Does the derivative of the antiderivative of a BV function $f$ agree with $f$ at all but countably many points of differentiability?

Question

Let $f: (a, b) \to \mathbb R$ be a function of bounded variation, and write

$$F(x) := \int_a^x f(t) \, dt$$

for the antiderivative. Is it true that at all but countably points of differentiability of $F$, we have $F’ = f$?

More precisely, let $D$ denote the set of differentiability of $F$. Does there exist some countable set $B \subset D$ of bad points such that $F’ = f$ on $D \setminus B$?

Answer 1

The answer is yes.

Indeed, since $f$ is of bounded variation, by changing values of $f$ at at most countably many points, without loss of generality assume that $f$ is right continuous. So, $f=g+h$, where $g$ is a continuous function and $h(x)=\sum_{y\in D\cap(a,x]}a_y$ for some at most countable subset $D$ of $(a,b)$, some real family $(a_y)_{y\in D}$ such that $\sum_{y\in D}|a_y|<\infty$, and all $x\in(a,b)$.

So, $D^c:=(a,b)\setminus D$ is the set of all points of continuity of $h$ and, equivalently, of $f$. Moreover, $D^c$ is the set of all points of differentiability of $F$, and $F'=f$ on $D^c$.

Answer 2

Also we may say: a BV function is a difference of two increasing functions whose set of discontinuity are disjoint; an increasing function has at most countably many discontinuity points; the integral function $F$ of a monotone function $f$ is differentiable at $x$ iff $f$ is continuous at $x$, in which case $F'(x)=f(x)$.