For $n\in\mathbb N$ we would like to have a function $\mu$ that assigns to each $E\subset\mathbb R^n$ a number $\mu(E)\in[0, \infty]$, the n-dimensional measure of $E$, such that $\mu(E)$ is given by the usual integral formulas when the latter apply. The function $\mu$ has the following properties:

1) If $E_1, E_2, ...$ is a finite or infinite sequence of disjoint sets, then $\mu(E_1\cup E_2\cup ...)=\mu(E_1)+\mu(E_2)+...$.

2) If $E$ is congruent to $F$ (that is, if $E$ can be transformed into $F$ by translations, rotations, and reflections), then $\mu(E)=\mu(F)$.

3) $\mu(Q)=1$, where $Q$ is the unit cube $Q=\{x\in\mathbb R^n: 0\le x_j\le 1$ for $j=1, ..., n\}$.

However, 1)-3) are mutually inconsistent. To see why, we can define an equivalence relation on $[0, 1)$ by declaring that $x~y$ iff $x-y\in Q$ is rational. Let $N$ be a subset of $[0, 1)$ that contains precisely one member of each equivalence class. Let $R=\mathbb Q\cap[0, 1)$, and for each $r\in R$ let $N_r=\{x+r: x\in N\cap[0, 1-r)\}\cup\{x+r-1: x\in N\cap[1-r, 1)\}$

To obtain $N_r$, shift $N$ to the right by $r$ units and then shift the part that sticks out beyond $[0, 1)$ one unit to the left. Then $N_r\subset[0, 1)$, every $x\in[0, 1)$ belongs to precisely one $N_r$. Indeed, if $y$ is the element of $N$ that belongs to the equivalence class of $x$, then $x\in N_r$ where $r=x-y$ if $x\ge y$ or $r=x-y+1$ if $x; on the other hand, if $x\in N_r\cap N_s$, then $x-r$ (or $x-r+1$) and $x-s$ (or $x-s+1$) would be distinct elements of $N$ belonging to the same equivalence class, which is impossible.

Suppose that now $\mu: \mathcal P(\mathbb R)\to [0, \infty]$ satisfies (i), (ii), and (iii). By (i) and (ii), $\mu(N)=\mu(N\cap[0, 1-r))+\mu(N\cap[1-r, 1))=\mu(N_r)$

for any $r\in R$. Also, since $R$ is countable and $[0, 1)$ is the disjoint union of the $N_r$‘s, $\mu([0, 1))=\sum_{r\in R}\mu(N_r)$

by (1) again. But $\mu([0,1))=1$ by 3), since $\mu(N_r)=\mu(N)$, the sum on the right is either 0 (if $\mu(N)=0$) or $\infty$ (if $\mu(N)>0$). Hence no such $\mu$ can exist.

But in 1924, Banach and Tarski proved the following amazing result:

Let $U$ and $V$ be arbitrary bounded open sets in $\mathbb R^n$, $n\ge 3$. There exist $k\in\mathbb N$ and subsets $E_1, ..., E_k, F_1, ..., F_k$ of $\mathbb R^n$ such that:

1) the $E_j$‘s are disjoint and their union is $U$;

2) the $F_j$‘s are disjoint and their union is $V$;

3) $E_j$ is congruent to $F_j$ for $j=1,..., k$.

The construction of sets $E_j$ and $F_j$ depends on the axiom of choice. But their existence precludes the construction of any $\mu: \mathcal P(\mathbb R^n)\to[0,\infty]$ that assigns positive, finite values to bounded open sets and satisfies (1) for finite sequences as well as (2).

“Don’t take anything Personally Nothing others do is because of you. What others say and do is a projection of their own reality, their own dream. When you are immune to the opinions and actions of others, you won’t be the victim of needless suffering” ~ Miguel Ruiz

To be continued tomorrow 🙂

References:

 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 19-21.

 Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.