Does “Riemann integrable $\Rightarrow$ discontinuity set is measure zero” need axiom of choice?

Question

My intuition is that it requires the axiom of choice to prove that if a bounded function $f$ on a bounded subset of $\mathbb{R}^n$ is Riemann integrable, the discontinuity set of $f$ is measure zero. Is this correct?

Answer 1

No, you don't need the axiom of choice. You can read the standard proof here.

This proof involves the sets $X_{1/n}$, defined as the points whose oscillation is greater than $1/n$, $n\in \mathbb{N}$. The set of discontinuities of $f$ is $\bigcup X_{1/n}$, since $f$ is continuous at a point iff the oscillation there is zero. For the shake of contradiction, if you suppose that the set of $f$'s discontinuities doesn't have measure zero, then there exists some $n_0\in\mathbb{N}$ such that $X_{1/n_0}$ isn't a null set (otherwise, since any countable union of null sets is null, the set of discontinuities will have zero measure!).

Maybe you got confused with the choice of $X_{1/n_0}$, thinking that the axiom of choice was needed. But $X_{1/n_0}$ isn't chosen, it must exist if the set of discontinuities isn't null.