Given a function $f$ that is Riemann integrable on $[a,b]$, does there exist a sequence of continuous functions $\{f_n\}_{n=1}^\infty$ that converges to $f$ pointwise everywhere on $[a,b]$?
If I just require pointwise almost everywhere, this follows from the fact that continuous functions are dense in $L^1[a,b]$ and norm convergence yields a subsequence that converges a.e. This is an exercise in Krantz, Real Analysis and Foundations (4th ed., p. 153); I have not been able to prove it, and when I queried the author he could not provide a proof either. However, I cannot find a counter-example, either.
Everywhere? no. Almost everywhere, yes.
A pointwise limit of a sequence of continuous functions is said to be a function of Baire class $1$. Baire proved many properties of such functions. In particular, if $E$ is a nonempty perfect set, then the restriction of $f$ to $E$ has a point of continuity.
Consider the following function $f$. Let $[a,b] = [0,1]$. Let $C$ be the middle-thirds Cantor set. So $C$ is a closed set of measure zero. Define $f: [0,1] \to \mathbb R$ as follows.
$\bullet \;f(x) = 0$ on $[0,1]\setminus C$.
$\bullet\;f(x) = 0$ on the endpoints of the open intervals in $[0,1]\setminus C$.
$\bullet\;f(x) = 1$ elsewhere, uncountably many remaining points of $C$.
First note that $f$ is continuous at every point of $[0,1]\setminus C$, a set of measure $1$, so $f$ is Riemann integrable.
But also note that the restriction of $f$ to the nonempty perfect set $C$ has no point of continuity: both $\{x \in C : f(x) = 0\}$ and $\{x \in C : f(x) = 1\}$ are dense in $C$. So $f$ is not of Baire class $1$.