Prove: $\operatorname{E}[X^2]<\infty\Longrightarrow \operatorname{E}[X]$ exists

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I don't know how to prove this. Lets assume that $X$ is a discrete random variable. I've just come this far: If we do a direct proof of the implication, then we start with the assumption: $$ \operatorname{E}[X^2]=\sum_{x\in Im(X)} x^2\cdot p_X(x) <\infty. $$ However, I don't know how I can follow from this that $\operatorname{E}[X]$ exists.

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We have $|X|\leq1+X^2$ and consequently $$\mathbb E|X|\leq1+\mathbb EX^2$$

So $\mathbb EX^2<\infty$ allows the conclusion that $\mathbb EX$ exists.

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An even more general statement, shown here, is if $$ \mathbb E(|X^r|)< \infty $$ then for any positive integer $s<r$, the $s$-th moment $\mathbb E[|X^s|]$ is also finite (and thus $\mathbb E[X^s]$ exists).

This can be done by noticing that $0<s<r \Longrightarrow |X|^s \le \max(1, |X|^r) $.

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Here's another; Use the Cauchy-Schwarz inequality.

$$E|X| = E|X|\cdot 1 \le \|1\|_2 \|X\|_2 = \left(E|X|^2\right)^{1/2}.$$

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I'm a big fan of Jenson's inequality for probability measures, for $f(x)=x^r$ with $r\geq 1$, implies:

$$f(E[X]])\leq E[f(X)],$$

which gives, after rearranging:

$$E[X]\leq E[X^r]^{1/r}$$