Let $f: E\rightarrow \bar{\mathbb{R}}$, $g: E\rightarrow \bar{\mathbb{R}}$ be two measurable functions define on a measurable set $E\subseteq R^d$, $ \bar{\mathbb{R}}=[-\infty, +\infty]$. Prove the set $\{x\in E\ | f(x)=g(x)\}$ is measurable.
My attempt: if $f, g$ are finite-valued, then let $h= f- g$, h then is measurable, we can get $\{x\in E\ | f(x)=g(x)\}=h^{-1}(\{0\})$ is measurable by definition. But now the ranges of $f, g$ are $\bar{\mathbb{R}}$, they are may not finite-valued. Now how to show the case? Any suggestion will be helpful.
$\{x: f(x) <g(x)\}=\cup_{r\in \mathbb Q} \{x: f(x) <r<g(x)\}$ which is measurable because $\{x: f(x) <r<g(x)\}=\{x: f(x) <r\}\cap \{x: r<g(x)\}$. Similarly, $\{x: f(x) >g(x)\}$ is measurable. Take the union of these two and then the complement.