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Vitali’s Impossibility Theorem

Claim: There is no measure function on the set of all subsets of real numbers. (To be understood somewhat intuitively for now, and made more formal later).

Proof: Suppose for contradiction that \( m:\mathcal P(\Bbb R)\to \Bbb R^* = \Bbb R\cup \{-\infty,\infty\}\) is a measure function on the set of all subsets of real numbers.

Define \(\sim\) to be a relation on the interval of real numbers \([0,1]\), by the condition \(x\sim y \) holds if and only if \(|x-y|\in\Bbb Q\).

Define \(F\) to be any complete set of representatives of the cells of the partition induced by \(\sim\). That is to say: For every cell \([\lambda]\) of the partition induced by \(\sim\), the set \(F\) has precisely one element (chosen arbitrarily) from the cell.

Define \(r:\Bbb N\to\Bbb Q\cap [-1,1]\) to be any arbitrary enumeration of the rational numbers in the interval \([-1,1]\). We will denote the mapping \(r(n) \) by the expression \(r_n\).

Define \(V=\bigcup_{n\in\Bbb N}(F+r_n)\).

Define \(m(F+r_n)=c\). Intuitively, these sets are mere translations of F, and so they should all share a common, constant measure which we can call c.

First we prove that \(1\le m(V)\le 3\) by showing that in fact \((0,1)\subseteq V\subseteq [-1,2]\).

Let \(x\in (0,1)\). Then x is in some cell of the partition induced by \(\sim\), call it \(x\in [\lambda]\), in which case we have that \(|x-\lambda|\in \Bbb Q\). This in turn entails that \(x=\lambda + s\) for some rational number s. Moreover, \(s\in[-1,1]\) which guarantees that s occurs in the enumeration r. That is to say, there must be some \(n\in\Bbb N\) such that \(s=r_n\). Then \(x\in F + r_n\subseteq V\). Therefore \((0,1)\subseteq V\).

Next let \(x\in V\) so that there is some \(F+r_n\) such that \(x\in F + r_n\). Then there is some \(\lambda\in F\) such that \(x=\lambda+r_n\). But we have \(0\le \lambda\le 1\) and \(-1\le r_n\le 1\) so that therefore \(-1\le x\le 2\). Therefore \(V\subseteq [-1,2]\).

The fact that \((0,1)\subseteq V\), now established, should intuitively imply that \(1\le m(V)\). This is because a reasonable measure of subsets of real numbers should assign the interval (0,1) a measure equal to the length, which is 1. And any set containing this interval, should then have measure at least 1. Likewise we should have that \(m(V)\le 3\) for a similar reason.

We now show that \(m(V)\in\{0,\infty\}\). Intuitively, if V is “made up of” a bunch of non-overlapping pieces, then its measure should be the sum of the measures of the pieces. (For a visualization, if you break a stick, then length of the stick was the sum of the lengths of the two pieces.) Therefore if we should that in fact \(V=\bigsqcup_{n\in\Bbb N}(F+r_n)\) is a disjoint union, it will follow that \(m(V)=\sum m(F+r_n)=\sum c\).

To prove disjointness, let \(x\in F+r_m\cap F+r_n\) so that there are \(\lambda,\mu\) such that \(x=\lambda+r_m = \mu+r_n\). In which case \(\lambda-\mu = r_n-r_m\), which also ensures that \(|\lambda-\mu|\in \Bbb Q\). By definition then \(\lambda\sim\mu\). But recall that F contains exactly one representative per cell of the associated partition, and therefore \(\lambda=\mu\). Using this in the earlier equations we also have \(r_m=r_n\) whence we further have \(F+r_m=F+r_n\). This proves that the collection of sets is disjoint (since if they share any element then they are equal).

We now have that \(m(V)=\sum c\). We then have two cases, \(c=0\) or \(c>0\). In the former case \(m(V)=0\) and in the latter case \(m(V)\). In either case, we obtain a contradiction with the earlier result \(1\le m(V)\le 3\).

Hence the proof by contradiction is now complete.

\(\blacksquare\)