$X\le n\;\; P. a.e$ and $\Bbb{E}(X)=n$ implies $X=n$ a.e?

Question

Let $X$ be a discrete random variable on a probability space $(\Omega,\mathcal{F},P).$

Suppose that $X\le n\;\; P. a.e$ and $\Bbb{E}(X)=n.$

Does it follow that $X=n$ a.e. ?

it's seems true but I don't know how to prove it.

Answer 1

The alternative is that $X < n$ on a set of positive probability. But then $$n= {\rm E}[X] = \int_{X<n}X\,{\rm dP} + \int_{X = n}X\,{\rm dP} < n\cdot \underbrace{{\rm P}[X < n]}_{\ne 0} + n\cdot {\rm P}[X = n] = n.$$

Answer 2

Define $X_n :=n-X\ge0$ a.s.. But $X_n$ has zero expectation, hence $X_n=0$ a.s.. (Note that "discreteness" of $X$ is not used in this proof).