Let consider rational numbers $\{r_n\}_{n=1}^{\infty}$ on [0, 1]. How to prove, that such sum


converges almost everywhere on [0, 1].

There are my thoughts about it.

I confined this task to considering only irrational points on [0, 1]. Then I extract $U_{\varepsilon}$ neighbourhood of any irrational point $x_0$ of [0, 1]; residual sum is evaluates by convergent series. Then I consider extracted neighbourhood and represent it as $\bigcup \limits_{n\in \mathbb{N}} (U_{1/n} \setminus U_{1/(n+1)})$. I divided that residual sum according to numbers appearance in one of the $U_{1/n} \setminus U_{1/(n+1)}$. This residual sum evaluates as series:

$$\cdots \le \sum_{k\in \mathbb{N}}\sqrt{k+1}\sum_{n_k \le n_{0,k}}^{\infty}\frac{1}{n_k^2}$$ where inner sum contains only numbers of rational points which appears in corresponding neighbourhood $U_{1/k} \setminus U_{1/(k+1)}$. There I come to a standstill - if I could evaluate $k \le n_{k}$ for every entry in the corresponding inner sum, I would evaluate that residual sum as convergent series: $\sum \limits_{n\in \mathbb{N}}\frac{\sqrt{n+1}}{n^2}$.

Do you know any better approach or how to finish my?

  • $\begingroup$ Would that sum still connerge almost everywhere even if $r_n$ were all the rations (The rationals are countable) $\endgroup$ – Amr Apr 16 '14 at 8:42
  • $\begingroup$ @Amr, $\{r_n\}$ is a set of all rations on [0, 1]. $\endgroup$ – Andrei Kulunchakov Apr 16 '14 at 8:53
  • $\begingroup$ A friend told me that if that sum converges a.e to let's say integrable function $f$ (which has been proven here), then for every open interval $I$ and for every $k\in \mathbb{N}$: $m\bigl(I\cap\{x|f(x)\geq r_k\}\bigl)>0$. Why is that true? $\endgroup$ – Gil Or Aug 16 '18 at 11:01

Let $f_n:[0,1]\to\Bbb{R}$ be defined, for $x\ne r_n$, by $f_n(x)=\dfrac{1}{\sqrt{\vert x-r_n\vert}}$.

Clearly, for every $n$ we have $$ \Vert f_n \Vert_1=\int_0^{r_n}\frac{dx}{\sqrt{r_n-x}}+ \int_{r_n}^1\frac{dx}{\sqrt{x-r_n}}=2(\sqrt{r_n}+\sqrt{1-r_n})\leq 2\sqrt{2} $$ Now the function $f:[0,1]\to [0,+\infty]$ defined by $f=\sum_{n\geq}\frac{1}{n^2}f_n$, belongs to $L^1([0,1])$ since this series converges normally in $L^1([0,1])$. But the fact that $f\in L^1([0,1])$ implies that the series $\sum_{n\geq}\frac{1}{n^2}f_n(x)$ is almost every where convergent.


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