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?
Best Answer
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. $y \in C$.