It have read multiple times online and on this forum that when $f:I\to \mathbb R$ and $\alpha :X\to \mathbb R$, with $I\subseteq X$ and $X$ is a closed interval, if $f$ and $\alpha$ are discontinuous at the same point $m\in I$, then $\int_If\;d\alpha$ does not exist.
But all these proofs make one assumption, which is that the partitions of the intervals $I$ never contain singletons. This follows directly from the definition they use for a partition, which I know is pretty standard, but my book I use to study analysis (Analysis I Terence Tao 4th ed) does not define it the exact same way and actually allow for a partition of an interval to contain a singleton (The exact definition of my book being: A partition of $I$ is a finite set $P$ of bounded intervals contained in $I$, such that every $x$ in $I$ lies in exactly one of the bounded intervals $J$ in $P$.
This being said I can't seem to prove that the aforementioned result still holds, and by considering partitions where we isolate the point of discontinuity of $f$ and $\alpha$ in some singleton, I can show that that the difference between the upper and lower Riemann Stieltjes sums approaches zero (since trivially $\sup_{x\in \{m\}} f(x)-\inf_{x\in \{m\}} f(x)=0$ where $f$ and $\alpha$ are discontinuous in $m$).
So does this definition of a partition allow for $\int_I f\;d\alpha$ to exist even if $f$ and $\alpha$ share a discontinuity?
