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Set of rational numbers $\mathbb{Q}$ is measure $0$.

I approach this question by two sides.

(First) Like here Showing that rationals have Lebesgue measure zero., $$ \mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0. $$

(Second) Using the definition of Lebuesgue measure. let's order $\mathbb{Q}=\bigcup_{i=1}^{\infty}\left\{ r_{i}\right\} $. For given $\epsilon>0$, cover $\mathbb{Q}$ by $\bigcup_{i=1}^{\infty}\left\{ r_{i}-\frac{\epsilon}{2^{i+1}},\ r_{i}+\frac{\epsilon}{2^{i+1}}\right\} $(Let this covers $E_{\epsilon}$). Then $\mu\left(\mathbb{Q}\right)\leq\sum_{i=1}^{\infty}\epsilon/2^{i}\leq\epsilon$. So $\mu\left(\mathbb{Q}\right)=0$.

Here, I have a question on the second proof. If we cover $\mathbb{Q}$ like the second proof, I think the covers $E_{\epsilon}$ cover all real numbers $\mathbb{R}$.

The reason is here : If we let $a\in\mathbb{R}\setminus\mathbb{Q}$, for any $\delta>0$, there exists $b\in\mathbb{Q}$ such that $|a-b|<\delta$ (by the density of $\mathbb{Q}$). So $a$ must be in some intervers in the covers $E_{\epsilon}$. So $\mu\left(E_\epsilon\right)=\infty$ $\forall\epsilon>0$.

What is my fault in my reason?

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kayak
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    How do you infer $a$ is covered from $|a-b|<\delta$? There is no guarantee that $\epsilon/2^{i+1}$ is greater than $\delta$. – Cave Johnson Dec 24 '16 at 04:40
  • @CaveJohnson There must be rational numbers next to $a$. So I thought so. – kayak Dec 24 '16 at 04:43
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    This sort of thing is tricky, it'll take a while to wrap your head around this sort of thing because the picture doesn't feel right to start with. This ends up not covering all of $\mathbb{R}$. – Amin Idelhaj Dec 24 '16 at 04:48
  • Suppose $\epsilon_i=|r_i-\pi|$ for each $i.$ Is $\epsilon_i\gt0$ for each $i?$ Does your argument prove that the intervals $(r_i-\epsilon_i,,r_i+\epsilon_i)$ cover all real numbers? All right, which one covers the real number $\pi$? – bof Dec 24 '16 at 05:06
  • @bof no cover covers $\pi$, but I think those covers cover $\mathbb{R}\setminus {\pi}$. – kayak Dec 24 '16 at 05:18
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    Sure they cover $\mathbb R\setminus\pi.$ But doesn't *your argument* show that they cover $\mathbb R$? For any $\delta\gt0$ there exists $b\in\mathbb Q$ such that $|\pi-b|\lt\delta$ (by the density of $\mathbb Q$). "So" $\pi$ must be in some interval . . . – bof Dec 24 '16 at 05:24
  • @bof Your covers differ from mine. – kayak Dec 24 '16 at 05:46
  • You wrote: "If we let $a\in\mathbb{R}\setminus\mathbb{Q}$, for any $\delta>0$, there exists $b\in\mathbb{Q}$ such that $|a-b| < \delta$ (by the density of $\mathbb{Q}$). So $a$ must be in some intervers in the covers $E_\varepsilon$. So $\mu(E_\varepsilon) = \infty$ $\forall \varepsilon>0$." But it can happen that for every rational number $r_i$, the radius $\varepsilon/2^i$ is less than the distance from $r_i$ to $a$. Thus $a$ will not get covered. $\qquad$ – Michael Hardy Dec 24 '16 at 05:59
  • Yep, my cover is different. Why can't I use the same argument you did, to show that my cover covers every real number? – bof Dec 24 '16 at 06:04

1 Answers1

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The fault in your reasoning is where you claim, without justification, "So $a$ must be in some . . ." Unless you come up with some reason for that "so", there is nothing more to be explained here.

Your conclusion is wrong, those intervals do not cover $\mathbb R.$ That is clear from measure theory, as you know. In order to give an explicit example of a real number which is not covered, we have to know how the rationals are enumerated. Since you did not assume anything special about the enumeration, I can use any enumeration I like. Let me define a special enumeration which makes it easy to exhibit an uncovered number. (Also I will take $\epsilon=1.$)

First, let $q_1,q_2,q_3,\dots$ be your favorite enumeration of the rationals. Now I define a new enumeration $\{r_i\}$ recursively. If $r_1,\dots,r_{i-1}$ have already been defined, then I define $r_i$ to be the first $q_j$ such that $q_j\notin\{r_1,\dots,r_{i-1}\}$ and $|q_j-\sqrt2|\gt\frac1{2^{i+1}}.$

Does the sequence $r_1,r_2,r_3,\dots$ contain all the rationals?

Since your sequence $q_1,q_2,q_3,\dots$ contains all the rationals, it will be enough to show that each $q_j$ occurs in the sequence $r_1,r_2,r_3,\dots.\ $ I will prove this by induction on $j.$ Suppose that each of the numbers $q_1,q_2,\dots,q_{j-1}$ occurs in the sequence $\{r_i\}.$ Since $q_j$ is rational while $\sqrt2$ is irrational, we know that $|q_j-\sqrt2|\gt0.$ Thus (with $j$ fixed) if we choose $i$ large enough, we will have both $\{q_1,\dots,q_{j-1}\}\subseteq\{r_1,\dots,r_{i-1}\}$ and $\frac1{2^{i+1}}\lt|q_j-\sqrt2|.$ For such an $i$ we will certainly have $q_j\in\{r_1,\dots,r_{i-1},r_i\}.$

Do the intervals $(r_i-\frac1{2^{i+1}},\ r_i+\frac1{2^{i+1}})$ cover $\mathbb R$?

No, none of those intervals covers $\sqrt2,$ because $|r_i-\sqrt2|\gt\frac1{2^{i+1}}.$

bof
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  • Can you please tell me why ${q_k}_{k=1}^\infty$ contains all rationals? – Kumar Oct 19 '20 at 19:00
  • I wrote thise answer 4 years ago, so I don't remember anything about it, but I can read it. You're asking about the line that says "Since your sequence $q_1,q_2,q_3,\dots$ contains all the rationals . . .". Four lines above that, I read: "First, let $q_1,q_2,q_3,\dots$ be your favorite enumeration of the rationals." So that's why. The sequence was assumed to contain all the rationals. The phrase "enumeration of the rationals" means "sequence containing all the rationals and nothing else." – bof Oct 19 '20 at 23:24