probability - show $mathbb{E} vert X vert = int_0^infty mathbb{P}({vert X vert>y})dy leq sum_{n=0}^inftymathbb P {vert X vert>y} $.

Looking for a hint to show $\mathbb{E} \vert X \vert = \int_0^\infty \mathbb{P}(\{\vert X \vert>y\})dy \leq \sum_{n=0}^\infty\mathbb P \{\vert X \vert>y\} $.

This is from Theorem 2.3.7 in Durrett (Probability: Theory and examples)

The first equality makes sense by Fubini and the definition of expectation (Durrett Lemma 2.2.8). I'm having a hard time showing the second, though. My gut intuition makes me feel like it should be the other direction.

Answer

As you mention, the first equation is essentially Fubini, along with rewriting $\mathbb{P}\{\lvert X\rvert > y\}$ as an integral.

For the inequality on the right: observe that
$$\begin{align}
\int_0^\infty \mathbb{P}\{ \lvert X\rvert > y\} dy
&= \sum_{n=0}^\infty \int_n^{n+1} \mathbb{P}\{ \lvert X\rvert > y\} dy \leq
\sum_{n=0}^\infty \int_n^{n+1} \mathbb{P}\{ \lvert X\rvert > n\} dy \\
&= \sum_{n=0}^\infty \mathbb{P}\{ \lvert X\rvert > n\}

\end{align}$$
using the fact that for $y \geq x$, $\mathbb{P}\{ \lvert X\rvert > y\} \leq \mathbb{P}\{ \lvert X\rvert > x\}$ since $\{ \lvert X\rvert > y\} \subseteq \{ \lvert X\rvert > x\}$.