The Vitali Paradox of Measure

STEM fields are often interested in measuring regions. For example, an astrophysicist might be interested in measuring the volume of a star. The concept of measure is intuitive, and some of its features are self-evident. For example,

(i) if a region can be divided into countably many nonoverlapping subregions, then the measure of the region should be equal to the sum of the measures of the subregions;

(ii) if two regions are identical up to rotations, translations, and reflections, then their measures should be equal; and

(iii) the measure of a square (cube, hypercube, etc.) with side $1$ should be $1$.


We could begin to formalize this concept by attempting to measure all subsets of the set of real numbers $\mathbb{R}$, but it turns out that this is impossible, as I will now demonstrate.

Let us say that two real numbers are equivalent if their difference is rational. Construct a subset $N$ of the half-open interval $[0,1)$ in $\mathbb{R}$ such that no two distinct elements are equivalent (this is possible by the axiom of choice), and define

\begin{equation}
    N_r = \{x+r : x \in N \cap [0,1-r)\} \cup \{x+r-1 : x \in N \cap [1-r,1)\},
\end{equation}

where $r\in\mathbb Q\cap[0,1)$.

It follows that all $x \in [0,1)$ belong to exactly one $N_r$, namely $r = x-y$ or $r = x-y+1$, where $y \in N$ such that $x$ is equivalent to $y$. Otherwise, if $x \in N_r \cap N_s$, there exist $y, z \in N$ such that $x$ is equivalent to both $y$ and $z$, implying by transitivity that $y$ is equivalent to $z$, which is a contradiction. Thus, the $N_r$ are disjoint.

Then, by (i) and (ii), the measures of $N$ and $N_r$ are equal for all $r$. Moreover, by (iii) and (i),

\begin{equation}
    1=\mu\left(\left[0,1\right)\right)=\mu\left(\bigcup_rN_r\right)=\sum_r\mu\left(N_r\right)=\sum_r\mu\left(N\right),
\end{equation}

where $\mu(N)$ is the measure of $N$. However, the rightmost sum is either $0$ or $\infty$, which is a contradiction.

This problem is a consequence of our ambition to assign a measure to all of $2^{\mathbb R}$. Nevertheless, we can circumvent it by focusing on a particular subset of $2^{\mathbb R}$ rather than the whole, and this is much of the subject of measure theory.