# Series of integrable functions converges pointwise almost everywhere

I need some help, solving the following problem I found in my textbook. QUESTIONS APPEAR IN BOLD CAPITALS.

Let $(X,\Sigma,\mu)$ be a measure space and $f_n \colon X \to \mathbb{C}$ ($n \in \mathbb{N}$) be a sequence of integrable functions such that $$\sum_{n=1}^\infty \int_X |f_n| \mathrm{d}\mu < \infty.$$

Then $\sum_{n=1}^\infty f_n$ converges almost everywhere to an integrable function $f \colon X \to \mathbb{C}$ and $$\int_X f \mathrm{d}\mu = \sum_{n = 1}^\infty \int_X f_n \mathrm{d}\mu.$$

I know that monotonous convergence gives me $$\sum_{n=1}^\infty \int_X |f_n| \mathrm{d}\mu =\int_X \sum_{n=1}^\infty |f_n| \mathrm{d}\mu,$$ i.e. $\sum_{n=1}^\infty |f_n|$ is integrable. Integrability of $\sum_{n=1}^\infty |f_n|$ gives me that this series converges almost everywhere. Hence, if $\sum_{n=1}^\infty |f_n(x)|$ converges, also $\sum_{n=1}^\infty f_n(x)$ converges. This implies that $\sum_{n=1}^\infty f_n$ converges almost everywhere to some $f$.

Furthermore, $\sum_{n=1}^\infty |f_n|$ a majorant for $|\sum_{n=1}^\infty f_n|$ and since $\sum_{n=1}^\infty f_n$ is measurable as the limit of measurable functions this gives the integrability of $\sum_{n=1}^\infty f_n$.

WHY IS $f$ MEASURABLE? (I think, this won't be the case if the measure space is not assumed to be complete. Am I right? But in case the measure space IS complete, I am good to go on.)

If I have this, then $f$ would be integrable as well, because it almost everywhere equals an integrable function.

This implies $$\int_X f\mathrm{d}\mu = \int_X \sum_{n=1}^\infty f_n \mathrm{d}\mu.$$ BUT WHY IS THIS EQUAL TO $\sum_{n=1}^\infty\int_X f_n \mathrm{d}\mu$?

• So in the first question, you want to know why a function is measurable if it's almost everywhere equal to a measurable function? Commented May 28, 2016 at 12:25
• @Timkinsella Yes, since the problem as far as I know does NOT assume completeness of the measure space. Commented May 28, 2016 at 12:27
• @Timkinsella: Well, in general this is not possible. See math.stackexchange.com/questions/1307044/…. But maybe the setting of this problem still allows this conclusion. Commented May 28, 2016 at 12:29
• I think sometimes people assume a completion has been taken without stating it explicitly. As you probably know, you can always complete a measure. Commented May 28, 2016 at 12:30
• For your other question, I think you can just use the dominated convergence theorem. Commented May 28, 2016 at 12:36

You can define $f(x)$ as $\lim_{N\to +\infty}\sum_{n=1}^N f_n(x)$ when the series $\sum_{n=1}^N f_n(x)$ converges and $0$ otherwise. In this case, the function $f$ is measurable since the set of convergence of a series of measurable functions is measurable and $\sum_{n=1}^N f_n\to f$ almost everywhere.
For the last question, as Tim kinsella suggests, you can use the dominated convergence theorem applied to the sequence $g_N$ defined by $\sum_{n=1}^N f_n$ and the dominating function $g:=\sum_{n=1}^N \left|f_n\right|$.