We know the following classic result. If $K\subset\mathbb{R}^m$ ($m>1$) is a compact set and $u$ is superharmonic on a neighborhood of $K$, then we can extend $u$ to a superharmonic function $\overline{u}$ on $\mathbb{R}^m$ that coincides with $u$ on $K$. For $|x|>$ some positive number $r$, we have $\overline{u}=a+b U(x)$, where $a$ is real, $b$ positive and $U(x)=-\log|x|$ if $m=2$ and $=|x|^{2-m}$ if $m>2$ ($|.|$ Euclidean norm) ( see "Classical Potential Theory", by Armitage and Gardiner, pg 192).
We notice that $\overline{u}$ is not superharmonic at $x=\infty$, but subharmonic. My question is: is there a way to take out one point $y_0$ from $K$, and then extend $u$ to a superharmonic function $\tilde{u}$ on $\mathbb{R}^m_\infty\setminus\{y_0\}$ such that on $K\setminus\{y_0\}$ $u$ and $\tilde{u}$ coincide? ($\mathbb{R}^m_\infty=\mathbb{R}^m\cup\{\infty\}$ one point compactification.) By the way $u$ is superharmonic at infinity, if it is lower semicontinuous at infinity and $u(\infty)$ is bounded below by the average of $u$ over any ball of sufficiently large radius (see Helms' "Introduction to Potential Yheory", chapter on the Dirichlet problem for unbounded regions).
