# For a real-valued random variable it holds: $E(|X|)<\infty\Leftrightarrow \sum_{n\in\mathbb{N}}P(|X|>n)<\infty$

Let $(\Omega,\mathcal{A},P)$ be a measurable space and $X$ be a real-valued random variable on $\Omega$. I want to show that it holds: $$E(|X|)<\infty\Leftrightarrow \sum_{n\in\mathbb{N}}P(|X|>n)<\infty$$

Let's try the first direction: From definition we've got $$E(|X|)=\int |X|\;dP=\sup_{g\in\mathbb{E}_+\;:\;g=|X|}\sum_{i=1}^n\alpha_iP(A_i)<\infty$$ where $$\mathbb{E}_+:=\left\{g=\sum_{i=1}^n\alpha_i1_{A_i} : \alpha_1,\ldots,\alpha_n>0,A_1,\ldots A_n\in\mathcal{A}\text{ pairwise disjoint}\right\}$$

While the statement sounds intuitively correct to me, I'm unable to figure out how I need to proceed from here.

Why are you trying to prove this from first-principles. Are you not allowed to assume the definition of an integral? The proof is trivial from the fact that for a non-negative random variable $Y$, $E[Y]=\int_0^\infty (1-F(y))dy=\int_0^\infty P(Y>x)$, where $F(y)$ is the cdf. You can prove this fact by writing $P(Y>x)=\int_x^\infty d\mu-P(Y=x)$ where $\mu$ is the push-forward of $P$ onto $\mathbb{R}$ and then (carefully!) switching order of integration. Then the result you want follows by bounding the integral by a sum, since $P(Y>x)$ is monotonically decreasing.
• The fist part, i.e. $$E(|X|)=\int_0^\infty P\left(|X|>t\right)dt$$ as well as the fact that $P\left(|X|>t\right)$, is absolutely clear to me. However, how can we bound the integral by a sum? – 0xbadf00d Nov 2 '14 at 19:08
• Or is $$\int_0^\infty P\left(|X|>t\right)dt=\sum_{n\in\mathbb{N}}\int_{n-1}^n P\left(|X|>t\right)dt$$ what you mean? – 0xbadf00d Nov 2 '14 at 19:12
• $P(|X|>t)$ is a monotonically decreasing function of $t$. By a series comparison test the integral is bounded above by the sum over $t\in\mathbb{N}$. en.wikipedia.org/wiki/Integral_test_for_convergence – Alex R. Nov 2 '14 at 19:14