$|X|$ is $\sigma(X^2)$-measurable I am looking for a proof of the statement in the title. $X$ is a random variable defined on some measure space.
My thinking was as follows. This statement is equivalent to saying  $\sigma(|X|) \subseteq \sigma(X^2)$. The definition is that $\sigma(|X|) = \{|X| \in B\}$ and $\sigma(X^2) = \{X^2 \in B\}$ for Borel sets $B$ of the real line. Here I assume the following. If I only look at Borel sets of the form $B = (a,\infty]$ where $a>0$, the proof will be complete since the I can generate the other Borel sets by taking unions of the ones of the form above. I don't know if this way of thinking is correct.
So then I state the following. $|X| \in (a,\infty]$ iff $X^2 \in (a^2,\infty]$. So their $\sigma$-algebras are the same. Therefore, one is measurable with respect to the other. Can someone check if this proof is solid?

I would prove the following more general fact:

Let $Y$ be any random variable and suppose $f : \mathbb{R} \to \mathbb{R}$ is Borel measurable.  Then $f(Y)$ is $\sigma(Y)$-measurable.

This is very easy to prove once you notice that $\{f(Y) \in B\} = \{ Y \in f^{-1}(B)\}$.
Then apply with $Y = X^2$ and $f(x) = \sqrt{|x|}$.
