Convergence of harmonic functions in $L^1$ implies uniform convergence on compact sets Resorting to an analog of what's done here, I'm trying to prove the following statement:

Let $u_m: \mathbb{R}^n \to \mathbb{R}$ be a sequence of harmonic
   functions and suppose there exists a continuous function $u:
 \mathbb{R}^n \to \mathbb{R}$ such that $u_m$ converges to $u$ in $L^1$
   in any compact subset, i.e., for every limited $A\subset
 \mathbb{R}^n$, we have  $$ \int_A \left\vert u_m(x) -
 u(x)\right\vert~dx \to 0, $$ as $m\to \infty$. 
Show such convergence is uniform in compact sets.

My attempt:
Let $\Omega \subset \mathbb{R}^n$ be an open set and $(u_m)_{m \in \mathbb{N}} \subset C^{\infty}(\Omega)$. If $\Omega$ is open, then $\Omega^c = \mathbb{R}^n \setminus\Omega$ is closed, so we may take $r = \frac{1}{2}dist(A, \Omega^c)$ and define $V = \cup_{x_0 \in A} B(x_0, r)$ such that $\tilde{A} = \overline{V}$, where $B(x_0, r)$ denotes the ball of center $x_0$ and radius $r$. 
At this point, I want to say there exists $n \in \mathbb{N}$ such that $\forall x_0 \in A$, with $r$ as defined above, so that we may apply the mean value-property to $u_m(x_0) - u(x_0)$ in the following manner:
$$
|u_m(x_0) - u(x_0)| \leq \frac{1}{|B(x_0, r)|} \int_{B(x_0, r)} |u_m(y) - u(y)| ~dy \leq \int_{\tilde{A}} |u_m(y) - u(y)| ~dy \to 0,
$$
as $m \to \infty$, by the $L^1$ convergence hypothesis. Since $y \in \tilde{A}$ is arbitrary on the RHS of the last inequality, we'd conclude the convergence is uniform and the proof would be finished. 
However, the given function $u$ is not harmonic, only continuous, so we're not able to apply the mean-value theorem the way I described and follow on accordingly with the proof above. Is there any way to correct this? If not, then how may I prove such uniform convergence?
 A: There are two steps to this problem, showing first that $u$ is harmonic and then the uniform convergence over compact sets.
To prove that $u$ is harmonic, it is enough to show that $u$ has the mean-value property. The latter coupled with continuity of $u$ will imply that $u$ is harmonic. To this end, fix $r>0$ and let $B(x_0, r)\subset \mathbb{R}^n$ be a ball of radius $r$ centered at $x_0$. 
Since each $u_m$ is harmonic, it has the mean-value property, hence
$$
\left|u_m(x_0) - \frac{1}{|B(x_0, r)|} \int_{B(x_0, r)} u(y) dy  \right|  = \left| \frac{1}{|B(x_0, r)|} \int_{B(x_0, r)} [u_m(y) - u(y)] dy  \right| \leq  \\ \frac{1}{|B(x_0, r)|} \int_{B(x_0, r)} |u_m(y) - u(y)| dy   \to 0  \text{ as } m \to \infty,
$$
where the last convergence follows from the condition of the problem.
We see that $u_m (x) $ converges pointwise to $\overline{u}_r(x) := \frac{1}{|B(x_0, r)|} \int_{B(x_0, r)} u(y) dy$ for all $x\in \mathbb{R}^n$. But recall that $|| u_m - u||_{L^1(K)} \to 0$ if $K\subset \mathbb{R}^n$ is compact. Hence there is a subsequence of $u_m$, call it $u_m$ after relabeling, which converges to $u$ almost everywhere on $K$. But then the two almost everywhere limits of $u_m$ must coincide (almost everywhere), hence $u(x) = \overline{u}_r(x)$ almost everywhere on any compact $K$ and hence almost everywhere on $\mathbb{R}^n$. Thanks to the continuity of $u$ the equality $u(x) = \overline{u}_r(x)$ holds everywhere. Since $r>0$ was arbitrary, $u$ is harmonic in $\mathbb{R}^n$.
Next, we show the uniform convergence of $u_m$ to $u$ over bounded sets. Fix any $K\subset \mathbb{R}^n$ bounded and let $\Omega \subset \mathbb{R}^n$ be a compact set containing $K$ with its $1$-neighborhood. Fix also $0<r<1$, then for any $x\in K$ using the mean-value property for  $u_m$ and $u$ we get
$$
|u_m(x) - u(x) | \leq \frac{1}{|B(x,r)|} \int\limits_{B(x, r)} | u_m(y) - u(y) | dy \leq \frac{1}{c_n r^n} \int\limits_{\Omega} | u_m(y) - u(y) | dy.
$$
The right-hand side converges to $0$ as $m\to \infty$ and does not depend on $x\in K$. Hence the convergence on $K$ is uniform and the proof is complete.
