Show that $f^{2}$ is measurable on the interval I

Question

Let f be a measurable function on the interval I. Show that $f^{2}$ is measurable on the interval I.

By definition, function $f$ is Lebesgue measurable on an interval $I$ if for every $s\in\mathbb{R}$ the set $\{x\in I\mid f(x)>s\}$ is a Lebesgue measurable set.

So I have to show that: $\{x\in I\mid f(x)^{2}>s\}$ .

Can anyone help?

Answer 1

$|f(x)| \leq s$ iff $-s \leq f(x) \leq s$. So $\{x: f(x)^{2} \leq s\}=\{x: |f(x)| \leq \sqrt s\}$ is measurable.

Answer 2

Let's consider a more general problem:

Let $f:X \rightarrow Y$ and $g:Y\rightarrow Z$ be measurable functions, where $(X,\mathcal{A}),(Y,\mathcal{B})$ and $(Z,\mathcal{C})$ are measurable spaces. Can we say that $g\circ f:X \rightarrow Z$ is a measurable function?

For any $C\in \mathcal{C}$ we have, that $$(g\circ f)^{-1} (C)= f^{-1}(g^{-1}(C)),$$ where $g^{-1}(C) \in \mathcal{B}$ by measurability of $g$ and therefore $f^{-1}(g^{-1}(C))\in \mathcal{A}$ by measurability of $f$, which proves that $g\circ f$ is measurable.

And how does this solve your problem? Well clearly, since $g(y)=y^2$ is continuous, it is also (Borel-)measurable and we can immediately conclude that $g\circ f = f^2$ is measurable as well.