Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(f_{n})_{n}$ a sequence in $\mathcal{L}^1(\mathbb{R})$ and $f_{n}<f_{n+1}$. Also $\int_{\mathbb{R}}f_{k}^-dm < \infty$ for some $k\in \mathbb{N}$. Show that $$\lim_{n\to\infty}\int_{\mathbb{R}}f_{n}dm=\int_{\mathbb{R}}\lim_{n\to\infty}f_{n}dm.$$

I think write this like a growing sequence for use the monotone Lebesgue theorem, some help?


share|cite|improve this question

If you assume that the limit exists almost-everywhere (otherwise the question doesn't even make sense), you basically spelled out the answer.

Without loss of generality, the negative part of $f_0$ is integrable (we can always forget finitely many elements of the sequence). Put $g_n=f_n+(f_0^-)$. Then $g_n$ are integrable, positive and increasing, so by monotone convergence $$\int f_n+\int f_0^-=\int g_n\to \int\lim g_n=\int(\lim f_n+f_0^-)=\int\lim f_n+\int f_0^-$$ from which you immediately get the result.

share|cite|improve this answer
The pointwise limit of an increasing sequence of functions always exists in the extended reals. – Michael Greinecker Jun 14 '12 at 5:43
You can use \left( and \right) to adjust parenthesis size. – Did Jun 14 '12 at 10:00
M. Greinecker: you're right. It still makes sense, because all the functions considered are (mostly) positive. Still, I hesitate to write an integral of a function infinite on a big set. – tomasz Jun 14 '12 at 13:03
did: I know, but doing that wouldn't improve readability much in this case, and would make the expression two-line, which would be bad. Instead, I changed the notation somewhat -- should be prettier now. – tomasz Jun 14 '12 at 13:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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