I want to show that the following function $$f:[0,2\pi]\times [0,2\pi] \rightarrow \mathbb{R}; (x,y)\mapsto y\sin(xy)$$ is measurable.
We had a theorem that says that for all $c\in \mathbb{R}$, $f^{-1}([c,\infty))$ has to be in the sigma algebra of the domain. In our case I would assume that this is $\mathfrak{B}([0,2\pi]^2)$ where $\mathfrak{B}$ denotes the borel sigma algebra. But then if I consider $f^{-1}([c,\infty))=\{(x,y): y\sin(xy)\in [c,\infty)\}$ I don't know how to conclude. So i would say that if
- $c>\max_{(x,y)}{f(x,y)}$ then $f^{-1}([c,\infty))=\emptyset\in \mathfrak{B}([0,2\pi]^2) $
- if $c<\min_{(x,y)}{f(x,y)}$ then $f^{-1}([c,\infty))=\emptyset\in \mathfrak{B}([0,2\pi]^2) $
- if $\min_{(x,y)}{f(x,y)}\leq c\leq \max_{(x,y)}{f(x,y)}$ then $f^{-1}([c,\infty))$ is an interval and a subset of $[0,2\pi]$ and thus clearly in $\mathfrak{B}([0,2\pi]^2) $
But does this argumentation works?
Thanks for your help