# $\int\lim_{n\to\infty}f_nd\mu = \lim_{n\to\infty}\int f_n d\mu$

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?

• What do you know about the sequence $(f_n + f_1^-)_{n\in\mathbb{N}\setminus \{0\}}$? – Daniel Fischer May 19 '16 at 20:34
• Hmm. $(f_n)$ is non-decreasing, so I don't know why it shouldn't be possible that $\int f_n + f_1{^-}d\mu = \infty$ for $n$ large enough? I probably don't understand your hint, sorry.. – Tesla May 19 '16 at 21:15
• That is possible. But what does Beppo Levi's theorem have to do with the sequence? – Daniel Fischer May 19 '16 at 21:18
• Since $\int f{_1}{^-} d\mu \lt \infty$ and $(f_n)$ non-decreasing, we obtain $\int f{_1}{^-} d\mu \ge \int f{_n}{^-} d\mu$ for all $n$ and therefore the sequence $(f_n+ f{_1}{^-})$ you gave me is non-negative, what is required in order to make use of Beppo Levi's theorem? – Tesla May 19 '16 at 21:26
• I didn't say $f_n + f_n^-$, I said $f_n + f_1^-$. What are the premises of the theorem? Are they satisfied? – Daniel Fischer May 19 '16 at 21:28

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?

• You don't need $f_1^- \geqslant f_2^- \geqslant \dotsc$. If $(f_n)$ is a nondecreasing sequence, then so is $(f_n + g)$ for every $g$. And with $f_1 + f_1^- = f_1^+ \geqslant 0$, we get $0 \leqslant f_1 + f_1^- \leqslant f_2 + f_1^- \leqslant f_3 + f_1^- \leqslant \dotsc$. So we can apply Levi's theorem to that sequence. Now note that $$\int_{\Omega} f_n + f_1^- \,d\mu = \int_{\Omega} f_n\,d\mu + \int_{\Omega} f_1^-\,d\mu\quad \text{and}\quad \int_{\Omega} f + f_1^-\,d\mu = \int_{\Omega} f\,d\mu + \int_{\Omega} f_1^-\,d\mu.$$ – Daniel Fischer May 20 '16 at 11:15
• Okay so $\lim_{n\to\infty}\int_\Omega f_n + f_1^-d\mu =\lim_{n\to\infty}(\int_\Omega f_nd\mu+\int_\Omega f_1^-d\mu) = \lim_{n\to\infty}\int_\Omega f_nd\mu+ \lim_{n\to\infty}\int_\Omega f_1^-d\mu = \int_\Omega \lim_{n\to\infty} f_nd\mu + \int_\Omega \lim_{n\to\infty} f_1^-\mu$, more particularly $\lim_{n\to\infty}\int_\Omega f_nd\mu = \int_\Omega \lim_{n\to\infty} f_nd\mu$ ? – Tesla May 20 '16 at 14:27
• Yes, since $\int_{\Omega} f_1^- \,d\mu < +\infty$, we can add $-\int_{\Omega} f_1^-\,d\mu$ to both sides of the equation. – Daniel Fischer May 20 '16 at 14:35