le \text{ess.sup}_\mu A\quad\mu\text{-a.s. for all }f\in A $$ and if another Borel measurable function $Y$ satisfies the condition above, then $\text{ess.sup}_\mu A\le Y$ $\mu$-a.s.

My question is: Is it true that $$\sup A=\text{ess.sup}_\mu A$$ $\mu$-almost surely?

  • $\begingroup$ Maybe I'm being naive, but is there any specific reason to suspect that $\sup A$ is even (essentially) measurable? $\endgroup$ – Jochen Glueck Oct 15 '19 at 20:31
  • $\begingroup$ $\sup A$ is even lower semicontinuous as a supremum of continuous functions, in particular Borel-measurable. $\endgroup$ – Dieter Kadelka Oct 15 '19 at 21:10
  • $\begingroup$ @DieterKadelka: Good point, thank you. $\endgroup$ – Jochen Glueck Oct 15 '19 at 21:55

The claim now is correct.

For the following proof we only need that $X$ is separable metric and $A$ a family of lower semicontinuous (l.s.c.) functions on $X$. (Of course these assumptions may be further weakened.) To simplify notation we may assume that $0 \leq f \leq 1$ for each $f \in A$. Let $g$ be an arbitrary representant of ess sup$_{f \in A} f$ and $h := \sup_{f \in A} f$, which is l.s.c. again, hence Borel-measurable.

By V.I.Bogachev, Measure Theory I (2007), 4.7.1 there is an at most countable subset $\{f_n\} \subset A$ such that $\sup_n f_n = g$ $\mu$-a.e., hence $g \leq h$ $\mu$-a.e. If not $h \leq g$ $\mu$-a.e., then there is $B \in \cal{B}(X)$ with $\mu(B) > 0$ and $g(x) < h(x)$ for $x \in B$. But then there are rational $0 \leq p < q \leq 1$ with $\mu(g \leq p, h \geq q) > 0$. Let $C := \{x \in X \colon g(x) \leq p, h(x) \geq q\}$. Since $X$ is separable metric there is $x \in C$ with $\mu(U \cap C) > 0$ for any open neighbourhood of $x$. Let $f \in A$ with $f(x) > \frac{p+q}{2}$, then $U := \{y \in X \colon f(y) > \frac{p+q}{2}\}$ is an open neighbourhood of $x$, hence $\mu(U \cap C) > 0$. But this contradicts the assumption that $g(y) \leq p$ for $\mu$-a.e. $y \in C$.

  • $\begingroup$ Very nice proof $\endgroup$ – Littlefield Oct 18 '19 at 19:58

In general no. Assume that $X$ is a set with atomless nontrivial probability measure on $\cal{P}(X)$, endowed with the discrete metric, and $A := \{1_{\{x\}} \colon x \in X\}$. Then ess sup $_\mu A \equiv 0$ and $\sup A \equiv 1$. The answer may be true with additional assumptions. (For the existence of such $X$ see f.i. Jech (1978), Set Theory.)

  • $\begingroup$ But, in general, $1_{\{x\}}$ is not continuous. Notice that I assume $A$ a set of continuous bounded functions. It is right that with the trivial metric $1_{\{x\}}$ is continuous. I wanted to include that the metric space is Polish. I have edit the question with a Polish metric space. $\endgroup$ – Littlefield Oct 15 '19 at 16:13
  • $\begingroup$ At least you need another assumption: $Y := supp~ \mu = X$. Otherwise let $f$ be any continuous function on $X$ with $f|Y \equiv 0$, $f \geq 0$ and not $f \equiv 0$. Take $A = \{f\}$. Then again ess sup$_\mu A = 0$ but $\sup A \not\equiv 0$. $\endgroup$ – Dieter Kadelka Oct 15 '19 at 16:43
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    $\begingroup$ I think that this is not a problem. Note that I want $\mu$-a.s. equality. So in your counterexample you have a.s. equality. $\endgroup$ – Littlefield Oct 15 '19 at 19:28
  • $\begingroup$ That's right. Since $X$ is polish $\mu$ is regular and you can assume that $X$ is compact metric and supp $\mu = X$. Maybe this simplifies the proof. Further you may assume that $0 \leq f \leq 1$ for all $f \in A$. $\endgroup$ – Dieter Kadelka Oct 15 '19 at 21:16

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