Square-integrable implies integrable on a probability space

$L^2 \subseteq L^1$ on a probability space. If $Z$ is measurable and $Z^2$ is integrable with respect to a probability measure, then $Z$ is integrable.

This is the simplest instance of the inclusion $L^q \subseteq L^p$ for $p \le q$ on a finite measure space. The finiteness of the measure is essential: on the line with Lebesgue measure, $Z(x) = 1/(1+|x|)$ is square integrable but not integrable.

Proof. From $(|z| - 1)^2 \ge 0$ one gets the pointwise bound $|z| \le (1 + z^2)/2$. The right-hand side, as a function of $\omega$, is integrable — the constant is integrable because the measure is finite, and $Z^2$ is integrable by hypothesis — so Integrable.mono applies.