Is it possible that $X>\operatorname E[X]$ for a random variable $X$?

Question

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $X\in\mathcal L^1(\operatorname P)$. I'm wondering whether or not we can prove that $\exists\omega\in\Omega$ with $$X(\omega)\le\operatorname E[X]\;.\tag 1$$ Assuming that $X>\operatorname E[X]$ doesn't immediately (by taking the expectation) yield a contradiction, while the statement itself seems to be trivially true. So, can we show $(1)$ and if so, how?

Answer 1

Suppose that $X\in L^1$ satisfies $X > \mathbb{E}[X]$ a.s. In particular, there exists $a> \mathbb{E}[X]$ such that $p\stackrel{\rm def}{=} \mathbb{P}\{X \geq a\} > 0$.

Then $$\begin{align} \mathbb{E}[X] &\geq a\cdot \mathbb{P}\{X \geq a\} + \mathbb{E}[X]\cdot \mathbb{P}\{X < a\} = a\cdot p + \mathbb{E}[X]\cdot (1-p) \\ &= \mathbb{E}[X]+\underbrace{(a-\mathbb{E}[X])}_{>0}\cdot p > \mathbb{E}[X] \end{align}$$ which is a contradiction.