Riemann-integrable iff pointwise limit of step functions

Question

Is the following true? Let $f : [a,b] \to \mathbb R$ be a bounded function:

The function $f$ is Riemann-integrable if and only if there exists a sequence of step functions $\varphi_n$ converging pointwise to $f$.

Answer 1

One of my students asked the same question, so I decided to post an answer here in case it helps.

Both sides are not correct. We now construct a Riemann integrable function which cannot be approximated pointwise by step functions.

1. Let $D$ be a non-Borel measurable subset of the Cantor set $C$, and let $f=1_D$ the indicator function of $D$. Then $f$ is Riemann integrable but not Borel measurable. The Riemann integrability of $f$ can either be proved by definition, or by the Riemann-Lebesgue theorem. In fact, since $f$ agrees with the indicator function $1_C$ on the open set $\mathbb{R}\setminus C$, we can deduce from the continuity of $1_C$ on $\mathbb{R}\setminus C$ that $f$ is continuous on $\mathbb{R}\setminus C$. Hence $f$ is a bounded function whose discontinuity has measure zero.

2. Also note that any step function is Borel measurable, and a pointwise limit of Borel measurable functions is Borel measurable. So $f$ can not be a pointwise limit of step functions.