Let $f$ a measurable function, then $f^2$ is a measurable function, $f:X\rightarrow\bar{\mathbb{R}}$

Question

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?

Answer 1

You can do this in a more streamlined way. In fact, we can prove something more general:

Let $h(x)=f(x)g(x)$ for two measurable functions $f$ and $g$. We can prove that $fg$ is measurable.

Let $U_a=\{(x,y):xy>a\}.\ U_a$ is open in $\mathbb R^2$ so it is a countable union of basis elements of the form $I_n=(a_n<x<b_n)\times (c_n<y<d_n).$

Now, the sets $\{x: a_n<f(x)<b_n\}$ and $\{x: a_n<g(x)<b_n\}$ are measurable, so we have that $\{x:(f(x),g(x))\in I_n\}=\{x:a_n<f(x)<b_n\}\cap \{x:a_n<g(x)<b_n\}$ are measurable.

By construction, $(f(x),g(x))\in \bigcup_n I_n\Leftrightarrow f(x)g(x)>a.$

Putting this together, we get $\{x:f(x)g(x)>a\}=\bigcup_n \{x:(f(x),g(x))\in I_n\}$ is measurable, being the countable union of measurable sets.