If the expected value is on the boundary of the range, then the random variable is a.s. constant

Question

Let $X$ be a real-valued random variable on $\Omega$, $I\subseteq\mathbb{R}$ be an interval, $X(\Omega)\subseteq I$ and $E[|X|]<\infty$.

Why does $E[X]\in\partial I$ imply that $X=E[X]$ almost surely?

---

$\partial M$ denotes the boundary of a subset $M$ of a topological space.

Answer 1

Let $I=[a,b]$ where $a<b$.There are two cases:

Case 1. $E[X]=a$. Consider random variable $Y=X-E[X]$. Because $X(\Omega) \subset I$ we have $Y(x)\geq 0$ for $x \in \Omega$. Next note that:

$$E[Y]=E[X-E[X]]=E[X]-E[E[X]]=E[X]-E[X]=0$$

Now using standard fact from measure theory (if $f \geq 0$ and $\int_{\Omega}fd\mu=0$ then $f=0$ almost surely) you have $Y=0$ almost sure, so $E[X]=X$ almost surely. ly Case 2. $E[X]=b$. Proof is similar to case 1, but for $Y=E[X]-X$.