Is this function borel-measurable?

Question

is the function $f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto\ \begin{cases} \frac{1}{\sin(x)} & x\notin \{z|z=k\pi,k\in\mathbb{Z}\}\ \\ 0 & x\in \{z|z=k\pi,k\in\mathbb{Z}\} \ \end{cases}$ borel-measurable? If so, which way can I show that? By now I only know that characterstic functions are measurable and linear combinations, products of measurable functions are... nothing about continuous functions. I also tried splitting it up by using characteristic functions which didn't work.

Answer 1

Note that every continuous function on $\mathbb{R}$ is Borel measurable. In this case, $f$ is continuous on $\mathbb{R}$ \ $A$, where $A$ = {$x : x= k \pi , k \in \mathbb{Z}$}.

By definition, $f$ is measurable iff {$x| f(x)>a$} is measurable for all $a \in \mathbb{R}$.

Claim:

{$x| f(x)>a$}= $(f^{-1}(a, \infty) \cap$ $\mathbb{R}$ \ $A$) $\cup A$.

Proof:

By continuity of $f$ on $\mathbb{R}$ \ $A$, {$x| f(x)>a$} = {$x| 1/sin(x)>a$}= $f^{-1}(a, \infty) \cap$ $\mathbb{R}$ \ $A$. When $f(x)= 0$ is taken into consideration, we have {$x| f(x)>a$} = $A$. In general,{$x| f(x)>a$} is the union of those two sets.

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$f^{-1}(a, \infty) \cap$ $\mathbb{R}$ \ $A$ is the intersection two Borel sets. Hence, it is Borel.$A$is Borel as well; so does the union of two Borel measurable sets. Therefore, {$x| f(x)>a$} is Borel. By defintion, $f$ is Borel measurable.

Note that, $f^{-1}(a, \infty)$ is Borel because of the continuity of $f$ on $\mathbb{R}$ \ $A$.