I was reading this proof and needed a clarification. The question is too old to post in it, hence this question.

The original link is here: Outer Measure of the set $E=(\mathbb Q\times \mathbb R) \cup (\mathbb R\times \mathbb Q)$

The proof writer mentioned that using an enumeration $\mathbb{Q} = \{x_1, x_2, \cdots\}$, for each $\epsilon > 0$ let us consider the following sets:

$$ \begin{gathered} A_{i,n} = (x_i - \epsilon 4^{-i-n}, x_i + \epsilon 4^{-i-n}) \times (-2^n, 2^n), \\ B_{i,n} = (-2^n, 2^n) \times (x_i - \epsilon 4^{-i-n}, x_i + \epsilon 4^{-i-n}). \end{gathered} $$

It is easy to see that $E \subseteq \bigcup_{i,n\geq 1} A_{i,n} \cup B_{i,n}$.

I don't understand why $(-2^n, 2^n)$ would cover the real part of the set. How should we define $n$ to make this happen?

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    $\begingroup$ For any $x\in\mathbb{R}$, there is $n\in\mathbb{N}$ such that $|x|<2^n$. $\endgroup$ Sep 23 at 5:20
  • $\begingroup$ With a fixed $i$, you only need to cover the lines $\{x_i\}\times\mathbb R$ and $\mathbb R\times\{x_i\}$. Do that by making increasingly longer and thinner rectangles. Say for the first line use the rectangles $A_{i,n}$, for every $n\in\mathbb N$. The area of it is $2\cdot\epsilon 4^{-i-n}\cdot 2\cdot 2^n=4\epsilon 4^{-i}2^{-n}$ so the total area (for all $n$) is $4\epsilon 4^{-i}$ (as$\sum_{n=1}^{\infty}2^{-n}=1$), which still decreases exponentially in $i$ and now you can sum it over $i$. $\endgroup$ Sep 23 at 5:30
  • $\begingroup$ @OliverDíaz I guess this follows from the Archimedean property of the reals. $\endgroup$
    – Note
    Sep 23 at 6:06
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    $\begingroup$ @Note: yes it does. At this point, all properties of the reals can be used (and are used) in the construction of measures and integral. $\endgroup$ Sep 23 at 13:44
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    $\begingroup$ If that is the problem, you can even revert to high school algebra and say, if $x\in\mathbb R$, and you want $|x|<2^n$, just pick an integer $n$ bigger than $\log_2 |x|$, e.g. $n=1+\lfloor\log_2|x|\rfloor$ (or, pick any integer if $x=0$). (The fact that you can pick that integer is a consequence of the Archimedean property, of course.) $\endgroup$ Sep 24 at 8:19


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