$X\in L^2$ implies $Y:=(X-E[X])^2 \in L^1$

Question

Given the random variable $X \in L^2$, we define a new variable $Y:=(X-E[X])^2$, where $E$ is the expectation. Why can we conclude that $Y\in L^1$?

Is it because for any function $f \in L^2$ the norm is $(\int_X|f|^2)^{\frac{1}{2}}$, so the norm squared would be a new function $g \in L^2$ with $(\int_X|g|)$?

Answer 1

$$E|Y| = E[(X - E[X])^2] = E[X^2] - E[X]^2 \le E[X^2] < \infty.$$