# A Measure Theory Surprise

· 6 minutes read

On the real line, the collection of Borel sets plays a central role in analysis and measure theory in particular, as it contains almost every set that comes up naturally in those contexts. The set of Borel sets is defined as the $\sigma$-field generated by the collection of finite intervals on the real line, and so by the extension theorem, all Borel sets are Lebesgue-measurable.

The definition can be extended to higher dimensions: the $k$-dimensional Borel sets are elements of the $\sigma$-field $\mathscr{R}^k$ generated by the collection of $k$-dimensional bounded rectangles

$\{(x_1,\ldots, x_k)\colon a_i\le x_i < b_i, 1\le i\le k\}.$Similar to the $1$-dimensional case, we can define the $k$-dimensional Lebesgue measure $\lambda_k$ on these Borels sets as the unique extension of the set function assigning volumes to the bounded rectangles. Thus, we see how the Lebesgue measure corresponds to and generalizes the concepts of length, area, and volume.

## The volume analogy

Intuitively, if we move a set in Euclidean space, its Lebesgue measure should stay the same. We say the Lebesgue measure has **translation invariance**:

For any set $A\in\msR^k$ and $x\in\R^k$, the set $A + x = \{a + x\colon a\in A\}$ is a Borel set, and $\lambda_k(A+x) = \lambda_k(A)$.

Similarly, if we scale a Borel set by a factor, its Lebesgue measure (volume) should be scaled accordingly, and if we rotate a Borel set, then its Lebesgue measure should be unchanged. This volume analogy can be distilled into the following theorem:

Let $T\colon \R^k\to\R^k$ be a linear transformation and $A\in\msR^k$, then $TA\in\msR^k$ and $\lambda_k(TA) = \abs{\det T} \cdot \lambda_k(A)$.

All good? Well, no, because this “theorem” is wrong.

It is almost correct though: the statement holds if $T$ is invertible.

Ah I see, if $T$ is not invertible, then it will map $A$ into a smaller subspace of $\R^k$, which means the volume of $TA$ is $0$, while $\abs{\det T}$ is … also $0$. Huh?

So what went wrong? The problem is not that $TA$ has measure zero, but that it *need not be a Borel set* at all. It turns out the collection of Borel sets is not closed under linear transformation! In fact, an example can be found with the simplest of transformations: let $\pi\colon\R^2\to\R$ be the projection $\pi(x_1,x_2)=x_1$. The rest of the article will focus on finding a set $A\subset [0,1]^2$ such that $\pi[A]$ is not a Borel set.

## A cross-sectional study

We first need some definitions. If $X$ and $Y$ are sets and $Z\subset X\times Y$, then for $y\in Y$, the **section** of $Z$ at $y$ is the subset of $X$ defined as $\{x\colon (x,y)\in Z\}$. This is like the cross-section of an object using a plane in 3D space.

Before we proceed, let us go on a slight tangent. Working with $\R$ directly is a bit cumbersome, so we will switch perspective a bit. For each $x\in\R$, associate $x$ with its continued fraction representation $[x_0; x_1, x_2, \ldots]$, a sequence of positive integers, except $x_0$ which can be any integers.

This sequence is finite for rational $x$ and infinite for irrational $x$. To compare two continued fractions $[a_0;a_1,a_2,\ldots]$ and $[b_0;b_1,b_2,\ldots]$, compare the first differing coefficients $a_k$ and $b_k$ (treat missing coefficients as $\infty$), and flip the order if $k$ is odd. Most importantly, every (finite or infinite) sequence of positive integers represents a continued fraction in $[0,1]$.

## Universal sets

Let $\msC$ be a family of subsets of $X$. A **universal set** for $\msC$ is a set $Z\subset X\times Y$ such that $\msC$ is *precisely* the collection of all sections of $Z$ at every point $y\in Y$. A universal set for $\msC$ can be thought as the “graph” of a surjection $f\colon Y\to \msC$, so it exists if and only if the cardinality of $\msC$ does not exceed that of $Y$.

Recall that a second-countable topological space $X$ is one having a countable base, i.e., a countable family $\mcU$ of open subsets in $X$ such that every open set in $X$ is a union of some subfamily of $\mcU$. Our good friend $\R^k$ is an example: the rectangles with rational endpoints form the required base. We have our first result.

Theorem. For any second-countable topological space $X$, there is an open set $G$ in $X\times [0,1]$ which is universal for the collection of all open sets in $X$, and a closed set $F$ in $X\times [0,1]$ which is universal for the collection of all closed sets in $X$.

It is not difficult to show that the topology of a second-countable space has cardinality not exceeding continuum, so a universal set for the topology of $X$ must exist. The novel part of this theorem is that we can force that universal set to be open.

*Proof.* Let $\{U_n\}_{n\ge 1}$ be a base of $X$.

For positive integers $k$ and $n$, let $A_k(n)$ be the set of $x$ with $x_k=n$ and $x_{k+1}$ exists. Given a prefix $[0;a_1,\ldots,a_{k-1}]$, $A_k(n)$ contains the interval $(p,q)$, where if $k$ is odd, then $p=[0;a_1,\ldots,a_{k-1},n+1]$, $q=[0;a_1,\ldots,a_{k-1},n]$, and if $k$ is even, then $p=[0;a_1,\ldots,a_{k-1},n]$, $q=[0;a_1,\ldots,a_{k-1},n+1]$. Moreover, every $x\in A_k(n)$ lies in such an interval (in fact, countably many such intervals), so $A_k(n)$ is open, being a union of open intervals in $\R$.

Let $G$ be the union of the open sets $U_n\times A_k(n)$, ranging over all $k$ and $n$. For $x$ in $[0,1]$, if $x=[0;x_1,x_2,\ldots,x_k]$, then $x$ lies in each of $A_1(x_1),\ldots,A_{k-1}(x_{k-1})$, so the section of $G$ at $x$ is $\bigcup_{i=1}^{k-1} U_{x_i}$; similarly, if $x=[0;x_1,x_2,\ldots]$, then the section of $G$ at $x$ is $\bigcup_{i=1}^\infty U_{x_i}$.

Hence, each section of $G$ is a union of some subfamily of $\{U_n\}$, and since the continued fractions include every finite and infinite sequence of positive integers, the converse also holds. In other words, the sections of $G$ are precisely the countable unions of sets in $\{U_n\}$, i.e., all the open sets of $X$. The complement $F$ of $G$ is then a closed set in $X\times\R$ universal for the closed sets of $X$. $\blacksquare$

The theorem still holds (and is easier to prove) if we replace $\R$ by the Baire space, the set of all infinite sequences of positive integers with the induced product topology. In fact, our preceding tangent and half of our proof is mostly for mapping the reals to a quasi-Baire space using continued fractions.

Couldn’t you have shown the proof for the Baire space, then map it to $\R$ for better clarity?

To be honest, I’m not sure how to derive the statement for $\R$ directly from the statement for the Baire space. Though, hopefully you will remember this theorem better, knowing that I have led you through a more annoying and laborious direction.