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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?

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  • $\begingroup$ where is $E$? is it compactly supported inside $D$? $\endgroup$ – mathworker21 Sep 22 '19 at 19:42
  • $\begingroup$ Yes. It belongs to $D$. I edited and corrected my question. $\endgroup$ – M. Rahmat Sep 22 '19 at 20:27
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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 $u(t)=|t|$, $D=(-1,1)$, $E=\{0\}$.)

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  • $\begingroup$ Would you please explain how $\mu{\epsilon}(E)$ is zero but not $ \mu(E)$, in the case $m=2$? $\endgroup$ – M. Rahmat Sep 24 '19 at 22:25
  • $\begingroup$ @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 $m=1$....) $\endgroup$ – David C. Ullrich Sep 25 '19 at 12:31

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