Extension of a $\delta$-subharmonic function that is subharmonic on a reduced domain

Suppose $$B$$ is a ball in $$\mathbb{R}^{m}$$ and $$u$$ and $$s$$ are subharmonic on $$B$$. Suppose there is a closed subset $$F$$ of the closure of $$B$$ with no interior such that $$v=u-s$$ is subharmonic on $$B\setminus F$$. Is there a way to prove that $$v$$ is subharmonic on $$B$$? Notice that if $$u$$ and $$s$$ are $$C^{2}$$-smooth, the solution is easy: for each $$x\in F$$ we may take a sequence $$(x_{n})$$ in $$B\setminus F$$ that converges to $$x$$. Then by taking the limit of $$\Delta v(x_{n})\geq0$$ we can prove the case. But what if $$u$$ and $$s$$ are not smooth?

• Something is missing in the statement. What if $F=B$, for example? – Alexandre Eremenko Jul 6 at 21:14
• Sorry! I corrected. – M. Rahmat Jul 6 at 21:45
• This is not correct: take $u(x)=0,s(x)=|x_1|$, where $x=(x_1,x_2)$ both subharmonic, $v(x)=-|x_1|$ is harmonic away from the line $F=\{ x:|x_1|=0\}$. – Alexandre Eremenko Jul 7 at 5:05