Q:

The integral with respect to the Riesz measure over a compact with empty interior

Suppose $D$ is a bounded domain of $\mathbb{R}^{n}$ ($n\geq2$), $u$ a subharmonic function on $D$ and $\mu$ the Riesz measure associated with $u$. Let $F$ be a closed set with empty interior in $D$. Can we say:
$$\int_{F}\phi(t) d\mu(t)= \int_{F}u(t)\Delta\phi(t) dt$$ for all test function $\phi\in \mathscr D(D)?$ ($\Delta$ is the Laplace operator and $dt$ the Lebesgue meaure on $D$).
Notice that the above equality holds if the integrals are taken over $D$ instead of $F$.

A:

It's been a while since I've studied this stuff, but I think Riesz measures can be point masses (i.e. there are subhamonic $u$ with $\Delta u = \delta_{x_0}$ in the sense of distributions). So taking $F$ to be that point immediately disproves your proposed equality, since integrating over a set of Lebesgue measure $0$ gives $0$.

