Let $\{q_n\}$ be an enumeration of the rational numbers, how can I show that $\sum_{n=1}^\infty \frac{2^{-n}}{|x-q_n|}<\infty$ almost everywhere?

Question

I saw this in the following question: Is there a function with infinite integral on every interval?

I already understood all other steps on the first answer, however, I don't know how to prove the following step:

Let $\{q_n\}$ be an enumeration of the rational numbers, how can I justify that $$\sum_{n=1}^\infty \frac{2^{-n}}{|x-q_n|}<\infty$$ for almost every $x\in\mathbb{R}$ (i.e. almost everywhere)?

I know it has something to do with the fact that $2^{-n}$ tend to zero exponentially while $|x-q_n|$ tends to zero linearly.

Also, there are some modification that I made that shouldn't change the result, which are using all the rational numbers instead of only those between 0 and 1, and removing the square root (since it is squared anyways)

Answer 1

I think you have to take $(q_n)$ to be an ennumeration of rationals in a finite interval instead of the whole line.

$\int_a^{b} \sum \frac {2^{-n}} {|x-q_n|} dx = \sum \int_a^{b}\frac {2^{-n}} {|x-q_n|} dx$ and $\int_a^{b} \frac 1 {|x-q_n|}dx=\int_{a-q_n}^{b-q_n} \frac 1 {\sqrt {|y|} } dy$. Since the integral here is bounded and $\sum \frac1 {2^{n}} <\infty$ it follows that $\int_a^{b} \sum \frac {2^{-n}} {|x-q_n|} dx <\infty$ which implies $\sum \frac {2^{-n}} {|x-q_n|} dx<\infty$ for almost all $x \in (a,b)$. Since $a$ and $b$ are arbitrary we see that the sum is finite for almost all real values of $x$.

Answer 2

You could define the sets

$A_{q_j , \epsilon} := \{ y |$ $ $ $|y-q_j| \leq \epsilon \cdot (1.5)^{-j} \}$

and the set $A_{\epsilon} := \bigcup_{j = 1}^{\infty} A_{q_j,\epsilon}$ and note that

$m(A_{\epsilon}) \leq 2\epsilon$ and on $A_{\epsilon}^c$; the value of

$\sum_{n=1}^{\infty} \frac{2^{-n}}{|x-q_n|}$ is atmost

$\sum_{n=1}^{\infty} 2^{-n}(1.5)^{n}\frac{1}{\epsilon}$ which is finite;

Now take the set $\bigcup_{\epsilon > 0} A_{\epsilon}^c$ and note that for $z \in \bigcup_{\epsilon > 0} A_{\epsilon}^c$ there is some $\beta_z > 0$ so that $z \in A_{\beta_z}^c$ for which $\sum_{n=1}^{\infty} \frac{2^{-n}}{|z-q_n|} \leq \sum_{n=1}^{\infty} 2^{-n}1.5^{n}\frac{1}{\beta_z} < \infty$.