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?