MT Lesson 1: Non-measurable set
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In this section, we show that we need some specified set to construct the definition of measure.
What is measure? Intuitively, the measure measures the entity. For example, we use length $b-a$ to measure the range $(a,b]$. Then we think, what a measure should be like? Define measure as function $\lambda: \mathcal{P}(\mathbb{R})\to\mathbb{R}_+\bigcup \{ +\infty \}$, where $\mathcal{P}(\mathbb{R})$ is the family of subsets of $\mathbb{R}$. We want measure to have the following properties:
- $\lambda\big((a,b]\big)=b-a$ for all $a<b$
- (invariant of transformation) $\lambda(A+x)=\lambda(A)$ for all $A\in \mathcal{R}(\mathbb{R})$, $x\in \mathbb{R}$.
- (sigma additive) $\lambda(\bigcup_{j\geq1}A_j)=\sum_{j\geq1} \lambda(A_j)$ for all countable family of disjoint subsets of $\mathbb{R}$.
Then we show that those properties can’t be satisfied.
Firstly we define the equivalence class. We say $x\sim y$ or $x$ is equivalent to $y$ or $y \in [x]$ if $x-y\in \mathbb{Q}$, where $\mathbb{Q}$ is the set of rational numbers.
Then we define $\Omega$ be the set of inequivalent numbers in $(0,1)$. Given $p,q\in \mathbb{Q}$, We claim that the relation of $\Omega+p$ and $\Omega+q$ can either be $\Omega+p=\Omega+q$ or $(\Omega+p)\bigcap(\Omega+q)=\emptyset$.
Proof: If there exist a $x\in (\Omega+p)\bigcap(\Omega+q)$, then we write $x=\alpha+p=\beta+q$, thus $\alpha-\beta=q-p\in\mathbb{Q}$. If $p=q$, $\alpha=\beta$, then $\Omega+p=\Omega+q$. If $p\ne q$, then $\alpha\neq \beta$ and $\alpha\sim \beta$, which contradicts the definition of $\Omega$. $\blacksquare$
Given $q\in \mathbb{Q}$ and $-1\leq q\leq 1$, we have $(\Omega+q)\subset (-1, 2)$, thus
\[\bigcup_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}} (\Omega+q)\subset (-1,2)\]We have $E\subset F \Longrightarrow \lambda(E)\leq \lambda(F)$, because
\[\lambda(F)-\lambda(E) \overset{sigma\,additive}{=} \lambda(F\backslash E)\geq 0\]Thus
\[\lambda\left(\bigcup_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}} (\Omega+q) \right) \leq \lambda((-1,2)) = 3\]Given sigma additive and invariant of transformation, we have
\[3\geq\lambda\left(\bigcup_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}} (\Omega+q) \right) = \sum_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}} \lambda(\Omega+q)\\ = \sum_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}} \lambda(\Omega)\\\]Thus $\lambda(\Omega)=0$.
On the other hand, we claim that
\[(0,1)\subset \bigcup_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}}(\Omega+q)\]Proof: For any $x\in (0,1)$, there exist $\alpha\in [x]\bigcap \Omega$ such that $x-\alpha=p\in \mathbb{Q}$, and $\alpha \in(0,1)$, thus $p\in(-1,1)$. Thus
\(x=\alpha+p\in \bigcup_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}}(\Omega+q)\) $\blacksquare$
Thus
\[\lambda\left(\bigcup_{ \begin{array}{c} q\in \mathbb{Q}\\ -1\leq q\leq 1 \end{array}} (\Omega+q) \right)\geq\lambda((0,1))=1\]which leads to a contradiction. Thus the properties can’t be met for all family of subsets of $\mathbb{R}$, that’s why we need to specify some sets in the next section.