If $A$ is right-continuous and of bounded variation and $f$ is integrable, what is the variation of $t\mapsto\int_{(0,\:t]}f\:{\rm d}A$?

Question

Let

• $T>0$
• $I:=(0,T]$
• $A:\overline I\to\mathbb R$ be right-continuous and of bounded variation
• $f:I\to\mathbb R$ be Borel measurable with $$\int|f|\:{\rm d}|A|<\infty\tag1$$ and $$g(t):=\int_{(0,\:t]}f\:{\rm d}A\;\;\;\text{for }t\in\overline I$$

Let $(\mu^+,\mu^-)$ denote the Jordan decomposition of ${\rm d}A$ and $$f^\pm:=\max(\pm f,0)\;.$$ Then, $$g(t)=\underbrace{\left(\int_{(0,\:t]}f^+\:{\rm d}\mu^++\int_{(0,\:t]}f^-\:{\rm d}\mu^-\right)}_{=:\:g^+(t)}-\underbrace{\left(\int_{(0,\:t]}f^+\:{\rm d}\mu^-+\int_{(0,\:t]}f^-\:{\rm d}\mu^+\right)}_{=:\:g^-(t)}\tag2$$ for all $t\in\overline I$. Since $g^\pm$ is nondecreasing, we see that $g$ is of bounded variation. By Lebesgue's dominated convergence theorem, we also obtain right-continuity of $g$.

Now, we know that $g$ is of bounded variation. But are we able to write down the variation function of $g$ in an explicit form?

Answer 1

In general, when $\nu$ is a signed measure with Jordan decomposition $(\nu^+,\nu^-)$, the variation function of $t\mapsto \nu((0,t])$ is $t\mapsto |\nu|((0,t])$ where $|\nu| = \nu^+ + \nu^-$. Not coincidentally, $|\nu|$ called the variation of $\nu$.

The reason is that the Jordan decomposition of the BV function obtained by integrating a measure corresponds to the Jordan decomposition of that measure, as Wikipedia notes.

In your case $\nu= f\mu$, so $\nu^+ = f^+\mu^++f^-\mu^-$ and $\nu^- = f^-\mu^++f^-\mu^+$, hence $|\nu| = |f| |\mu|$. In other words, the variation function is
$$t\mapsto \int_{(0,t]} |f|\:{\rm d}|A|$$ which is precisely the quantity you assumed finite.