How can I prove $$\int\lim_{n\to\infty}f_nd\mu = \lim_{n\to\infty}\int f_n d\mu$$ given a measure space $(\Omega,\mathfrak A, \mu)$, a non-decreasing sequence $(f_n)$ of measurable functions on $\Omega$ and $\int f{_1}{^-} d\mu \lt \infty$?
I have tried to get in into a form so that I could maybe make use of the theorem of Beppo Levi, but I've failed. Can someone give me a hint where I should start?
Ok so this is my result so far:
Since $(f_n(x)+f_1^-(x))_{n\in\mathbb N}$ is non-decreasing and $$f{_1}{^-}(x) \ge f{_2}{^-}(x)\ge...$$ for all $x \in\Omega $ , we obtain $$0\le f_1(x)+f{_1}{^-}(x)\le f_2(x) + f{_1}{^-}(x) \le...$$ for all $x$ and $$f(x) + f{_1}{^-}(x)=\lim_{n\to\infty}f_n(x)+f_1^-(x).$$ Beppi Levi's theorem now tells us that $$\lim_{n\to\infty}\int_\Omega f_n+f_1^-d\mu =\int_\Omega \lim_{n\to\infty} f_n+f_1^-d\mu=\int_\Omega f+f_1^-d\mu.$$
But what I wanted to prove is $$\int_\Omega \lim_{n\to\infty} f_nd\mu=\lim_{n\to\infty}\int_\Omega f_nd\mu,$$so how do I get there? Is it right what I did so far? I know that $f{_1}{^-}(x)$ does not depend on $n$, is that the decisive for the last step?