Does the integration by parts formula $\int f'g = - \int fg'$ hold true for compactly supported, continuous functions of bounded variation?

Question

Assume that $f:\mathbb R \to \mathbb C$ is a continuous function of bounded variation with support in $[-T,T]$. We know that $f$ is differentiable almost everywhere. Further, let $g \in C^\infty(\mathbb R)$ be bounded. Does the integration by parts formula $$ \int f'(x)g(x) \,dx = -\int f(x)g'(x) \, dx $$ hold true?

Answer 1

Let $T > 1$ and let $f = \chi_{[0,1]}$ be the characteristic function of $[0,1]$. Then $$ \int_{-T}^T f g' = \int_0^1 g' = g(1) - g(0). $$ On the other hand, since $f' = 0$ a.e., $$ \int_{-T}^T f' g = 0. $$

If we understand $f'$ as the measure-valued derivative $Df$ of a BV function, then $Df = \delta_0 - \delta_{-1}$, and $$ \int_{-T}^T g\, Df = g(0) - g(1), $$ that coincides with $-\int_{-T}^T f g'$.