Convergence of harmonic functions in $L^1_{loc}$ implies uniform convergence on compact subsets I came across this question when reading the proof of Lemma 6.19 p246 in the book A Course in Complex Analysis and Riemann Surfaces by Schlag. In the proof, they make the following reasoning.
Take $V\subset \mathbb{C}$ open and $\{u_n \}_n \subset C^{\infty}(V)$ a sequence of harmonic functions on $V$ which converges in the sense of $L^1_{loc}$ to $u_{\infty}$, meaning that for any compact $K\subset V$, we have 
$$ \iint_K |u_n(z)-u_{\infty}(z)|dz \rightarrow 0 $$
as $n\rightarrow \infty$. Now, by the mean value property (MVP), we have for each disk $D(z_0,r)$:
$$u_n(z_0)= \frac{1}{\pi r^2}\iint_{D(z_0,r)} u_n(z)dz$$
which implies that $\{u_n \}_n$ is a Cauchy sequence in $C(V)$ and therefore converges uniformly on compact subsets of $V$ to $u_{\infty}$.
My questions are:
1) What do they mean with '$\{u_n \}_n$ is a Cauchy sequence in $C(V)$'? I am pretty sure that this means $\sup_{z\in V}|u_n(z)-u_m(z)|\rightarrow 0$ as $n,m\rightarrow \infty$ but I just want to make sure this interpretation is correct. 
2) How does this alternative way of writing the MVP imply the uniform convergence on compact subsets of $V$ to $u_{\infty}$? I can only show that this implies pointwise convergence, but this is not as strong as their claim.
Any help would be very much appreciated.
 A: I'll answer question 2, which will hopefully clarify question 1 a bit. (I don't have the book so I am not sure about what they use as the notational convention.)
Let $K\subset V$ be compact. I claim first that there exists $K' \subset V$ and an open set $W \subset K'$ such that $K \subset W$. 
This follows by noting that $K$ has positive distance to $V^C$ which is closed. So let $r = \frac12 \mathrm{dist}(K, V^C)$ and you can set $W = \cup_{p\in K} D(p,r)$ and $K'$ the closure of $W$. 
Then for any $p\in K$, using the same $r$ as in the above construction, we have that using the mean value property
$$ |u_n(p) - u_\infty(p)| \leq \frac{1}{\pi r^2} \iint_{D(p,r)} | u_n(z) - u_\infty(z)| ~\mathrm{d}z \leq \frac{1}{\pi r^2} \iint_{K'} |u_n(z) - u_\infty(z)| ~\mathrm{d}z $$
Note that on the far right the domain of integration is $K'$ independent of $p\in K$. Hence the convergence of $u_n\to u_\infty$ is uniform on $K$.  (But note that the "$\epsilon$" in the convergence is determined by the $L^1$ convergence on a somewhat bigger $K'$, not just $K$ itself.)
