Let $X$ be a random variable and $c$ be a fixed constant, $c>0$. Then $\mathscr{E}|X|<\infty$ if and only if $\sum_{n=1}^\infty \mathscr{P}\{|X|\geq cn\}$ converges.
$\bullet$ Proof.
Given $c>0$, define series of sets $\Lambda_n=\{cn\leq|X|<c(n+1)\}$, and since $\Lambda_n$'s are disjoint, we have, by additivity property $$\mathscr{E}(|X|)=\int_{\cup_n\Lambda_n}|X|d\mathscr{P}=\sum_n\int_{\Lambda_n}|X|d\mathscr{P}.$$Thus, by mean value theorem, for each $\Lambda_n$, $$\sum_{n=0}^\infty cn\mathscr{P}(\Lambda_n)=c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n)\leq\mathscr{E}(|X|)\leq\sum_{n=0}^\infty (cn+c)\mathscr{P}(\Lambda_n)=c+c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n).$$Then we need to show $$c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n)=\sum_{n=0}^\infty\mathscr{P}(|X|\geq cn).$$
For $N\geq1$ and finite, $$\begin{array}{rl} c\sum_{n=0}^N n\mathscr{P}(\Lambda_n)
& = c\sum_{n=0}^N n\{\mathscr{P}(|X|\geq cn)-\mathscr{P}(|X|\geq c(n+1))\} \\
& = c\sum_{n=1}^N (n-(n-1))\mathscr{P}(|X|\geq cn) - cN\mathscr{P}(|X|\geq c(N+1)) \\
& = c\sum_{n=1}^N \mathscr{P}(|X|\geq cn) - cN\mathscr{P}(|X|\geq c(N+1)), \end{array}$$since for $k=0$, $k\{\mathscr{P}(|X|\geq ck)-\mathscr{P}(|X|\geq c(k+1))\}=0$ and for $1\leq k\leq N$, $$\begin{array}{rl}
k = 1 &\rightarrow \{\mathscr{P}(|X|\geq c)-\mathscr{P}(|X|\geq 2c)\} \\
k = 2 &\rightarrow 2\{\mathscr{P}(|X|\geq 2c)-\mathscr{P}(|X|\geq 3c)\} \\
\vdots & \\
k = N-1 &\rightarrow (N-1)\{\mathscr{P}(|X|\geq c(N-1))-\mathscr{P}(|X|\geq cN)\} \\
k = N &\rightarrow N\{\mathscr{P}(|X|\geq cN)-\mathscr{P}(|X|\geq c(N+1))\}. \end{array}$$Hence, $$c\sum_{n=0}^N n\mathscr{P}(\Lambda_n)\leq c\sum_{n=1}^N \mathscr{P}(|X|\geq cn)\leq c\sum_{n=0}^N n\mathscr{P}(\Lambda_n)+cN\mathscr{P}(|X|\geq c(N+1)).$$And $$\begin{array}{rl} cN\mathscr{P}(|X|\geq c(N+1))
& = cN\int_{\{|X|\geq c(N+1)\}}d\mathscr{P} \\
& \leq c(N+1)\int_{\{|X|\geq c(N+1)\}}d\mathscr{P} \\
& \leq \int_{\{|X|\geq c(N+1)\}}|X|d\mathscr{P}. \end{array}$$We have if $\mathscr{E}(|X|)<\infty$, then $ cN\mathscr{P}(|X|\geq c(N+1)) \rightarrow 0$ as $N\rightarrow\infty$ since $\{|X|\geq c(N+1)\}\rightarrow\emptyset$, then $c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n)=\sum_{n=0}^\infty\mathscr{P}(|X|\geq cn).$ If $\mathscr{E}(|X|)=\infty$, then $c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n) = \infty$ leads to $\sum_{n=0}^\infty\mathscr{P}(|X|\geq cn)=\infty.$
For $N\geq1$ and finite, $$\begin{array}{rl} c\sum_{n=0}^N n\mathscr{P}(\Lambda_n)
& = c\sum_{n=0}^N n\{\mathscr{P}(|X|\geq cn)-\mathscr{P}(|X|\geq c(n+1))\} \\
& = c\sum_{n=1}^N (n-(n-1))\mathscr{P}(|X|\geq cn) - cN\mathscr{P}(|X|\geq c(N+1)) \\
& = c\sum_{n=1}^N \mathscr{P}(|X|\geq cn) - cN\mathscr{P}(|X|\geq c(N+1)), \end{array}$$since for $k=0$, $k\{\mathscr{P}(|X|\geq ck)-\mathscr{P}(|X|\geq c(k+1))\}=0$ and for $1\leq k\leq N$, $$\begin{array}{rl}
k = 1 &\rightarrow \{\mathscr{P}(|X|\geq c)-\mathscr{P}(|X|\geq 2c)\} \\
k = 2 &\rightarrow 2\{\mathscr{P}(|X|\geq 2c)-\mathscr{P}(|X|\geq 3c)\} \\
\vdots & \\
k = N-1 &\rightarrow (N-1)\{\mathscr{P}(|X|\geq c(N-1))-\mathscr{P}(|X|\geq cN)\} \\
k = N &\rightarrow N\{\mathscr{P}(|X|\geq cN)-\mathscr{P}(|X|\geq c(N+1))\}. \end{array}$$Hence, $$c\sum_{n=0}^N n\mathscr{P}(\Lambda_n)\leq c\sum_{n=1}^N \mathscr{P}(|X|\geq cn)\leq c\sum_{n=0}^N n\mathscr{P}(\Lambda_n)+cN\mathscr{P}(|X|\geq c(N+1)).$$And $$\begin{array}{rl} cN\mathscr{P}(|X|\geq c(N+1))
& = cN\int_{\{|X|\geq c(N+1)\}}d\mathscr{P} \\
& \leq c(N+1)\int_{\{|X|\geq c(N+1)\}}d\mathscr{P} \\
& \leq \int_{\{|X|\geq c(N+1)\}}|X|d\mathscr{P}. \end{array}$$We have if $\mathscr{E}(|X|)<\infty$, then $ cN\mathscr{P}(|X|\geq c(N+1)) \rightarrow 0$ as $N\rightarrow\infty$ since $\{|X|\geq c(N+1)\}\rightarrow\emptyset$, then $c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n)=\sum_{n=0}^\infty\mathscr{P}(|X|\geq cn).$ If $\mathscr{E}(|X|)=\infty$, then $c\sum_{n=0}^\infty n\mathscr{P}(\Lambda_n) = \infty$ leads to $\sum_{n=0}^\infty\mathscr{P}(|X|\geq cn)=\infty.$
$\Box$
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