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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)

AlephZero
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  • In the post you have referred to it does not say that the sum is finite for every irrational number $x$. It only says that the sum is finite almost everywhere. – Kavi Rama Murthy Dec 08 '19 at 05:24
  • I know, however, based on the steps, I understand that the set where it is not finite are the rational number, unless I understood something wrong. – AlephZero Dec 08 '19 at 05:28
  • 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) – AlephZero Dec 08 '19 at 05:31
  • The claim that the sum is fin iet for every irrational nmber $x$ is too strong and I am not suer oif it si true. Even if it is true the proof will involve a good amount of Number Theory. The fact that the sum is finite almost everywhere is trivial and it is already proved in the other post. – Kavi Rama Murthy Dec 08 '19 at 05:33
  • ok, I will change that to almost everywhere, since I'm not completely sure of that – AlephZero Dec 08 '19 at 05:35
  • In that post $(q_n)$ is an ennumeration of rationsal in $[0,1]$, not all rational numbers. That makes a difference too to the proof. – Kavi Rama Murthy Dec 08 '19 at 05:45
  • If the sum is convergent for all rationals $q_n$, shouldn't it be convergent for every sub sequence $g_n$ of rationals? But if we take $x=\pi$ and $g_n=\frac{1}{2^{n+1}}$, then the sum $\sum_{n=1}^\infty \frac{2^{-n}}{|x-g_n|}$ is not convergent. – mathisgood Dec 08 '19 at 06:36
  • it is convergent: https://www.wolframalpha.com/input/?i=sum+2%5E%28-n%29%2F%28%7Cpi-1%2F2%5E%28n%2B1%29%7C%29+1+to+infinite – AlephZero Dec 08 '19 at 06:40

2 Answers2

1

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 $\sum_{n=1}^{\infty} \frac{2^{-n}}{|z-q_n|} \leq \sum_{n=1}^{\infty} 2^{-n}1.5^{n}\frac{1}{\beta_z} < \infty$.

acreativename
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  • What do you mean by "the result"? because your limit tends to infinite as $\epsilon\rightarrow 0$, so do you mean the summation can't be finite? – AlephZero Dec 08 '19 at 06:32
  • 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 $\sum_{n=1}^{\infty} \frac{2^{-n}}{|z-q_n|} \leq \sum_{n=1}^{\infty} 2^{-n}1.5^{n}\frac{1}{\beta_z} < \infty$ – acreativename Dec 08 '19 at 07:06
  • I should edit the proof now. – acreativename Dec 08 '19 at 07:07
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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 $x$.

Kavi Rama Murthy
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  • ok, but say I have both, an enumeration of the rational numbers between $(a,b)$ and an enumeration of the rational numbers outside $(a,b)$, then, the value $\frac{1}{x-q_n}$ of the rationals outside is bounded above by $\max{\frac{1}{q_n-a},\frac{1}{q_n-b}}$ Hence, the sumation or the integral from $a$ to $b$ should also be finite as well, isn't it? – AlephZero Dec 08 '19 at 05:52
  • @GuadalupeAnimation Boundedness of each term does not give you finiteness of the infinite sum. In the proof I have presented what is needed is boundedness of the integral of $\frac 1 {\sqrt {|y|}}$ from $a-q_n$ to $b-q_n$ and for this, we need boundeness of $(q_n)$. – Kavi Rama Murthy Dec 08 '19 at 05:56