Limit of bounded harmonic functions is harmonic I am trying to solve this old qual problem:

Suppose $\{u_n\}_{n=1}^\infty$ is a sequence of functions harmonic on an open set $U \subset \mathbb{C}$ and uniformly bounded by 1.  Suppose there is a function $u : U \to \mathbb{R}$ such that $u_n \to u$ pointwise.  Show $u$ is harmonic on $U$.

I would like to solve this using only elementary complex analysis.  My idea is this:
Pick a point $z \in U$ and a disk $D_r(z) \subset U$.  Since the disk is simply connected, each $u_n$ is the real part of some holomorphic $f_n$.  Now, we would like to show that $f_n$ converges uniformly on compact subsets...
I haven't gotten any further than this.
Will this approach work?  Other ideas which use complex analysis are appreciated.
 A: This answer does not use complex analysis per se, but a basic property of the Lebesgue integral, the dominated convergence theorem. Sorry! I thought you might find it useful nonetheless. 
To show that $u$ is harmonic, choose a point $z_0\in U$ and observe that each $u_n$ satisfies the mean-value property at $z_0$:
$$
u_n(z_0) = \frac{1}{2\pi r}\int_{|z-z_0|=r} u_n(z), \text{ for all sufficiently small $r>0$.}
$$
Since each $u_n$ is dominated in modulus by $1$, the dominated convergence theorem allows us to let $n$ approach infinity and interchange limit with integral. Then
$$
u(z_0) = \frac{1}{2\pi r}\int_{|z-z_0|=r} u(z), \text{ for all sufficiently small $r>0$.}
$$
A: As you indicate, we can assume that $U$ is a disk, since being harmonic is a local property, and so there exists a sequence of holomorphic functions $f_n$ with $u_n = \Re f_n$. In order to elevate pointwise convergence to locally uniform convergence you need something like boundedness, and one common trick to pass from a bounded real part to a bounded function is to exponentiate, i.e., consider $g_n = \exp f_n$. Then $(g_n)$ is a sequence of holomorphic functions with $|g_n| = \exp u_n \le e$. By Montel's theorem every uniformly bounded sequence of holomorphic functions is normal, i.e., there exists a locally uniformly convergent subsequence $g_{n_k} \to g$, where $g$ is holomorphic. Then $|g_{n_k}|  = \exp u_n \to \exp u$ pointwise, so $|g| = \exp u$, and $u = \log |g|$ is harmonic.
