How to prove that if $E(X^2) < \infty$ then $E(X) < \infty$?

Question

How to prove that if $E(X^2) < \infty$ then $E(X) < \infty$?

Here is my attempt:

It's easy to show that if $E(X^2) < \infty$ then $E(X) < \infty$ when $X^2 \ge X$ (by using the monotonicity of expectation).

But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as $$ Eg(x) = \int |g(x)|f(x) < \infty? $$

Answer 1

The one reached onto you by the comment of Did is my favourite.

Another one is: $$\mathbb EX^2-(\mathbb EX)^2=\mathbb E(X-\mathbb EX)^2\geq0$$implying that: $$(\mathbb EX)^2\leq\mathbb EX^2<\infty$$

Answer 2

By Jensen inequity for $g(x) = x^ 2$, $$ g(\mathbb{E}(X)) \le E( g(X) ), $$ hence, $$(\mathbb{E}X )^2 \le \mathbb{E}X^2 < \infty, $$ thus $$ \mathbb{E}(X) < \infty $$

Answer 3

Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says $$ E(X)^2\le E\!\left(X^2\right)E(1)=E\!\left(X^2\right) $$