# How can I prove that one function is a Borel-measurable function of another?

I am referring to Borel-measurability as it relates to probability theory. The definition used: A Borel-measurable function $f$ from $R\to R$ is a function such that $f^{-1}(B)\in \mathscr{B}$ for all $B \in \mathscr{B}$, where I believe $\mathscr{B}$ represents the Borel algebra.

For example, I am told $X^2$ is a Borel-measurable function of $X$, but $X$ is not a Borel-measurable function of $X^2$. I don't know why either is true. I sense that $X^2$ being continuous is the reason for the first.

Would $X$ or $X^2$ be Borel-measurable functions of $|X|$? What happens with constants are added or multiplied (subtracted or divided)? Such as $\frac {X+3} 6$ or $X-c$?

We say that a random variable $Y$ is a Borel-measurable function of $X$ if there exists a Borel measurable function $f\colon\mathbb R\to\mathbb R$ such that $Y=f\left(X\right)$.
In order to see that $X^2$ is a Borel-measurable function of $X$, consider $f\colon x\mapsto x^2$.
If $X=f\left(X^ 2\right)$ and $X$ takes the values $1$ and $-1$, we get a contradiction, since if $\omega$ is such that $X\left(\omega\right)=1$, then $1=f\left(1\right)$ and if $\omega'$ is such that $X\left(\omega'\right)=-1$, then $-1=f\left(1\right)$.
• So then for something like $\frac {x-3}6$ being or not being a borel measurable function of $|X|$, I would do it like this?: If we considered $f: |x| \mapsto \frac {x-3}6$, and if $\frac {x-3}6 = f(|X|)$ where $\frac {x-3}6$ takes on the values... I can't come up with two values that come from the same $|X|$ but I know this function is supposed to not be borel-measurable. I could easily get one $\frac {x-3}6$ value from two $|X|$, but that's not the goal. Commented Dec 7, 2016 at 4:45
• Your function $f$ is not well defined (what would be $f(-3)$? And $f(-3)$? Commented Dec 7, 2016 at 20:39
• $f(-3)$ and $f(-3)$? Those are the same aren't they? Commented Dec 7, 2016 at 22:00
• Sorry I meant $f(3)$ and $f(-3)$. Commented Dec 7, 2016 at 22:03
• What about them? $f(3)=0$ and $f(-3)=-1$? Commented Dec 7, 2016 at 23:05