Does convergence of a sequence of subharmonic functions imply the vague convergence of their Riesz measures?

Suppose $$D$$ is a bounded domain of $$\mathbb{R}^m$$ for $$m>1$$ and $$\{u_n\}_{n\geq1}$$ is a sequence of subharmonic functions on $$D$$. Assume $$u_n\to u_0$$ pointwise on $$D$$ and $$u_0$$ is subharmonic on $$D$$. Let $$\mu_n$$ be the Riesz measure associted to each $$u_n$$ for $$n\geq0$$. Suppose also that for a compact set $$K\subset D$$ we have $$\mu_n(K)=0$$ for all $$n>0$$. It is well-known that the sequence of measures $$\{\mu_n\}$$ has a subsequence that is vaguely convergent, and so $$\int_Kf(x)d\mu_{n_k}(x)\to \int_Kf(x)d\nu(x),$$ as $$k\to\infty$$, for all continuous functions $$f$$ and for some measure $$\nu$$.

My question is: can we conclude that the restrictions of $$\nu$$ and $$\mu_0$$? to $$K$$ coincide? In particular, do we have also $$\mu_0(K)=0$$?

• I don't understand. $\mu(K) = 0$ for each compact $K \subset D$ implies $\mu \equiv 0$ by inner regularity. What am I missing? Oct 3 '20 at 10:02
• I think you can use Lebesgue Dominated convergence theorem Oct 3 '20 at 10:04
• Plus, you should replace "compact $K$" by "open and bounded $K$", I believe. Oct 3 '20 at 11:20
• Vague convergence would not imply that $\nu(K)=0$. Oct 3 '20 at 11:50

The answer is no. Example (I use complex notation in dimension 2) $$u_n(z)=\max\{\log|z|-1/n,0\}\to\max\{\log|z|,0\}$$ $$K=\{ z:|z|\leq 1\}$$ is compact. $$\mu_n(K)=0$$ but he limit measure is supported on $$K$$. In fact, in this example, $$\mu_n$$ is the uniform measure on the circle $$|z|=e^{1/n}>1$$ while the limit is the uniform measure on the unit circle.

• Sorry, but I am confused! I have two problems. 1) How did you get the measures of $K$? 2) According to my understanding, the Riesz measure $\mu_n$ of $u_n$ is defined as $$\int_D \Psi d\mu_n =a\int_Du_n\Delta \Psi d\lambda$$ for all test function $\Psi$ in $D$ ($C^\infty$-smooth compactly supported in $D$). Now if you take for $D$ the ball of, say, radius 2 centered at the origin, then $u_n$'s are bounded and by dominated convergence theorem we get, as $n\to\infty$, $$\int_Du_0 \Psi d\lambda=\int_D \Psi d\mu_0=\int_D \Psi d\nu.$$ Isn't this a contradiction with what you are saying? Oct 3 '20 at 21:00