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 $?



Is this a superharmonic function?
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.


