Proving that the function $x^2$ is a Borel function

Question

How should I prove that the function $x^2$ where $x^+=\max(x,0)$, $x^-=-\min(x,0)$ and $|x|=x^+ + x^-$ is a Borel function?

I know that a Borel function is a random variable $\mathbb{R} \rightarrow \mathbb{R}$. But how to mathematically prove it? I have no idea... May be I should use indicators? But how?

Answer 1

$f: \mathbb{R} \rightarrow \mathbb{R}$ is borel $\iff \; \forall B \in \mathcal{B}(\mathbb{R}) : f^{-1}(B) \in \mathcal{B}(\mathbb{R}) \iff \; \forall b \in \mathbb{R} : f^{-1}(-\infty,b) \in \mathcal{B}(\mathbb{R}).$

$$ f^{-1}((-\infty,b)) = \{x : x^2 \in (-\infty,b) \} = \{ x : x^2 < b \}$$

If $b < 0 :$ then $\{ x : x^2 < b \} = \varnothing.$

If $b \geq 0$ then

$$ x^2 < b \iff - \sqrt b < x < \sqrt b.$$

So $$ f^{-1}((-\infty,b))= \begin{cases} \varnothing &\in \mathcal{B}(\mathbb{R}) & b < 0 \\ (-\sqrt b, \sqrt b)& \in \mathcal{B}(\mathbb{R}) & b \geq 0\end{cases}$$ and $f$ is borel.

Consider $f(x) = \vert x \vert.$ Let $b \in \mathbb{R}$:

1. If $b < 0$:

$$ f^{-1}(-\infty ,b) = \{x \in \mathbb{R} : f(x) = \vert x \vert < 0\} = \varnothing$$

Since the empty set is an open (closed) it is a borel set so $ f^{-1}(-\infty ,b) \in \mathcal{B}(\mathbb{R}) \in \forall b < 0$.

2. if $ b \geq 0:$

$$ f^{-1}(-\infty ,b) = \{x \in \mathbb{R} : f(x) = \vert x \vert < b\} = (-b,b)$$

Since $(-b,b)$ is an open set it is a borel set and $ f^{-1}(-\infty ,b) \in \mathcal{B}(\mathbb{R}) \in \forall b \geq 0.$

Answer 2

If you really want to run the details with $x \to x^2$, I suggest that you take a look at this Show that continuous functions on $\mathbb R$ are Borel-measurable

Probably though, it would be smarter to really understand the answer given in this link, and thus to be able to show that this holds for any continuous function on $\mathbb{R}$, which includes the case you are looking for.