# $(t_k)_{k\geq 1}$ secuence in $[0,1]$ , $\sum a_k <\infty$ implies $\sum \frac{a_k}{\sqrt{|t_k-x|}}$ a.e

I've read a solution for the following problem but I'm not convinced if solution it's correct.

The problem: Let $$(t_k)_{k\geq 1}$$ be a sequence in $$[0,1]$$ and $$(a_k)_{k\geq 1}$$ be a sequence of nonnegative real numbers such that $$\sum_{k=1}^{\infty} a_k < \infty$$. Prove that $$\sum_{k=1}^{\infty}\frac{a_k}{\sqrt{|x-t_k|}}$$ converges for almost all $$x \in [0,1]$$.

The solution: Consider $$f_n(x)=\sum_{k=1}^{n} \frac{a_k}{\sqrt{|x-t_k|}}\geq 0$$. Since for each $$t \in [0,1]$$ we have

$$\int_{0}^1 \frac{dx}{\sqrt{|t-x|}}=\int_{0}^t \frac{dx}{\sqrt{t-x}}+\int_{t}^1 \frac{dx}{\sqrt{x-t}}=\frac{\sqrt{t}}{2}+\frac{\sqrt{1-t}}{2}\leq 1$$ we infer that $$\int_{0}^1 f_n = \sum_{k=1}^{n} \int_{0}^{1} \frac{a_k}{\sqrt{|x-t_k|}}\leq \sum_{k=1}^{n} a_k$$ Hence $$(f_n)_n$$ is convergent in $$L^{1}([0,1])$$. In particular by a $$\textbf{ familiar theorem}$$, the numerical sequence $$(f_n(x))_n$$ will have to converge for almost all $$x \in [0,1]$$

My question is. What familiar theorem? In general convergence in $$L^{1}[0,1]$$ doesn't imply almost everywhere pointwise convergence. Just imply the existence of a subsequence convergent pointwise a.e. Please, let me know if that solution is wrong or another aproach to solve it.

• You do see that $f_1 \le f_2 \le f_3 \le \dots$, don't you? – GEdgar Dec 6 '18 at 23:41
• @GEdgar Yes. Do you mean Monotone Convergence Theorem? – Pablo Herrera Dec 7 '18 at 2:43

## 1 Answer

First we should note that $$f(x):=\lim_{n \rightarrow \infty} f_n(x) = \sum_{k=1}^\infty a_k |x-t_k|^{-1/2}$$ exists in $$[0,\infty]$$. Strictly speaking, this function is not defined in all points $$t_k$$. But since we have only countable many of them, the corresponding set is a $$\lambda$$-nullset.

Now we can apply the monotone convergence theorem in order to get $$\int_0^1 f(x) \, dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n a_k \int_0^1 |x-t_k|^{-1/2} \, dx \le \sum_{k=1}^\infty a_k<\infty,$$ where in the last step we have used your argument. Thus $$f$$ is integrable and this implies that $$f(x) < \infty$$ for almost all $$x \in [0,1]$$ - in other words, the series converges for almost everywhere. In fact, note that $$C:=\int_0^1 f(x) dx \ge k \lambda\{f >k\},$$ i.e. $$\lambda \{f >k\} \le C/k$$. Thus letting $$k \rightarrow \infty$$, we get $$\lambda\{f= \infty\} =0$$.

• Good solution. I think that another aproach it would be have a monotone sequence such that it has a convergent subsequence almost everywhere then it is convergent almost everywhere. What do you think about that? – Pablo Herrera Dec 7 '18 at 19:26