how to show that $\mathbb{E}[|X|] = 0 \implies \mathbb{P}(X = 0) = 1$?

Question

if $X$ was a discrete random variable then :

$$\mathbb{E}[|X|] = \sum |x_i|\mathbb{P}(X=x_i) = 0\cdot\mathbb{P}(X =0) + \sum_{x_i \neq 0} |x_i|\mathbb{P}(X=x_i) = 0$$

would imply that $\mathbb{P}(X=x_i) = 0, \; \forall x_i \neq 0$

and since $\sum \mathbb{P}(X=x_i) = 1$ we have to have $\mathbb{P}(X = 0) = 1$

but how would you prove this if $X$ was any kind of random variable ?

Answer 1

By Markov's inequality, for all $\epsilon > 0$ $$P(|X| > \epsilon) \le \frac{E[|X|]}{\epsilon} = 0$$ Take limit as $\epsilon \downarrow 0$. $P(|X| > 0) = 0$, so $P(X = 0) = P(|X| = 0) = 1$.