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I believe so, and was trying to prove it. Here's my attempt by contradiction:

Let $A=\{\omega:|X(\omega)|>0\}$, and let $P(A)>0$. Then

$$ E[|X|]=\int_{\Omega}|X|dP=\int_{A}|X|dP $$ But I am not sure how to proceed further, I'd like to bound $X$ by infimum on $A$, but this infimum can be zero.

Andrey Carl
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  • Remember $L^1$ is a complete metric space, so if the norm is $0$, the function is in the equivalence class of $0$, i.e. it is $0$ a.s. If you don't have enough functional analysis, approximate $X$ by simple functions from below. – Adam Hughes Jun 07 '16 at 02:41
  • Keep in mind also that this is a general result for Lebesgue integration, i.e. it is valid for all measures, and not just probability measures http://math.stackexchange.com/questions/1270639/prove-that-the-lebesgue-integral-of-f-chi-equals-to-0-indicates-that-f-0-a – Chill2Macht Jun 13 '16 at 14:57

1 Answers1

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You are right that if $E[ |X| ] = 0$, then $X = 0$ a.s. (actually, it's an "if and only if").

Suppose $E[|X| ] = 0$. We want to show $P(A) = 0$, where $A = \{ \omega \mid X(\omega) \neq 0 \}$. Note that $\{ \omega \mid X(\omega) \neq 0\} = \{ \omega \mid |X(\omega)| \neq 0 \}$.

Now, $P(\{ \omega \mid |X(\omega)| \neq 0 \}) = P( \bigcup \limits_{n = 1}^{\infty} \{ \omega \mid |X(\omega)| \geq \frac{1}{n} \}) \leq \sum \limits_{n = 1}^{\infty} P(\{\omega \mid |X(\omega)| \geq \frac{1}{n}\})$

$ = \sum \limits_{n = 1}^{\infty} \int \limits_{\{\omega \mid |X(\omega)| \geq \frac{1}{n}\}} \,dP \leq \sum \limits_{n = 1}^{\infty} \int \limits_{\{\omega \mid |X(\omega)| \geq \frac{1}{n}\}} n |X(\omega)|\,dP \leq \sum \limits_{n = 1}^{\infty} n\int \limits_{\Omega} |X(\omega)|\,dP = \sum \limits_{n = 1}^{\infty} nE[|X|] = 0.$

layman
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