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?

| cite | improve this question | | | | |
  • 3
    $\begingroup$ What do you know about the sequence $(f_n + f_1^-)_{n\in\mathbb{N}\setminus \{0\}}$? $\endgroup$ – Daniel Fischer May 19 '16 at 20:34
  • $\begingroup$ 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.. $\endgroup$ – Tesla May 19 '16 at 21:15
  • $\begingroup$ That is possible. But what does Beppo Levi's theorem have to do with the sequence? $\endgroup$ – Daniel Fischer May 19 '16 at 21:18
  • $\begingroup$ 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? $\endgroup$ – Tesla May 19 '16 at 21:26
  • $\begingroup$ I didn't say $f_n + f_n^-$, I said $f_n + f_1^-$. What are the premises of the theorem? Are they satisfied? $\endgroup$ – 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?

| cite | improve this answer | | | | |
  • 1
    $\begingroup$ 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 $$\int_{\Omega} f_n + f_1^- \,d\mu = \int_{\Omega} f_n\,d\mu + \int_{\Omega} f_1^-\,d\mu\quad \text{and}\quad \int_{\Omega} f + f_1^-\,d\mu = \int_{\Omega} f\,d\mu + \int_{\Omega} f_1^-\,d\mu.$$ $\endgroup$ – Daniel Fischer May 20 '16 at 11:15
  • $\begingroup$ 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 $\lim_{n\to\infty}\int_\Omega f_nd\mu = \int_\Omega \lim_{n\to\infty} f_nd\mu$ ? $\endgroup$ – Tesla May 20 '16 at 14:27
  • 1
    $\begingroup$ Yes, since $\int_{\Omega} f_1^- \,d\mu < +\infty$, we can add $-\int_{\Omega} f_1^-\,d\mu$ to both sides of the equation. $\endgroup$ – Daniel Fischer May 20 '16 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.