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

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?

• You are almost there. Now use the fact that $f$ is measurable – Eduardo Longa Jun 23 '19 at 1:15
• (Small remark: You're assuming $c\ge 0$ here. What happens if $c<0$?) Hard to know why you're stuck, though. From your other post, it appears you know the definition of measurability. – Ted Shifrin Jun 23 '19 at 1:17
• @TedShifrin i know this set is measurable $\{x:f(x)>\sqrt{c}\}$ but how can i know $\{x:f(x)<-\sqrt{c}\}$ is measurable? – Bvss12 Jun 23 '19 at 1:19
• If you only know one sort of set works, how do you write the second set in a different way? – Ted Shifrin Jun 23 '19 at 1:24
• oh, you have reason. $\{x:f(x)<-\sqrt{c}\}=\{x:-f(x)>\sqrt{c}\}$ and this is a measurable @TedShifrin – Bvss12 Jun 23 '19 at 1:30

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