The maximum principle for subharmonic functions

Question

Let $V$ be a bounded open set in $\mathbb{R}^{n}$ with $n\geq2$, and $u$ a subharmonic function on an open set containing the closure $ \overline{V} $ of $V$. Suppose we have $u<M$ on $V$ ($M$ is a constant). Can we conclude that $u\leq M$ on $ \overline{V} $?

Answer 1

I have a partial answer to this question: if $u$ is continuous at each point of the boundary $\partial V$ of $V$, the answer is yes! In fact, we have by continuity of $u$ and the maximum principle, $$u(\zeta)=\limsup_{\substack{x\rightarrow\zeta\\(x\in V)}}u(x)\leq M$$ for all point $\zeta\in\partial V$, which proves the claim!