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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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I don't think that the statement is true. Let $\Omega\subset\mathbb{R}^{2}$ be a bounded open set, and let $(x_{n})_{n\geq1}$ be an enumeration of the 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, hence can take the value $+\infty$ only on a polar set. This set contains in particular the points $x_{n}$, $n\geq1$. Hence, $U^{\mu}$ is discontinuous at each point of $\Omega$ where it is finite that is quasi-everywhere in $\Omega$.

However, a weaker 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. He gives the proof for the case $m\geq3$.

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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 at 5:51

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