Suppose I have $f \in BV_{loc}(\mathbb{R})$ and some open set $K \subset \mathbb{R}$ with $f = 0$ a.e. on $K$. Then is it true that $\partial_{x}f$ must necessarily be a measure (and not $L^p$) if $f$ is not identically $0$ everywhere that it is defined?
Intuitively, it seems like the result should be true since there should be a jump discontinuity. For example if $K = (-1,1)$ you could have $f = \mathbb{1}_{\{x=0\}}$ and so $f = 0$ a.e. and $\partial_{x}f = \delta_{\{x=0\}}$. I find it hard to believe there could be an example where $f = 0$ a.e. but $\partial_{x}f$ is not a measure, but I am not sure how to prove it.
I tried to argue as follows. Let $E \subset K = (-1,1)$ be such that $|E| = 0$ and $f = 0$ a.e. on $E$. Then let $y \in E$ be a point where $f$ is defined and such that $f(y) \ne 0$. Then one idea would be to show that $lim_{x \to y^{-}} f(x) \ne lim_{x \to y^{+}} f(x)$. But because $f$ is a Lebesgue function, I don't think this pointwise evaluation within the limit makes sense so I got stuck here.
I would appreciate any help.