Where $\{q_n\}=\mathbb Q$ and $f_n:[q_n-2^{-n-1},q_n+2^{-n-1}]\to[0,\infty)$ with $\int f_n\,d\lambda=1$, show $\sum_{n=1}^\infty f_n<\infty$ a.e. That is: Let $\mathbb Q=\{q_n\}_{n\in\mathbb N}$ be an enumeration of the rationals. Let $f_n$ be a nonnegative Borel measurable function supported on $q_n\pm 2^{-n-1}$ with $\int f_n\,d\lambda =1$, where $\lambda$ is the Lebesgue measure. Show that $\sum_{n=1}^\infty f_n(x)<\infty$ a.e. (with respect to $\lambda$).
This is part of an old preliminary exam in Analysis I'm using to prep for my prelim. I'm at a bit of a loss here; so far I've focused on thinking about $A\cap S_{n}$, where $S_n$ is the support of $f_n$ and $A$ is the set on which $\sum_{n=1}^\infty f_n(x)=\infty$. I've wondered if there is some way to show that (for any subsequence $S_{n_k}$) the size of this intersection shrinks much faster than $f_n$ grows. But I can't seem to flesh this idea out. Am I on the right track?
 A: In fact each $f_n$ need not be integrable, which means that growth of $f_n$ will not play much role here. What is important here is the following observations:


*

*For a.e. $x \in \Bbb{R}$ only finitely many of supports $\operatorname{supp}(f_n)$ overlap.

*Each $f_n$ is finite a.e.


Using this, you can reduce the infinite sum $\sum_{n=1}^{\infty} f_n(x)$ to a finitely many sum of finite numbers for a.e. $x$, which is enough to prove the claim.

Let $I_n = [q_n - 2^{-n-1}, q_n + 2^{-n-1}]$ and notice that
$$ \int \sum_{n=1}^{\infty} \mathbf{1}_{I_n}(x) \, \lambda(\mathrm{d}x) < \infty. $$
This shows that the set
$$A = \{ x :  \textstyle \sum_{n=1}^{\infty} \mathbf{1}_{I_n}(x) < \infty \}$$
satisfies $\lambda(\Bbb{R}\setminus A) = 0$. (Notice that this is the standard Borel-Cantelli argument.) On the other hand, we know that each $f_n$ is finite a.e. Thus the set
$$B = \{ x : f_n(x) < \infty \text{ for all } n \geq 1 \}$$
also satisfies $\lambda(\Bbb{R}\setminus B) = 0$. Now for each $x \in A \cap B$,
$$ \sum_{n=1}^{\infty} f_n(x) = \sum_{n \, : \, x \in I_n} f_n(x) $$
is a finitely many sum of finite numbers, and hence finite.
