TITLE: A short proof that the set of discontinuities of a regulated function is countable
QUESTION [0 upvotes]: Does a short proof exist for the following theorem?

The set of discontinuities of a regulated function on $[a, b]$ is countable.

REPLY [5 votes]: Let $f$ be a regulated function, and by the definition there are a sequence of step functions $f_n$ converging uniformly to $f$. Let $D_n$ for any $n$ be the set of discontinuities of $f_n$ and $D=\displaystyle\bigcup_n D_n$. If $x \notin D$, it follows that $f$ is continuous at $x$ (why?), and therefore the set of discontinuities of $f$ is included in $D$, which is a countable union of finite sets, since step functions have finitely many discontinuities. It follows that $D$ is countable.
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