# Null sets and the Riesz measure of a subharmonic function

Let $$D$$ be a bounded domain of $$\mathbb{R}^{m}$$ with $$m>1$$, and $$u$$ a subharmonic function on $$D$$. Let $$u_{\epsilon}$$ be a sequence of smooth subharmonic functions on $$D_{\epsilon}$$ (the set of elements of $$D$$ having a distance bigger than $$\epsilon$$ from the complement of $$D$$) that decreases to $$u$$ pointwise. Let $$\mu$$ be the Riesz measure associated to $$u$$ and $$\mu_{\epsilon}$$ the Riesz measure associated to $$u_{\epsilon}$$. Let $$E\subset D$$ be a Borel set.

My question is: suppose $$E$$ is a $$\mu_{\epsilon}-$$null set for all $$\epsilon>0$$, can we conclude that $$E$$ is $$\mu-$$null set?

• where is $E$? is it compactly supported inside $D$? – mathworker21 Sep 22 '19 at 19:42
• Yes. It belongs to $D$. I edited and corrected my question. – M. Rahmat Sep 22 '19 at 20:27

No. For example $$m=2$$, $$D$$ the unit disk, $$u(x,y)=|y|$$, $$E=\{(t,0):|t|\le1/2\}$$.
(That's really just an example for $$m=1$$ with an extra variable added since you specified $$m>1$$; for $$m=1$$ take $$u(t)=|t|$$, $$D=(-1,1)$$, $$E=\{0\}$$.)
• Would you please explain how $\mu{\epsilon}(E)$ is zero but not $\mu(E)$, in the case $m=2$? – M. Rahmat Sep 24 '19 at 22:25
• @M.Rahmat Well first, $u_\epsilon$ is smooth. So $d\mu_\epsilon=\Delta u_\epsilon dm$, where $m$ is Lebesgue measure; hence $\mu_\epsilon(E)=\int_E\Delta u_\epsilon=0$ since $m(E)=0$. For $\mu(E)>0$, first you should figure out what $\mu$ is. (Do that first in the case $m=1$....) – David C. Ullrich Sep 25 '19 at 12:31