# Convergence of Lebesgue integral with negative part finite

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?

Thanks!

-

## 1 Answer

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.

-
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