Hi everyone: Let $ \Omega $ be a bounded open set of $ \mathbb{R}^{N} $, $ N\geq2 $, and $ F\subset \Omega $ with empty interior. Suppose there exists a superharmonic function $ u $ on $ \Omega\setminus F $ such that $$ \lim_{x\rightarrow y}u(x)=+\infty $$ for all $ y\in F $. Now, we define $ w(x) $ to be equal to $ u(x) $ for $ x\in \Omega\setminus F $, and $ +\infty $ for $ x\in F $. Is $ w $ a superharmonic function on $ \Omega $?


1 Answer 1


First you have to assume that $F$ has measure zero, since a superharmonic function is locally integrable.

Under this assumption, setting $u=\infty$ on $F$ results in a superharmonic function. To see this, note that by the limit assumption $$\lim_{x\to y}u(x)=\infty$$ for all $y\in F$, the resulting function is still lower semi-continuous.

Also, the mean value property $$u(x)\geq \frac{1}{|B(x,r)|} \int_{B(x,r)} u$$ remains valid if $F$ has measure zero.

These imply that $u$ is superharmonic.

  • $\begingroup$ Thanks for your answer. All I know is that the harmonic measure of $F$ with respect to $\Omega\setminus F$ is zero. But Inerrability is not a part of the definition of a superharmonic function. $\endgroup$
    – M. Rahmat
    Nov 26, 2016 at 20:55
  • $\begingroup$ You are right, it is not part of the definition but it follows from the mean value property. The proof is essentially a connectedness argument that uses the fact that superharmonic functions attain a minimum on compact sets. $\endgroup$ Nov 26, 2016 at 21:08
  • $\begingroup$ I agree with what you have proved: if $F$ is a null set, then for sure $u$ is superharmonic. What I want to make sure is this: is it possible that $F$ have only empty interior without $u$ being superharmonic? $\endgroup$
    – M. Rahmat
    Nov 26, 2016 at 21:36
  • 2
    $\begingroup$ @Rahmat: The answer to your question in comment is no. If $F$ has positive measure than it has positive capacity. And a superharmonic function can be infinite only on a set of zero capacity. $\endgroup$ Nov 26, 2016 at 21:39
  • $\begingroup$ I was just thinking that the continuity of the function near $F$ makes it impossible that the measure of $F$ be positive. $\endgroup$
    – M. Rahmat
    Nov 26, 2016 at 23:08

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