# Math4121 Lecture 31

## Chapter 3: Lebesgue Integration

### Non-measurable sets

#### Definition: Vitali's construction

Step 1. Define an equivalence relation on $\mathbb{R}$ as follows:

Recall a relation is an equivalence relation if it is reflexive, symmetric, and transitive.

1. Reflexive: $x\sim x$ for all $x\in\mathbb{R}$
2. Symmetric: $x\sim y$ implies $y\sim x$ for all $x,y\in\mathbb{R}$
3. Transitive: $x\sim y$ and $y\sim z$ implies $x\sim z$ for all $x,y,z\in\mathbb{R}$

Say $x\sim y$ if $x-y\in\mathbb{Q}$.

This is an equivalence relation, easy to show by the properties above.

We denote the equivalence class of $x$ by $\mathbb{R}/\sim$, where $[x]=\{x+q:q\in\mathbb{Q}\}$.

If $z\in [x]$, then so is the fractional part of $z$, i.e. $z-\lfloor z\rfloor\in [x]$. So in every equivalence class $[x]$ we can find an element in $[x]\cap (0,1)$. Take one such real number from every equivalence class, and call the set of all such numbers $\mathcal{N}$.

Step 2. Show that $\mathcal{N}$ is not Lebesgue measurable.

We defined the translation of $S$ as follows:

Given a set $S\subseteq\mathbb{R}$ and a real number $a\in\mathbb{R}$, the translation of $S$ by $a$ is defined as

$$
S+a=\{x+a:x\in S\}
$$

Outer measure is translation invariant, i.e. $m_e(S+a)=m_e(S)$ for all $S\subseteq\mathbb{R}$ and $a\in\mathbb{R}$, which also holds for inner measure.

Properties of $\mathcal{N}$:

1. $(0,1)\subseteq\bigcup_{q\in \mathbb{Q}\cap (-1,1)} (\mathcal{N}+q)\subseteq (-1,2)$
2. $\{\mathcal{N}+q:q\in\mathbb{Q}\cap (-1,1)\}$ is pairwise disjoint.

Suppose $\mathcal{N}$ is measurable. Then by (1) 
$$
\begin{aligned}
    1&\leq \sum_{q\in\mathbb{Q}\cap (-1,1)} (\mathcal{N}+q)\\
    &=\sum_{q\in\mathbb{Q}\cap (-1,1)} m(\mathcal{N})
\end{aligned}
$$
So $m(\mathcal{N})\neq 0$.

By (2), we have

$$
\begin{aligned}
3&\geq \sum_{q\in\mathbb{Q}\cap (-1,1)} m(\mathcal{N}+q)\\
&=\sum_{q\in\mathbb{Q}\cap (-1,1)} m(\mathcal{N})\\
&=m(\mathcal{N})\sum_{q\in\mathbb{Q}\cap (-1,1)} 1\\
&=\infty
\end{aligned}
$$

This is a contradiction. So $\mathcal{N}$ is not Lebesgue measurable.

QED

Appendix:

(1) $I\subseteq\bigcup_{q\in\mathbb{Q}\cap (-1,1)} (\mathcal{N}+q)$

Let $x\in I$. We need to find $q\in\mathbb{Q}\cap (-1,1)$ such that $x-q\in\mathcal{N}$. $\exists y\in\mathcal{N}$ such that $y\in (0,1)\cap [x]$. Then $x-y=q\in \mathbb{Q}$ and since $x,y\in I$, we have $q\in (-1,1)$.

(2) $\{\mathcal{N}+q:q\in\mathbb{Q}\cap (-1,1)\}$ is pairwise disjoint.

Suppose $\mathcal{N}+q_1=\mathcal{N}+q_2$ for some $q_1,q_2\in\mathbb{Q}\cap (-1,1)$. We want to show $q_1=q_2$.

Take $x$ in the intersection, then  this means $y=x-q_1, z=x-q_2\in\mathcal{N}$.

But $y\sim z$, this contradicts the fact that $\mathcal{N}$ contains only one element from each equivalence class. So $q_1=q_2$.

#### Axiom of choice

Given a set $S$, $\exists \psi:\mathscr{P}(S)\to S$ such that $\psi(T)\in T, \forall T\subseteq\mathscr{P}(S)$.

For any set $S$, there exists a map that maps every non-empty subset of $S$ to an element of that subset.

This leads to some weird results, e.g. Banach-Tarski paradox.

_Godel showed that the axiom of choice is not contradictory to ZF set theory._ You have ZFC

_Cohen showed that the negation of the axiom of choice is not contradictory to ZF set theory._ You have ZF

You can choose the axiom or not.