Is $f$ a Borel measurable function?

Question

I have define $f$ by $f=\frac{1}{b}$ if $x \in \mathbb{Q} \cap [0,1]$ and $x=\frac{a}{b}$ in lowest terms. $f=34$ if $x \in [0,1] \backslash \mathbb{Q}$. Is $f$ Borel measurable? I know I need to show that $f^{-1} ((a,\infty))$ is a Borel set but I'm struggling to do so. Any help would be much appreciated.

Answer 1

We have that $\text{Im}(f)=\{\frac{1}{n}:n\in\Bbb N\}\cup\{34\}=\left(\bigcup_{n\in\Bbb N}\{\frac{1}{n}\}\right)\cup\{34\}$.

If $a\ge 34$ then $f^{-1}((a,+\infty))=\emptyset$, which is a Borel set.

If $1\le a<34$ then $f^{-1}((a,+\infty))=[0,1]\setminus\Bbb Q$, which is a Borel set.

If $a<0$ then $f^{-1}((a,+\infty))=[0,1]$, which is a Borel set.

If $a\in[0,1]$ then $[0,1]\setminus\Bbb Q\subset f^{-1}((a,+\infty))$, since $34\in(a,+\infty)$. The rest of $f^{-1}((a,+\infty))$ is a subset of $\Bbb Q\cap[0,1]$, which is countable, so we can write it as a countable union of singletons: $\bigcup_{x\in\Bbb Q\cap[0,1]\land f(x)\in(a,+\infty)}\{x\}$. Therefore $f^{-1}((a,+\infty))=\left(\bigcup_{x\in\Bbb Q\cap[0,1]\land f(x)\in(a,+\infty)}\{x\}\right)\cup\left([0,1]\setminus\Bbb Q\right)$, which is a Borel set.

We conclude $f$ is Borel measurable.