7. The Lebesgue measure Ⅱ

Approach from topology

(待修改! ) We will give another proof about the property that the *Lebesgue* measure $\mu$ is $\sigma$-additive. Define a function $$\mu :\mathcal{A}⟶\overline{{\mathbb{R}}_{+}}$$where $\mathcal{A}$ is generated by the interval.

Theorem 1

$\mu$ is a measure.

$S=\left\{\varnothing ,\mathbb{R},(a,b],(a,+\infty ),(-\infty ,b]|a,b\in \mathbb{R}\right\}$ is a semi-algrbra over $\mathbb{R}$. $\mathcal{A}=\mathcal{A}(\mathcal{S})$. Note that $\mu$ is a unique extension of $\mu {|}_{\mathcal{S}}:\mathcal{S}⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$ ($\text{interval}\mapsto \mu (\text{interval})$).

We prove that $\mu$ is a measure on $(-N,N]$, it sufficient to show that $\mu$ is continuous from above at $\varnothing$. It equals to prove the following property $${\left\{E_k\subset (-N,N]\right\}}_{k⩾1}⟶\varnothing ⟹\mu (E_k)\to 0$$Let ${\left\{E_k\subset (-N,N]\right\}}_{k⩾1}⟶\varnothing$ be an decreasing converges sequence such that $\mu (E_k)\nrightarrow 0$. In other words, $\mu (E_k)⩾2\delta >0$ for some $\delta >0$ and $\forall$ $k⩾1$.

##### Step 1

$E_1\in \mathcal{A}$, $E_1={\sum }_{j=1}^{n_1}(a_{1j},b_{1j}]$, we construct $$F_1=\sum _{j=1}^{n_1}(a_{1j}+{\epsilon }_1,b_{1j}]\stackrel{\text{obviously}}{\in }\mathcal{A}$$Observation $F_1\subset \overline{F_1}\subset E_1$.

We set $G_1=\overline{F_1}$ and choose the positive number ${\epsilon }_1$ small enough such that $\mu (E_1\backslash F_1)⩽\delta /2$.

##### Step 2

$2\delta ⩽\mu (E_2)=\mu (E_2\bigcap F_1)+\mu (E_2\backslash F_1)$, but $\mu (E_2\backslash F_1)⩽\mu (E_1\backslash F_1)⩽\delta /2$, it implies that $\mu (E_2\bigcap F_1)⩾$$2\delta -$$\delta /$$2>0$ and therefore we can write $E_2\bigcap F_1={\sum }_{j=1}^{n_2}(a_{2j},b_{2j}]$, we construct $$F_2=\sum _{j=1}^{n_2}(a_{2j}+{\epsilon }_2,b_{2j}]$$Then $F_2\subset \overline{F_2}\subset E_2\bigcap F_1$ and $F_2\subset F_1$ and $\overline{F_2}\subset E_2$, and we conclude that $\overline{F_2}\subset \overline{F_1}$. We set $G_2=\overline{F_2}$ and choose the positive number ${\epsilon }_2$ small enough such that $\mu ((E_2\bigcap F_1)\backslash F_2)⩽\delta /4$. Then we obtain $$\begin{array}{rl}\mu (E_2\backslash F_2)& =\mu ((E_2\bigcap F_1)\backslash F_2)+\mu ((E_2\backslash F_1)\backslash F_2)\\ & ⩽\delta /4+\mu (E_2\backslash F_1)\\ & ⩽\delta /4+\mu (E_1\backslash F_1)\\ & ⩽\delta /4+\delta /2\end{array}$$Suppose that we have constructed $(F_k{)}_{1⩽k⩽n}$ such that $$\begin{array}{rl}\overline{F_k}\subset E_k,1& ⩽k⩽n\\ F_{k+1}\subset F_k,1& ⩽k⩽n+1\end{array}$$$\mu (E_n\backslash F_n)⩽\delta /2+\delta /4+\cdots +\delta /2^n⩽\delta$. Consider the construction of $F_{n+1}$, $$\begin{array}{rl}2\delta ⩽\mu (E_{n+1})& =\mu (E_{n+1}\bigcap F_n)+\mu (E_{n+1}\backslash F_n)\\ & ⩽\mu (E_{n+1}\bigcap F_n)+\mu (E_n\backslash F_n)\\ & ⩽\mu (E_{n+1}\bigcap F_n)+\delta /2+\delta /4+\cdots +\delta /2^n\\ & ⩽\mu (E_{n+1}\bigcap F_n)+\delta \end{array}$$So $\mu (E_{n+1}\bigcap F_n)⩾\delta >0$ and therefore we can write $E_{n+1}\bigcap F_n={\sum }_{j=1}^{k_n}(a_{(n+1)j},b_{(n+1)j}]$ and we construct $$F_{n+1}=\sum _{j=1}^{k_n}(a_{(n+1)j}+{\epsilon }_{n+1},b_{(n+1)j}]$$Then $$\overline{F_{n+1}}\subset E_{n+1}\text{ and }{F}_{n+1}\subset F_n$$We can choose the positive number ${\epsilon }_{n+1}$ small enough to obtain the inequality $\mu ((E_{n+1}\bigcap F_n)\backslash F_{n+1})⩽\delta /2^{n+1}$, and $$\begin{array}{rl}\mu (E_{n+1}\backslash F_{n+1})& =\mu ((E_{n+1}\bigcap F_n)\backslash F_{n+1})+\mu ((E_{n+1}\backslash F_n)\backslash F_{n+1})\\ & ⩽\delta /2^{n+1}+\mu (E_{n+1}\backslash F_n)\\ & ⩽\delta /2^{n+1}+\mu (E_n\backslash F_n)\\ & ⩽\delta /2^{n+1}+\delta /2+\delta /4+\cdots +\delta /2^n\end{array}$$Then we complete the induction of $F_n$ and obtain a family $(F_n{)}_{n⩾1}$ which satisfy the following properties

- $\overline{F_k}\subset E_k$, $k⩾1$ - $F_{k+1}\subset F_k$, $k⩾1$ - $\mu (E_n\backslash F_n)⩽\delta /2+\cdots +\delta /2^n$

We set $G_k=\overline{F_k}$, and ${\left\{G_k\right\}}_{k⩾1}$ is an decreasing sequence since $G_{k+1}=\overline{F_{k+1}}\subset \overline{F_k}=G_k$ and which $G_k$ are all compact.

We claim that $G_k\ne \varnothing$. $$\begin{array}{rl}2\delta ⩽\mu (E_k)& =\mu (E_k\backslash F_k)+\mu (E_k\bigcap F_k)\\ (F_k\subset E_k)& ⩽\delta +\mu (F_k)\end{array}$$$⟹$ $\mu (F_k)⩾\delta$ $⟹$ $F_k\ne \varnothing$. Consequently, $G_k=\overline{F_k}\ne \varnothing$. But $$\bigcap _{k⩾1}{G}_k\subset \bigcap _{k⩾1}{E}_k=\varnothing$$​ $⟶⟵$

Remark (*Bolozano*-*Weierstrass* property)

The limit point of infinite subset of compact space exist.

Proof Let $X$ be a compact space, $S$ is a subset without limit point. We prove that $S$ is finite.

For $x\in X$, we can find an open neighborhood $O(x)$ such that $$\begin{array}{r}O(X)\bigcap S=\{\begin{array}{l}\varnothing ,x\notin S\\ \left\{x\right\},x\in S\end{array}\end{array}$$Since $X$ is compact, the collection $$\left\{O(X)|x\in X\right\}\stackrel{\text{cover}}{\to }X$$can be reduce to finite ${\bigcap }_{j=1}^n{O}_j(x)\stackrel{\text{cover}}{\to }X$. But every $O_j(x)$ contain most one point in $S$, then $S$ is finite. $\square$

For ${\left\{G_k\right\}}_{k⩾1}$ an decreasing sequence which $G_k$ is compact for all $k$, we can construct a sequence ${\left\{p_n\in G_n\right\}}_{n⩾1}$, it is a infinite subset of the compact set $G_1$, then it has the limit point $p$. If $p\notin G_1$, then there exist a neighborhood $U_p$ such that $U_p\bigcap {\left\{p_n\right\}}_{n⩾1}=\varnothing$ $⟶⟵$

Similarly, we can obtain the fact that $p\in G_n$ for all $n$, so $p\in {\bigcap }_{\alpha }{G}_{\alpha }\ne \varnothing$.