Question: Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is measurable. Prove $\{(x,y)\in\mathbb{R^2}:f(x)\geq f(y)\}$ is measurable.
My thoughts: I was hoping that there would be a way to "pull back" into an open set, since open sets in $\mathbb{R}$ are measurable. I was trying to just use the definition of a measurable function, but I am just getting thrown off by the $f(x)\geq f(y)$ bit, and how that affects everything. Also, I was wondering if you could provide any insight as to dealing with problems like this. In particular, if given a measurable function $f$, what is a good way of going about dealing with questions regarding a set with elements depending on $f$...
Any thoughts, suggestions, etc. are greatly appreciated! Thank you!
we have that the function $\phi:\mathbb{R}^2\to \mathbb{R}$, where $\phi(x,y)=x-y$ is measurable since it is continuous. So is the function $F(x,y)=(f(x),f(y))$ (from $\mathbb{R}^2\to \mathbb{R}^2$), since $f$ is measurable. Then $\phi\circ F:\mathbb{R}^2\to \mathbb{R}$ is measurable. Now, \begin{align} \{(x,y)\in\mathbb{R^2}:f(x)\geq f(y)\} &= \{(x,y)\in\mathbb{R^2}:f(x)-f(y)\geq 0\} \\&= \{(x,y)\in\mathbb{R^2}: \ \phi(f(x),f(y))\geq 0 \} \\&= \{(x,y)\in\mathbb{R^2}:\ \phi\circ F(x,y)\geq0\} \\&= (\phi\circ F)^{-1}\big([0,+\infty)\big) \end{align} and thus it is a measurable subset of $\mathbb{R}^2$.