Lebesgue Measures, Vitali Sets, and Banach-Tarski Paradox

Introduction

A fundamental problem in geometry is to determine the area or volume of a region in the plane or in three-dimensional space. The tools of integral calculus provide an effective means of solving this problem for regions bounded by sufficiently “nice” curves or surfaces. However, these methods become inadequate when dealing with more complicated sets—even in one dimension.

Ideally, for each $n \in \mathbb{N}$, we seek a function \(\mu : \mathcal{P}(\mathbb{R}^n) \to [0, \infty]\) that assigns to every subset $E \subset \mathbb{R}^n$ a number $\mu(E)$, interpreted as the $n$-dimensional measure of $E$, in such a way that $\mu(E)$ coincides with the usual integral formulas whenever those are applicable. Naturally, such a function $\mu$ should satisfy the following properties:

1. Countable additivity: If $E_1, E_2, \dots$ is a finite or countable collection of disjoint sets, then

   \[\mu(E_1 \cup E_2 \cup \cdots) = \mu(E_1) + \mu(E_2) + \cdots\]

2. Invariance under congruence: If $E$ and $F$ are congruent (that is, $E$ can be transformed into $F$ by translations, rotations, or reflections), then \(\mu(E) = \mu(F)\)

3. Normalization: The unit cube

   \[Q = { x \in \mathbb{R}^n : x_j < 1 \text{ for } j = 1, \dots, n }\]

   should have measure one, i.e., $\mu(Q) = 1$.

Unfortunately, these conditions are mutually inconsistent, as the following examples demonstrate.

Vitali Sets

Define an equivalence relation on $[0, 1)$ by declaring that $x \sim y$ if and only if $x - y$ is rational. Let $N$ be a subset of $[0, 1)$ containing exactly one representative from each equivalence class. Next, set $R = \mathbb{Q} \cap [0, 1)$, and for each $r \in R$ define

In other words, $N_r$ is obtained by shifting $N$ to the right by $r$ units and then translating any portion that extends beyond $[0, 1)$ one unit to the left. Thus, each $N_r$ is contained in $[0, 1)$, and every $x \in [0, 1)$ lies in exactly one of the sets $N_r$.

Now, suppose $\mu : \mathcal{P}(\mathbb{R}) \to [0, \infty]$ satisfies properties (1), (2), and (3). By (1) and (2), for any $r \in R$,

Moreover, since $R$ is countable and $[0, 1)$ is the disjoint union of the sets $N_r$, we have by (1) again,

But $\mu([0, 1)) = 1$ by (3), and since $\mu(N_r) = \mu(N)$ for all $r$, the series on the right is either $0$ (if $\mu(N) = 0$) or diverges to $\infty$ (if $\mu(N) > 0$). Therefore, no such function $\mu$ can exist.