convergence of a series - lebesgue

Question

Let $\xi_n$ a sequence in $[0,1]$. Prove that the series $$\sum_{n=1}^{+ \infty} \frac{1}{{n^2|x-\xi_n|}^{1/2}}$$ converge for almost all $x\in [0,1]$ (Lebesgue measure)

Answer 1

For each $n$,

\begin{align}\int_0^1 \frac{dx}{|x - \xi_n|^{1/2}} &= \int_0^{\xi_n} \frac{dx}{(\xi_n - x)^{1/2}} + \int_{\xi_n}^1 \frac{dx}{(x - \xi_n)^{1/2}}\\ &= \int_0^{\xi_n} \frac{dx}{x^{1/2}} + \int_{\xi_n}^1 \frac{dx}{(x - \xi_n)^{1/2}}\\ &= 2\sqrt{\xi_n} + 2\sqrt{1 - \xi_n}.\\ \end{align}

Since $$\sum_{n = 1}^\infty \frac{\sqrt{\xi_n} + \sqrt{1 - \xi_n}}{n^2} \le \sum_{n = 1}^\infty \frac{2}{n^2} < \infty,$$ it follows that the series $\sum_{n = 1}^\infty \frac{1}{n^2|x - \xi|^{1/2}}$ is integrable on $[0,1]$, so it is convergent almost everywhere in $[0,1]$.