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Suppose we define an equivalence relation on $[0,1)$ by saying that $x \sim y$ iff $x-y$ is rational. Let $N \subset [0,1)$ which contains exactly one member from each equivalence class. Also let $R = \mathbb{Q} \cap [0,1)$ and for each $r \in R$, $$N_{r} = \{x+r: x \in N \cap [0,1-r) \} \cup \{x+r-1: x \in N \cap [1-r,1) \}$$

Does $N$ consist of all the rational numbers in $[0,1)$? What is the motivation in defining $N_r$?

Source: Real Analysis: Modern Techniques and Their Applications by Folland

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Note that $x \in R$ implies $x \sim 0$ and so $R \subset [0]$, hence, clearly $R \not\subset N$. (This is equivalent to the arguments present in many of the answers. Perhaps it's slightly more concise. Can you prove that, in fact, $R = [0]$?) – cardinal Jun 18 '11 at 22:19
up vote 9 down vote accepted

No, $N$ does not contain all rational numbers in $[0,1)$. In fact, it contains one and only one rational number from $[0,1)$ (though we don't know which, since $N$ is defined non-constructively via the Axiom of Choice).

As is mentioned, $N$ contains exactly one member from each equivalence class under $\sim$. That means that:

(i) For every $x\in[0,1)$, there exists $y\in N$ such that $x\sim y$, i.e., such that $x-y\in\mathbb{Q}$; and

(ii) For every $x,y\in N$, if $x\sim y$ (that is, if $x-y\in\mathbb{Q}$), then $x=y$.

In particular, we know that there exists $x\in N$ such that $0\sim x$; since this requires that $-x\in\mathbb{Q}$, that means that $x$ is rational. But if $y\in N$ is rational, then $x-y\in\mathbb{Q}$, hence $x\sim y$, so by (ii) we conclude that $x=y$. So $N$ contains one and only one rational from $[0,1)$.

The motivation fro the definition of $N_r$ is that it is a "wrapped around" translate of $N$ by $r$. That is, take every number in $N$, add $r$ to it; if it is still in $[0,1)$, keep it there; otherwise, shift it back by $1$ (so that it falls inside $[0,1)$). These are just rational translates of $N$, adjusted to make sure they stay in the interval.

The reason you want the $N_r$ is that you will show that these countably many translated copies of $N$ will "cover" a nontrivial interval; they are pairwise disjoint (that is, if $r\neq s$, then $N_r\cap N_s=\emptyset$), so the fact that countably many disjoint sets cover a nontrivial interval will imply that, if they are all measurable, then their measure must be positive (by $\sigma$-additivity). Since they are all just translates of the same set $N$, they are each measurable if and only if $N$ itself is measurable, and they will all have the same measure. But then measurability of $N$ would imply that the union of the countably many disjoint translated copies of $N$ will have infinite measure, which will also be impossible (since they are all contained in $[0,1)$, so by monotonicity, their measure, if it exists, will be finite). These things put together will imply that $N$ cannot possibly be measurable.

(As to how one would come up with it in the first place, you'll have to ask Vitali...)

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A better way to state the definition would be $$ N_r = \{x + r \bmod{1} : x \in N\}. $$ Since $N$ contains exactly one member from each equivalence class, the sets $$ N_r, \, r \in \mathbb{Q} \cap [0,1) $$ partition $[0,1)$.

If $x,y \in \mathbb{Q}$ then $x-y \in \mathbb{Q}$ and so $N$ cannot contain both $x$ and $y$. In other words, $N$ contains at most one rational. On the other hand, the rationals in $[0,1)$ form an equivalence class (same argument), and so $N$ contains exactly one rational.

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$[0,1)=\cup N_r$; each $N_r$ is a translated copy of $N$ (the definition is so that it wraps around to stay in $[0,1)$). so if $N$ (and hence $N_r$) is measureable then $[0,1)$ has measure $0$ if the measure of $N$ is zero, or measure infinity if $N$ has some non-zero measure (assuming translation invariance etc).

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It is not true that $N=[0,1)\cap\mathbb{Q}$, because in fact, any two $a,b\in\mathbb{Q}$ (regardless of whether they are in $[0,1)$ or not) are equivalent under $\sim$, because $a-b\in\mathbb{Q}$; and $N$ consists of one choice of representative for each equivalence class of $\sim$, so $N$ will only have one rational number in it. In fact, $N$ is not defined explicitly; for each equivalence class of $\sim$, we need to choose a single representative in $[0,1)$, and then the set of these representatives is $N$, but the choice is not unique. Here is an analogy: if we needed to pick a representative of each equivalence class of $\equiv\pmod 4$ in $[0,67)$, we could choose $\{0,1,2,3\}$, or $\{4,9,18,63\}$, or lots of other things.

As to the motivation for the $N_r$'s, I'm not sure.

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To summarize, the construction is used to show, by counterexample, that there is no translation-invariant, countably-additive measure that can be defined on every subset of the reals; monotonicity and translation-invariance are assumed, and a contradiction is reached. The construction gives you a countably-infinite collection of disjoint subsets (disjointness is needed for additivity), all of equal measure, since all the sets are translates of one set (this uses translation-invariance), so that the total measure $m(N)$ is the sum of the measure of the translates, and then $m(N)=\lim_{n \rightarrow \infty}n\cdot m(t_i)$, where $t_i$ is a translate. Your choices then are that $m(t_i)$ is not zero, and then $m(N)=\infty$, or you can assign measure zero to each, i.e., $m(t_i)=0$. Then, having used the previous two conditions, as Arturo said, monotonicity is used in one of the cases (the one where $m(t_i) \neq0 $ reaching the contradiction that a (measurable) subset of $[0,1]$ has infinite measure (and, in particular, has measure larger than $1=m([0,1])$), i.e., by assuming countable additivity and translation-invariance of the measure, you have constructed a subset of $[0,1]$ that violates monotonicity.

Maybe a rewrite will help other readers; I just hope that if I get another Couric, it will be explained/justified.

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Please call them downvotes, not Courics (people will find it inappropriate or baffling, depending on whether or not they watch South Park). – Zev Chonoles Jun 18 '11 at 22:07
Sorry, Zev, I was living inside my head; I will call them downvotes from now on. – gary Jun 18 '11 at 22:26

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