Let $f$ be a real valued function then the set where $f$ is continuous is measurable.
Note the statement does not even requires that $f$ is measurable.
Let $f$ be a real valued function then the set where $f$ is continuous is measurable.
Note the statement does not even requires that $f$ is measurable.
Recall that the oscillation of $f$ at $x$ is defined by $osc_f(x)=\inf\{diam(f(-\frac 1 n+x,x+\frac 1 n)): n\in\mathbb N\}$. Let $O_n=\{x: osc_f(x)<\frac 1 n\}$. Note that $f$ is continuous at $x$ iff $osc_f(x)=0$ and $O_n$ is open (thus measurable) for all $n\in O_n$. Therefore $\bigcap_n O_n$ is the set of points where $f$ is continuous and it is measurable since is a countable intersection of measurable sets.