Measure of vanishing set for BV function

Question

Suppose $\Omega$ is open in $\mathbb{R}^N$ and $u \in BV(\Omega)$ is such that $u = 0$ on an open subset $E \subseteq \Omega$. If $D_i u$ denotes the $i$-th partial derivative measure associated to $u$, should it hold that $D_i u(E) = 0?$ I've been trying to make an argument based on approximation by test functions, but the fact $D_i u$ is a signed measure is causing complications. Thanks for any advice.

Answer 1

First of all, we have, for any Radon measure $\mu$, if $$ \int_F \phi d\mu=0 $$ for all $\phi \in C_0(F)$, then $|\mu|(F)=0$, where we use $|\cdot|$ to denote the total variation of Radon measure.

Now, if we have your set $E$ is open connected, then for any $\varphi\in C_0^1(E)$, we have $$ \int_E\varphi\,d Du_i = \int_E u\partial_i\phi = 0 \tag 1 $$ that is, $|Du_i|(E)=0$.

The problem is now we really need to test all $\varphi\in C_0(E)$ but not just $C_0^1(E)$. To overcome this problem, I suggest you to read this post, especially @PhoemueX's answer.

If your $E$ is not connected, you have to do it on each connected components.

PS: this is also hold if your $u$ is constant on each connected components.