smooth approximation of subharmonic functions Let $\Omega\subset\mathbb R^n$ open and $u\in\mathrm C^0(\Omega)$ subharmonic.
Why does $\Omega$ contain 
$(i)$ a sequence of smooth subharmonic functions $u_i$ cum $u_i\rightarrow u$ locally uniformly convergent and 
$(ii)$ $\forall i\in\mathbb N$ is $\Omega_i$ open cum $\bar\Omega_i\subset\Omega_{i+1}$ and $\Omega=\cup_i \ \Omega_i$
? Thanks for helping!
 A: Sketch:
$u$ is subharmonic in $U$ iff, for every $x\in U$ and $r>0$ such that $B_r(x) \subset U $ the following mean value inequality is true:
$$u(x) \le \frac{1}{|B_r(x)|} \int_{B_r(x)} u(y) dy $$ 
($|.|$ denoting Lebesgue volume). Let $\phi_k(x) = \psi_k(||x||) \ge 0 $ a sequence of smooth mollifiers such that $\mbox{supp} \,\phi_k  \subset B_{1/k}(0)$ and such that $\int \phi_k = 1$. Define
$$u_k(y):=\int u(y-x) \phi_k(x) dx$$
It is known (and I assume you do know) that $u_k\rightarrow u$ uniformly and $u_k$ is smooth on all open domains $V$ such that $V+B_{1/k}(0)\subset U$ for large $k$ (this is the reason why you will need ii)). Now
$$\frac{1}{|B_r(x)|} \int_{B_r(x)} u_k(y) dy = \frac{1}{|B_r(x)|} \int_{B_r(x)} \int u(y-z) \phi_k(z) dzdy$$
The inner integral may be taken over any open set the closure of which contains the support of $\phi_k$, that is, $B_{1/k}(0)$. In particular, for large $k$, you may take the ball of radius $r$ around $0$
Now transform (simply by translation) the inner integral so that you are integrating over $B_r(y)$, and use the mean value inequality in the inner integral to show that the expressions in this formula are $\ge u_k(y)$. That is, $u_k$ satisfies the mean value inequality and therefore $u_k$ is subharmonic. (There are still some gaps where the details need to be filled which I leave to you :-).
