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.

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

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 $\lambda\{f= \infty\} =0$.

  • $\begingroup$ 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? $\endgroup$ – Pablo Herrera Dec 7 '18 at 19:26

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