Measure Theoretic Probaibility - Video 5.1 - Vitali's Proof of the Impossibility of a Measure of all Real Numbers

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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 $m:\mathcal{P}(\mathbb{R})\to {\mathbb{R}}^{\ast }=\mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ 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 $|x-y|\in \mathbb{Q}$.

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 $r:\mathbb{N}\to \mathbb{Q}\cap [-1,1]$ 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 $V=\bigcup _{n\in \mathbb{N}}(F+r_n)$.

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 $|x-\lambda |\in \mathbb{Q}$. 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 $n\in \mathbb{N}$ 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 $m(V)\in \{0,\mathrm{\infty }\}$. 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 $V=\underset{n\in \mathbb{N}}{⨆}(F+r_n)$ 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 $|\lambda -\mu |\in \mathbb{Q}$. 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.

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