Lebesgue's criterion for Riemann-integrablity:
A function $f\colon[a,b]\to\mathbb R$ is Riemann-integrable if and only if it is bounded and continuous almost everywhere.
There are several proofs. I know one using du Bois-Reymond's criterion: a bounded function is Riemann-integrable if and only if for each $\epsilon>0$, there exists a positive number $\delta>0$ and a partition such that the total length of intervals on which the oscillation $>\epsilon$ is less than $\delta$. However, I would like to see whether we have a proof using the idea of maximal functions.
Extend the function from $[a,b]$ to $\mathbb R$ by $0$, we can reformulate the continuity as $\lim_{h\to0}(f(x+h)-f(x))=0$ for a.e. $x\in\mathbb R$.
We have seen many examples of convergence a.e. of which the proof takes advantage of a maximal function. For example, Lebesgue differentiation theorem follows from the fact that the Hardy-Littlewood maximal function is a bounded sublinear operator from $L^1$ to $L^{1,\infty}$. Similarly, Carleson's theorem follows from, say, the boundedness of the maximal Carleson operator.
I wonder whether we can prove the Lebesgue's criterion in a similar fashion.
Of course even if the question were more precisely formulated it's hard to imagine how one might prove the answer is no. But offhand it's hard to see how a proof via a maximal function is going to work. Because, by analogy with the results you mention that are proved using maximal functions, it seems like the relevant maximal function here is just $$Mf(x)=\sup_h|f(x+h)|,$$but that makes $Mf$ constant.