Supremum of continuous functions and essential supremum of continuous functions

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?

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

• Very nice proof – 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.)

• 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. – Littlefield Oct 15 '19 at 16:13
• 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 $\sup A \not\equiv 0$. – Dieter Kadelka Oct 15 '19 at 16:43
• 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. – Littlefield Oct 15 '19 at 19:28
• 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 $f \in A$. – Dieter Kadelka Oct 15 '19 at 21:16