# Sequence of integrable function with $\sum_{n=1}^\infty \|f_n\|_1<\infty$. Show that $\sum_{n=1}^\infty f_n$ converges a.e. and is integrable.

Let $\{f_{n}\}$ be a sequence of functions in $L^1(\mathbb{R})$ such that $\displaystyle \sum_{n=1}^\infty\|f\|_{1}<\infty.$ Show that $$f(x): = \sum_{n=1}^\infty f_n(x)\text{ converges a.e., }\, f\in L^{1}(\mathbb{R}), \text{ and } \int_{\mathbb{R}} f = \sum_{n=1}^\infty \int_{\mathbb{R}}f_n$$

Here's my solution:

First, recall that $(\mathbb{N},\mathcal{P}(\mathbb{N}),c)$ and $(\mathbb{R},\mathcal{M},m)$ are $\sigma$-finite measure spaces. From Tonelli's Theorem, $$\int_{\mathbb{N}\times\mathbb{R}}|f_{n}(x)| \, d(c\times m) = \int_{\mathbb{N}} \int_{\mathbb{R}}|f_{n}(x)| \, dc\times dm = \sum_{n=1}^\infty \|f_{n}\|_1<\infty.$$ Since $f_{n}(x)$ is integrable with respect to the product measure of $c$ and $m$, Fubini's Theorem gives $$\int_{\mathbb{R}}f(x)\,dm = \int_{\mathbb{R}}\sum_{n=1}^\infty f_n(x)\,dm =\sum_{n=1}^\infty \int_{\mathbb{R}}f_n(x) \, dm.$$

Now, are we able to say that $f$ is integrable, since the RHS of the above is finite? How can we say the sum converges a.e.? Does that follow immediately from the integrability of $f$?

Is there a simple way to prove this using either the Monotone Convergence Theorem or Lebesgue Dominated Convergence Theorem, without Tonelli or Fubini?

• Note that $||f|| ||g||$ looks different from $\|f\|\|g\|$. The former is coded as ||f|| ||g|| and the latter as \|f\|\|g\|. I changed it to the latter, which is standard. $\qquad$ – Michael Hardy Jul 24 '16 at 0:48
• I'm always looking to improve my TeX. Thanks for the edit! – Michael Shumate Jul 24 '16 at 0:49

I don't know how familiar with Banach spaces you are. It's a standard fact that "absolutely summable series converge" in a Banach space (in the same way this is true for series of numbers). More precisely, if $(v_{n})_{n \in \mathbb{N}}$ is a sequence in a Banach space $B$ and $\sum_{n = 1}^{\infty} \|v_{n}\|_{B} < \infty$, then $\sum_{n =1}^{N} v_{n}$ is a Cauchy sequence in $B$ and thus converges to an element we denote $\sum_{n = 1}^{\infty} v_{n}$. The limit satisfies $\left\|\sum_{n = 1}^{\infty} v_{n} \right\| \leq \sum_{n = 1}^{\infty} \|v_{n}\|_{B}.$ This is another way to approach your question.

Turning to your proof, recall that in Tonelli's theorem one of the conclusions is that the map $x \mapsto \int_{\mathbb{N}} |f_{n}(x)| c(dn)$ is measurable and integrable. In other words, since $$\int_{\mathbb{N}}\int_{\mathbb{R}} |f_{n}(x)| \, m(dx) c(dn) < \infty,$$ as you showed, we can conclude that $$x \mapsto \int_{\mathbb{N}} |f_{n}(x)| \, c(dn) = \sum_{n =1}^{\infty} |f_{n}(x)|$$ is measurable and integrable over $\mathbb{R}$. Since it's integrable, it's also finite a.e.

As to your other question, we could indeed use monotone convergence theorem instead of Fubini/Tonelli. (Some books call this result a corollary of the monotone convergence theorem.) Consider the sequence of functions $g_{N} = \sum_{n = 1}^{N} |f_{n}|$. Notice that $(g_{N})_{N \in \mathbb{N}}$ is a monotone sequence of non-negative functions converging pointwise a.e. (though possibly to infinity at some points). The monotone convergence theorem yields $$\int_{\mathbb{R}} \sum_{n = 1}^{\infty} |f_{n}(x)| \, dx = \lim_{N \to \infty} \sum_{n = 1}^{N} \int_{\mathbb{R}} |f_{n}(x)| \, dx = \sum_{n = 1}^{\infty} \int_{\mathbb{R}} |f_{n}(x)| \, dx < \infty.$$
This shows $\sum_{n = 1}^{\infty} |f_{n}(x)|$ is finite a.e. and thus $\sum_{n = 1}^{\infty} f_{n}(x)$ converges a.e. Using the triangle inequality, we also get integrability from the previous computation.

EDIT: Using the previous paragraph, we can now argue $\sum_{n = 1}^{\infty} \int_{\mathbb{R}} f_{n}(x) \, dx = \int_{\mathbb{R}} \sum_{n = 1}^{\infty} f_{n}(x) \, dx$, as follows: $$\left|\int_{\mathbb{R}} \left(\sum_{n = 1}^{\infty} f_{n}(x) - \sum_{n = 1}^{N} f_{n}(x) \right) \, dx \right| \leq \int_{\mathbb{R}} \sum_{n = N + 1}^{\infty} |f_{n}(x)| \, dx = \sum_{n = N + 1}^{\infty} \int_{\mathbb{R}} |f_{n}(x)| \, dx$$ and the right-hand side goes to zero as $N \to \infty$ by the previous paragraph (being the tail of a convergent series). This shows $$\int_{\mathbb{R}} \sum_{n = 1}^{\infty} f_{n}(x) \, dx = \lim_{n \to \infty} \sum_{n = 1}^{\infty} \int_{\mathbb{R}} f_{n}(x) \, dx$$ by the definition of limit.

• Thanks for the thoughtful answer. I have seen a similar question as a corollary to the monotone convergence theorem, but they assume that the sequence of functions was nonnegative from the start. I guess I wasn't sure that if I can swap the limit with the integral with regards to the absolute value of the functions, that I still can without the absolute value. – Michael Shumate Jul 24 '16 at 1:36
• I think I see what you're saying. I will add to my answer. – fourierwho Jul 24 '16 at 1:40
• See the paragraph starting with "EDIT." If anything else is unclear, feel free to say so. – fourierwho Jul 24 '16 at 1:48
• Yes, that makes a lot of sense. Thanks for the clarification. – Michael Shumate Jul 24 '16 at 1:50