Let $X \geqslant 0$ be a random variable on a probability space $\left( {\Omega ,\mathcal{F},\mathbb{P}} \right)$. Show that $\mathbb{E}\left[ X \right] < + \infty $ if and only if $\sum\limits_{n \in \mathbb{N}} {n\mathbb{P}\left( {n - 1 \leqslant X < n} \right)} < + \infty $.
My solution: $$\mathbb{E}\left[ X \right] = \mathbb{E}\left[ {\sum\limits_{n \in \mathbb{N}} {X{1_{n - 1 \leqslant X < n}}} } \right] = \sum\limits_{n \in \mathbb{N}} {\mathbb{E}\left[ {X{1_{n - 1 \leqslant X < n}}} \right]} \leqslant \sum\limits_{n \in \mathbb{N}} {\mathbb{E}\left[ {n{1_{n - 1 \leqslant X < n}}} \right]} = \sum\limits_{n \in \mathbb{N}} {n\mathbb{P}\left( {n - 1 \leqslant X < n} \right)} < + \infty $$
where the second equality follows from Lebesgue monotone convergence theorem and the inequality from the monotonicity of expectation. How to show that the converse also holds?
$\mathbb{E}\left[ X \right] < + \infty \Leftrightarrow \mathbb{E}\left[ {X + 1} \right] < + \infty $ and $$\mathbb{E}\left[ {X + 1} \right] = \mathbb{E}\left[ {\sum\limits_{n \in \mathbb{N}} {\left( {X + 1} \right){1_{n - 1 \leqslant X < n}}} } \right] = \sum\limits_{n \in \mathbb{N}} {\mathbb{E}\left[ {\left( {X + 1} \right){1_{n - 1 \leqslant X < n}}} \right]} \geqslant \sum\limits_{n \in \mathbb{N}} {n\mathbb{P}\left( {n - 1 \leqslant X < n} \right)} $$
Again, the second equality follows from the LMCT and the inequality from the monotonicity of expectation and the fact that $X \geqslant 0$. So, $\mathbb{E}\left[ X \right] < + \infty \Rightarrow \mathbb{E}\left[ {X + 1} \right] < + \infty $ which implies $$\sum\limits_{n \in \mathbb{N}} {n\mathbb{P}\left( {n - 1 \leqslant X < n} \right)} \leqslant \mathbb{E}\left[ {X + 1} \right] < + \infty $$ by the inequality above.