# Construction of a non-measurable set

A while ago I went through the construction of a non-measurable set in $\mathbb{R}$. That construction involved taking a quotient by an equivalence relation. At the time this might have been cause for concern. It occurs to me that since we have recently gone over this, perhaps now would be a good time to revisit the construction.

### Equivalence Relations and Quotients

Definition. An equivalence relation $\sim$ on a set $S$ is a binary operation that is,

• (Reflexive) $a \sim a$
• (Symmetric) $a \sim b \Rightarrow b \sim a$
• (Transitive) $a \sim b, b \sim c \Rightarrow a \sim c$

for all $a, b, c \in S$.

Definition. Given a set $S$ with an equivalence relation $\sim$, we can define the equivalence class of $a \in S$, denoted $[a]$, to be the set $\{ b \in S : a \sim b \}$.

Example. Let $S = \{ (a, b) : a, b \in \mathbb{Z}\}$ and define $(a,b) \sim (c, d)$ if and only if $ad-bc=0$. Let’s denote the equivalence class of $(a, b)$ by $\displaystyle\frac{a}{b}$. For example $(2, 6) \sim (3,9)$, so they are in the same equivalence class, which we will denote $\displaystyle\frac{1}{3}$. We see then that the equivalence classes are in bijection with the rationals $\mathbb{Q}$ (easy exercise!).

Definition. Given a set $S$ and an equivalence relation $\sim$, define the quotient $S/\!\sim$ to be the set of all equivalence classes of $\sim$.

### The Construction

We will construct a non-measurable set in $\mathbb{R}$ as the quotient of $[0,1] \subseteq \mathbb{R}$ by an equivalence relation $\sim$. Say $x \sim y$ if and only if $x-y \in \mathbb{Q}$ for all $x, y \in [0,1]$.

Exercise. Show that this is an equivalence relation.

Consider the quotient $[0,1]/\sim$, that is, the set of equivalence classes $\{ [x]: x \in [0,1]\}$.

Question. How are we going to represent equivalence classes? Is it necessarily okay to pick a representative from each class? How many classes are there? If, like me, you believe that this should be possible, then you believe in the axiom of choice. Let’s pretend for a moment that we believe that it is okay to pick representatives from each equivalence class.

Let $\mathcal{N}=[0,1]/\sim$. Let’s ask ourselves if this is measurable. One property of measure is translation invariance. That is, if I translate a set then I don’t change its measure. Let’s translate our set $\mathcal{N}$ around. Let $\{r_i\}$ be an enumeration of the rationals, and define

$\mathcal{N}_k = \mathcal{N}+r_k$                           (exercise: this is well-defined)

So, by translation invariance, $m(\mathcal{N}) = m(\mathcal{N}_k), \forall k$. We also observe that

1. The $\mathcal{N}_k$‘s are disjoint.

Indeed, suppose for a contradiction that $\exists r_k \neq r_l$ such that $\mathcal{N}_k \cap \mathcal{N}_l \neq \varnothing$. That is, there exists $x_\alpha, x_\beta$ such that $x_\alpha+r_k=x_\beta+r_l$ (hence $\alpha \neq \beta$). But then $x_\alpha - x_\beta = r_l-r_k \in \mathbb{Q}$ so $x_\alpha \sim x_\beta$, and that’s our contradiction. (Remember, we have chosen $\mathcal{N}$ to consist of unique representatives of equivalence classes of $\sim$)

2. $[0,1] \subseteq \bigcup_{k=1}^\infty \mathcal{N}_k \subseteq [-1,2]$.

If you don’t see this right away, consider it an exercise.

Since the measure of a disjoint union of sets is the sum of the measure of the sets, together with translation invarience, we have that $1 \leq \sum_{k=1}^\infty m(\mathcal{N}) \leq 3$. There are two options,

• $m(\mathcal{N}) = 0$ then $1 \leq 0 \leq 3$, contradiction!
• $m(\mathcal{N})>0$ then $1 \leq \infty \leq 3$, contradiction!

This shows that $\mathcal{N}$ is non-measurable.