I see these two:

How to show that a Riemann integrable function is bounded

If a function $f(x)$ is Riemann integrable on $[a,b]$, is $f(x)$ bounded on $[a,b]$?
The first uses a very different definition of Riemann integrable functions. The second post offers casual intuition, not a formal proof.
The definitions I’m working with:
A Riemann Sum is defined for a partition $mathcal{P}$ of $[a,b]$ as:
begin{align*}
mathcal{R}(f, mathcal{P}) &= sumlimits_{j=1}^k f(s_j) Delta_j \
end{align*}
The function is Riemann integrable if Riemann sums converge to a number $ell$ as the mesh sizes of the partitions approach zero. A function $f$ is Riemann integrable if for any $epsilon > 0$, there must exist some $delta > 0$ and some partition $mathcal{P}$ such that:
begin{align*}
m(mathcal{P}) < delta &implies mathcal{R}(f, mathcal{P})  ell < epsilon \
end{align*}
Intuitively, if $f$ is unbounded, it looks like that Riemann sum will not converge, but I can’t see how to formally demonstrate that.