Suppose $u$ is a subharmonic function on a bounded open set $V$ of $\mathbb{R}^m$ ($m\geq2$). Let $u_n$ be a sequence of smooth subharmonic functions that decreases pointwise to $u$ (obtained by convolution with a smoothing test function). Suppose $\mu$ is the Riesz measure associated to $u$ and $\mu_n$ the Riesz measure associated to $u_n$. We know that since $u_n$ is smooth, $\mu_n=\Delta u_n d\lambda$, where $\lambda$ is the $m-$ dimensional Lebesgue measure. I have two question: We know that for any any smooth function $\phi$ with compact support in $V$ (i.e. $\phi\in C^\infty_0(V)$, $$\lim_{n\to\infty}\int_V\phi(x)d\mu_n(x)=\int_V\phi(x)d\mu(x)?$$
My question is: Can we say that for any open set $W$ in $V$ the above equation holds? i.e. $$\lim_{n\to\infty}\int_W\phi(x)d\mu_n(x)=\int_W\phi(x)d\mu(x)?$$
As I said, I haven’t studied Riesz measures for subharmonic functions, so my answer could not be 100% appropriate or canonical.
Concerning your question, the answer is no. The idea is that you are basically multiplying $\phi$ by a characteristic function, so the function you test $\mu_n$ against is not continuous. That is in general a problem for converging measures.
If you take a general succession of Radon measures $\mu_n$ which converges (in measure) to $\mu$, the thesis is not true. For instance, take a radial $\phi$, $\mu$ to be the dirac delta in the origin, $\mu_n$ a standard regularization of it (so, basically, just the standard mollifier), $V=B_1(0)$ and $W=V\cap\{x_1>0\}$. We have
$$\lim_{n\to\infty}\int_W\phi(x)d\mu_n(x)=1/2 \lim_{n\to\infty}\int_V\phi(x)d\mu_n(x)=1/2\int_V\phi(x)d\mu(x),$$
which is in general nonzero, while
$$\int_W\phi(x)d\mu(x)=0.$$
You can realize the measures above with subharmonic functions taking $u=-\Gamma$ and $u_n$ a standard regularization of $u$, where $\Gamma$ is the fundamental solution of the Laplace operator in $\mathbb R^m$, s.t. $\Delta\Gamma=-\delta$ distributionally.
My only doubt is that I don’t know if that $u$ is regular enough for your setting, or it doesn’t count as it is unbounded. If not ok, you could take instead $u=\max(-\Gamma,-c)$ (with $c>0$ large enough for $u$ not to be constant on $V$), that is subharmonic, bounded, and the Laplacian is supported on a spherical surface inside $V$ that is basically the set of points where the gradient of $u$ is discontinuous. Finally, taking $W$ as the complement of that surface should do an analogous trick.