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There is a result saying that the set where a subharmonic function defined on an open set of $\mathbb{R}^{m}$ ($m\geq2$) is discontinuous is a polar set. Could someone give me a reference for this result?

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Let $\Omega\subset\mathbb{R}^{2}$ be a bounded open set, and let $(x_{n})_{n\geq1}$ be a sequence of all points of $\Omega$ with rational coordinates. Consider the discrete measure of finite mass, $$\mu=\sum_{n\geq1}\frac{1}{n^{2}}\delta_{x_{n}}.$$ Its logarithmic potential $$ U^{\mu}(z)=\int\log\frac{1}{|z-t|}d\mu(t) $$ is superharmonic. It takes the value $+\infty$ on a polar set which contains the set of points $x_{n}$, $n\geq1$, dense in $\Omega$. Hence, $U^{\mu}$ is discontinuous at each point of $\Omega$ where it is finite, that is quasi-everywhere in $\Omega$.

However, a Lusin-type property holds : Consider a potential (or a subharmonic function) in $\mathbb{R}^m$. For any $\epsilon$ there exists an open set $G_\epsilon$ with capacity less than $\epsilon$ such that the restriction of the potential to the complement of $G_\epsilon$ is continuous. This is Theorem 3.6 on p.185 of Landkof's book. The proof is given for the case $m\geq3$ only.

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  • $\begingroup$ Steel the set where this potential fails to be continuous is a countable set and is polar. Here continuity is taken in the sense that outside a polar set the function is continuous, not necessarily on an open set $\endgroup$ – M. Rahmat Aug 14 '19 at 5:51

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