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

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)

So $m(\mathcal{N})\neq 0$.

By (2), we have

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.