Suppose that $g:[a,b]\to \Bbb R$ is a function of bounded variation, and $f$ is Borel measurable with $|f|\le M$ on $[a,b]$. Define $g^+,g^-$ to be the Jordan decomposition so that these are increasing functions and $g=g^+-g^-$. Let $\nu^+$ be the measure such that $g^+$ is the CDF of $\nu^+$, and likewise for $\nu^-$. Take $\nu=\nu^+-\nu^-$, and define $\int_E f \ dg = \int_E f \ d\nu^+-\int_E f \ d\nu^-$.
I am trying to prove that $\left|\int_{[a,b]}f\ dg\right| \le MT$ where $T$ is the total variation of $g$.
I am primarily struggling to identify what $T$ is, as related to the measures. It makes some amount of intuitive sense to think that $T=|\nu|[a,b]$ where $|v|=\nu^++\nu^-$, however I am not able to make this connection. I know that
$$ \nu^+[a,x] = g^+(x)-g^+(a) $$ $$ \nu^-[a,x] = g^-(x)-g^-(a) $$
so $|\nu|[a,b]$ is
$$ \nu^+[a,b]+\nu^-[a,b] = g^+(b)-g^+(a) + g^-(b)-g^-(a)$$
I also have that $g^+(x) = g(x)+TV(g,[a,x])$ and $g^-(x) = TV(g,[a,x])$, and so $g^+(b) = g(b)-T$ and $g^+(a) = g(a)$ and so on. But plugging this into the result above doesn't seem to reach the desired conclusion, since it becomes
$$ g(b)-T-g(a)+(0-T) = g(b)-g(a)-2T $$
I tried to parse this answer: If $A$ is right-continuous and of bounded variation and $f$ is integrable, what is the variation of $t\mapsto\int_{(0,\:t]}f\:{\rm d}A$?
However, I'm not sure what $f\mu$ means here. I might assume that it is the measure such that $f$ is the CDF of $f\mu$, but then I don't understand why $(f\mu)^+ = f^+\mu^+$ and in general I'm not familiar with any theorems about generally manipulating an object like $f\mu$ assuming I've correctly identified what it is.