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?

  • 1
    $\begingroup$ You are almost there. Now use the fact that $f$ is measurable $\endgroup$ – Eduardo Longa Jun 23 '19 at 1:15
  • $\begingroup$ (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. $\endgroup$ – Ted Shifrin Jun 23 '19 at 1:17
  • $\begingroup$ @TedShifrin i know this set is measurable $\{x:f(x)>\sqrt{c}\}$ but how can i know $\{x:f(x)<-\sqrt{c}\}$ is measurable? $\endgroup$ – Bvss12 Jun 23 '19 at 1:19
  • $\begingroup$ If you only know one sort of set works, how do you write the second set in a different way? $\endgroup$ – Ted Shifrin Jun 23 '19 at 1:24
  • $\begingroup$ oh, you have reason. $\{x:f(x)<-\sqrt{c}\}=\{x:-f(x)>\sqrt{c}\}$ and this is a measurable @TedShifrin $\endgroup$ – Bvss12 Jun 23 '19 at 1:30

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.


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