The Weierstrass theorem from complex analysis The Weierstrass theorem from complex analysis states the following:
Suppose $f_n$ is a sequence of analytic functions converging uniformly on an any compact subset of its domain to $f$. Then $f$ is analytic. Stein gives a tentative proof of it using the Morera's theorem. Given any triangle $T$ in the domain, we want to show that $\int_T f = 0$, which is true if $\int_T f = \int_T \lim_n f_n$. I want to ask whether this argument really works or not since I am not able for now to come up with a justification for the exchange of the limit and the integration. DCT always comes in handy in such a situation, but I do not see how it might be applied here.
 A: All it takes is that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$ on any compact set and, in particular, on $T$. Now, you just use the theorm that says than if $(g_n)_{n\in\mathbb N}$ is sequence of functions from $[a,b]$ to $\mathbb C$ which converges uniformly to $g\colon[a,b]\longrightarrow\mathbb C$, then$$\lim_{n\to\infty}\int_a^bg_n(t)\,\mathrm dt=\int_a^bg(t)\,\mathrm dt.$$
A: You can use DCT if you like: Simply note that $|f_n| \le |f| + 1$ in $T$ for all sufficiently large $n$. Since the uniform limit of continuous functions on a compact set is continuous, $f$ is continuous and therefore $\int_T\big( |f| + 1\big) < \infty$. By DCT $\int_T f_n \to \int_T f$, so $\int_T f = 0$ and $f$ is analytic by Morera's theorem.
It's good to note that you really don't need DCT. With uniform convergence, we could write
$$
\left|\int_T f_n - \int_T f\right| \le \int_T |f_n-f| \le \int_T \sup_T|f_n-f|,
$$
and the last integral above is just the perimeter of the fixed triangle $T$ times $\sup_T|f_n-f|$, which goes to $0$ by hypothesis because $T$ is compact.
