Prove $f(x)$ is finite a.e. on $x \in [0,1]$

Question

Let $\{r_k\}=[0,1] \cap \mathbb{Q}$ and $$ f(x)=\sum_{k=1}^{\infty} \frac{1}{k^2 \sqrt{|x-r_k|}} $$ Prove that $f(x)<\infty$ a.e. $x \in [0,1]$.

I tried to show $f(x)$ is integrable on $[0,1]$ but could not find a solution. Many thanks for your help!

Answer 1

Let us set $I_k=\int_{0}^{1}\frac{1}{\sqrt{|x-r_k|}}dx.$ It can be easily computed that $I_k=2\left(\sqrt{r_k}+\sqrt{1-r_k}\right)\le 4.$

Now justify the interchange of integral and summation (This can be done via MCT or simply applying Tonelli’s theorem) to obtain that $$\int\limits_{0}^{1}f(x)dx=\sum_{k=1}^{\infty}\frac{1}{k^2}I_{k}\leq 4\sum_{K=1}^{\infty}\frac{1}{k^2}<\infty.$$

It follows therefore that $f(x)<\infty$ for almost every $x.$