Suppose $V$ is an open set of $\mathbb{R}^{n}$ $(n\geq2)$ and $u$ a bounded subharmonic function on $V$. Let $F$ be a closed subset of $V$ with empty interior. Suppose $u$ is continuous on $F$ quasi-everywhere (so everywhere outside a polar set). Can we conclude that $u$ is continuous everywhere on $F$?
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I don't see why you expect that. A typical example (in two dimensions) would be $$ u(z) = \max\left(-2, \sum_{n=1}^\infty 2^{-n} \log|z-2^{-n}|\right) $$ which is bounded and subharmonic on the unit disk, and continuous everywhere except at $0$. The discontinuity arises because $u(0) = -2\log2 > -2$, while $u(2^{-n}) = -2$ for all $n$.
As the set $F$, one can take $[0, 1/2]$ for example.