Suppose $f \in BV_{loc}(\mathbb{R})$ and $g \in C^{1}(\mathbb{R})$. Also assume that $f$ has compact support (so I guess $f \in BV(\mathbb{R})$ in fact).
Then do we have the integration by parts formula:
$$\int_{\mathbb{R}} f(x)g'(x)~dx = - \int_{\mathbb{R}} g(x)f'(x)~dx $$ if $f'$ is understood as a measure? I know that since $f \in BV_{loc}$ then its distributional derivative is a signed Radon measure. I only know that this formula holds for absolutely continuous functions, I'm not sure about a weaker case like this.