Let $X$ be a measurable function on a probability space $(\Omega,\mathcal A,P)$ valued in a metric space $(F,d)$ and let $1_A (\cdot)$ denote the indicator function on a set $A$. Denote by $B(x,h)$ the closed ball centered at $x$ with radius $h$ w.r.t. the metric $d$.
For a compact set $S\subseteq F$, I know that $E(\sup_{x\in S }1_{B(x,h)}(X))\geq \sup_{x\in S }E(1_{B(x,h)}(X))$. However, I have an intuition that both are equal.
Question: Does $E(\sup_{x\in S }1_{B(x,h)}(X))= \sup_{x\in S }E(1_{B(x,h)}(X))$ hold?
Consider $X : ([0, 1], \mathcal{B}, m) \rightarrow [0, 1]$ such that $X(\omega) = \omega$. Let $S = [0, 1]$ and $h = \frac{1}{4}$. Then $\sup_{x \in S} 1_{B(x, h)}(X) = 1$, but $E(1_{B(x, h)}(X)) = m(X \in B(x, h)) \leq \frac{1}{2}$ for all $x \in [0, 1]$.