# Continuity of subharmonic functions

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?

## 1 Answer

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 $$m\geq3$$.

• 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 – M. Rahmat Aug 14 at 5:51