In this old post, a sketch of the proof was given. My question is about some details in the proof(which are not given in the books).
Let me simply copy what t.b. wrote in the solution:
First of all, it follows from the uniform boundedness theorem that the sequence $(f_{n})$ is bounded in $L^{1}$.
Next, we can define for a measurable set $E$ a quantity $$\nu(E) = \lim_{n \to \infty} \int_{E} f_{n}$$ because the characteristic function of $E$ belongs to $L^{\infty}$.
Then one can verify that $\nu$ is a (signed) measure which is absolutely continuous with respect to $\mu$. By the Radon-Nikodym theorem it follows that $\nu(E) = \int_{E} f\,d\mu$ for a unique function (class) $f \in L^{1}$.
Now by definition we have $\displaystyle \int f g = \lim_{n \to \infty} \int f_{n} g$ for all characteristic functions $g$. But as the characteristic functions span a dense subspace of $L^{\infty}$ we conclude that this must hold for all $g \in L^{\infty}$.
The sketch is very clear, now here is my question:
(1) How to verify that the $\nu$ is a signed measure, i.e. $\nu$ is countably additive? Take a sequence of mutually disjoint measurable sets $(E_m)_{m\ge 1}$, I want to show that $$\nu(\cup_1^\infty E_m)=\lim_{n \to \infty} \int_{\cup_{m=1}^\infty E_m}f_{n}= \lim_{n \to \infty} \sum_{m=1}^{\infty} \int_{E_m}f_{n} =\sum_{m=1}^{\infty} \lim_{n \to \infty} \int_{E_m} f_{n} =\sum_{m=1}^\infty \nu(E_m)$$ My question is, why is it legal to switch $\sum_{m=1}^{\infty}$ and $\lim_{n \to \infty}$?
(2) In order to show that $f_n$ converges to $f$ weakly, we let $(g_m)$ be a norm convergent simple functions in $L^\infty$ with limit $g$, then we need to verify: $$\int f g=\lim_{m \to \infty}\int f g_m =\lim_{m \to \infty}\lim_{n \to \infty} \int f_{n} g_m=\lim_{n \to \infty}\lim_{m \to \infty} \int f_{n} g_m=\lim_{n \to \infty} \int f_{n} g$$ The first and the last equalities follows from DCT. The second one is also easy to verify, but I don't know why the third equality holds(switch $\lim_{n \to \infty}$ with $\lim_{m \to \infty}$).
