Show that $\sum_{n=1}^{\infty}f_n(x)$ is convergent a.e.

Question

I'm currently studying for my master's qualifying exam and stumbled upon this problem.

Let $\{r_1,r_2,\ldots \}$ be an enumeration of the rationals in $[0,1].$ Let $f_n:[0,1]\rightarrow\mathbb{R}$ be defined as:

$f_n(x) = \begin{cases} \frac{1}{2^n\sqrt(|x-r_n|)} & x \neq r_n \\ 0 & x=r_n \end{cases}$.

Show that $\sum_{n=1}^\infty f_n(x)$ is convergent a.e.

I know that for the case of $x=r_n$ we would be working over a set of measure zero since rationals are countable, but I don't know how to tackle the other part of this function sequence.

Any hints on how to tackle this problem will be greatly appreciated!

Answer 1

$\displaystyle\int_I\frac{1}{\sqrt{|x-r_n|}}dx=\int^{r_n}_0\frac{1}{\sqrt{|x-r_n|}}dx+\int^1_{r_n}\frac{1}{\sqrt{|x-r_n|}}dx=2\sqrt{r_n}+2\sqrt{1-r_n}\le 4$. Therefore,

$\int^1_0f_n(x)dx\le \frac{4}{2^n}\tag 1$

Now, $(1)$ and monotone convergence theorem imply that $\sum^{\infty}_{k=1}f_k(x))<\infty$ almost everywhere because

$\int^1_0(\sum^{\infty}_{k=1}f_n(x))dx=\sum^{\infty}_{k=1}\int^1_0f_n(x)dx<\infty\tag2 $

Answer 2

$2^{n}\int f_n(x) dx=\int_{-r_n} ^{1-r_n} \frac 1 {|\sqrt y|} dy=2(\sqrt {1-r_n} -\sqrt {r_n})=2\frac {1-2r_n} {\sqrt {1-r_n} +\sqrt {r_n})}$. Check that this is bounded. [Hint: $\sqrt {1-x} +\sqrt x \geq \sqrt2 $ for $0 \leq x \leq 1$]. Hence $\int f_n(x) dx \leq \frac C {2^{n}}$ for some $C<\infty$ which implies $\sum_n \int f_n(x) dx <\infty$. Hence $\sum f_n(x) <\infty$ almost everywhere.