measure theory - Let $f$ a measurable function, then $f^2$ is a measurable function, $f:Xrightarrowbar{mathbb{R}}$

Let $f$ a measurable function, then $f^2$ is a measurable function, $f:X\rightarrow\bar{\mathbb{R}}$ and
$\mathbb{A}$ a sigma-algebra of sets.

My attempt

Note $x\in(f^2)^{-1}(c,\infty)=\{x:f^2(x)>c\}=\{x:f(x)>\pm\sqrt{c}\}=\{x:f(x)>\sqrt{c}\}\cup\{x:f(x)<-\sqrt{c}\}$

Here i'm stuck. Can someone help me?