1.1Introduction

Suppose we want to be able to find the length, area, volume, or generally the “measure” of some region in $\bR^n$. If such a region is bounded by nice curves and functions, Riemann integration against the length differential $\dd x$, area differential $\dd A$, or volume differential $\dd V$ produces a reasonable value. The failures of Riemann integration have been discussed intensively, however; we seek to abandon it. Let us build from scratch a new mechanism of “measuring” a geometric area in $\bR^n$.

Ideally, for any $n \in \bZ_+$, we would like to have a function $\mu: \mathcal{P}(\bR^n) \to [0, \infty]$ that assigns, to each $E \subseteq \bR^n$, some value $\mu(E) \in [0, \infty]$, the $n$-dimensional measure of $E$. Such a function $\mu$ should also satisfy our intuition for how measuring things work.

1. If $E_1, E_2, \dots$ is a countably infinite sequence of disjoint sets, then

   $$\mu(E_1 \cup E_2 \cup \cdots) = \mu(E_1) + \mu(E_2) + \cdots.$$

2. If $E$ is congruent to $F$ (that is, a series of translations, rotations, and reflections can transform $E$ into $F$), then $\mu(E) = \mu(F)$.

3. $\mu([0, 1]^n) = 1$, where $[0, 1]^n$ is the unit cube.

Unfortunately, we cannot define a map $\mu: \mathcal{P}(\bR^n) \to [0, \infty]$ that achieves all three conditions. For simplicity, take $n = 1$, and we show such a measuring function cannot exist. We will adapt the argument showing that the outer measure is not additive. Indeed, for any two elements $a, b \in [0, 1)$, we define the equivalence relation $a \sim b$ if $a - b \in \bQ$. We construct the subset $N \subseteq [0, 1)$ that contains precisely one element from each equivalence class. Next, let $R = [0, 1) \cap \bQ$, and for each $r \in R$ define $N_r$ by

That is, to obtain $N_r$, we shift $N$ $r$ units to the right, and whatever overflows from $[0, 1)$ is shifted to the left $1$ unit. Then $N_r \subseteq [0, 1)$ and every $x \in [0, 1)$ belongs to precisely one $N_r$. Indeed, if $y \in N \cap [x]_\sim$, then $x \in N_r$ where $r = x - y$ if $x \ge y$, or otherwise $r = x - y + 1$; on the other hand, if $x \in N_r \cap N_s$, then without loss of generality $x - r$ and $x - s$ would be distinct elements in $N$, which is impossible.

Suppose now that $\mu: \mathcal{P}(\bR) \to [0, \infty]$ satisfies the requirements aforementioned. Then

for any $r \in R$. Also, since $R$ is countable and $[0, 1)$ is the disjoint union of each $N_r$, note

But the sum on the right is either $0$ (if $\mu(N) = 0$) or $\infty$ (if $\mu(N) > 0$). Thus $\mu$ cannot exist.

One might consider weakening the first requirement to work only for finitely many disjoint sets. Not only would this lose properties that make limiting processes work with respect to the measure, this would still be inconsistent for $n \ge 3$ by a proof by Banach and Tarski. Instead, we look to restrict $\mu$ to only certain subsets of $\bR^n$ instead of all, and we show that we do not need to give up these properties if we restrict the measure.

Note that measures will extend far past $\bR^n$ and lend itself to abstract spaces. Instead of measuring length/area/volume, we could instead measure the mass distribution of an object, a probability distribution for certain events occurring, and more.