A sequence of harmonic functions on $\mathbb{R}^{2}$ converging in distribution must necessarily converge locally uniformly to an harmonic function.

Question

I am self studying the Rudin's book of Functional Analysis and I stumbled upon this problem. Given a sequence $\{ f_{j} \}_{j}$ of harmonic functions on an open set $\Omega$ of $\mathbb{R}^{2},$ if $\{ f_{j} \}_{j}$ converges to a distribution $\Gamma$ then $\Gamma$ has to be an harmonic function on $\Omega.$ My attempted solution was the following. The main point was to prove that $\{ f_{j}\}_{j}$ is a Cauchy sequence in $L^{\infty}(K)$ for every compact $K.$ Since the function $f_{i}$ is harmonic for every $i$ given a compact set $K$ we have that $$ f_{i}(x)-f_{j}(x)=\int_{B(x,r)} \frac{1}{\vert B(x,r) \vert} (f_{i}(y)-f_{j}(y))\, dy, $$ where $B(x,r)$ is a ball of radius $r,$ with $r$ suitably small and $x$ is a point of $K$. Maybe if I could write this last integral using a test function then I would have concluded, but since now I did not get any idea. Once this point is done the conclusion follows easily: the pointwise limit of the sequence $\{f_{j}\}_{j}$ fulfills the mean value property.