Let $$u\colon B^n(0,1)\to \mathbb{R}$$ be a subharmonic function in the open unit ball in $\mathbb{R}^n$. The crucial assumption is that $u$ never equals $-\infty$. 



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> Is it true that $|u|$ is bounded on any compact subset of the unit ball?
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Remark. (1) $u$ is bounded above on any compact subset of the unit ball since it is upper semi-continuous.


(2) If the answer has the negative answer in the above generality, then I would like to ask it for $u$ being a plurisubharmonic function in the unit ball in $\mathbb{C}^{n/2}=\mathbb{R}^n$.



Boundedness of a finite subharmonic function
No. Take a zequence $x\_k\to 0$ and a sequence of positive numbers $\epsilon\_k$
such that $\sum\_k\epsilon\_k|K\_n(x\_k)|<\infty$.
Then the function 
$$\sum\_k\max\{\epsilon\_k K\_n(x-x\_k),-k\},$$
where $K\_n$ is the fundamental solution of the Laplace equation ($K\_n(x)=-|x|^{-n+2}$ when $n\geq 3$ and $K\_2=\log|x|$).
This function is finite everywhere, so your polar set and your $U$ are empty,
but the function is evidently unbounded from below.


When $n=2$ "subharmonic" and "plurisubharmonic" are the same, so it also gives a counterexample for plurisubharmonic functions (in any dimension, just let it be
independent of the rest of the complex variables).


