6. The Lebesgue measure Ⅰ

(待修改! ) We define a semi-algebra $\mathcal{S}\subset \Im (\mathbb{R})$ $$\mathcal{S}:=\left\{\varnothing ,\mathbb{R},(a,b],(a,+\infty ),(-\infty ,b)|a,b\in \mathbb{R}\right\}$$A function $$\mu :\mathcal{S}⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$$satisfy

- $\mu (\varnothing )=0$ - $\mu (\mathbb{R})=+\infty$ - $\mu ((a,b])=b-a$ - $\mu ((a,+\infty ))=+\infty$ - $\mu ((-\infty ,b])=+\infty$

Claim

$\mu$ is $\sigma$-additive.

- $\mu (\varnothing )=0$

By definition.

- $(A_j{)}_{j⩾1}\in \mathcal{S}$ and $A={\sum }_{j⩾1}{A}_j\in \mathcal{S}$ $⟹$ $\mu (A)={\sum }_{j⩾1}\mu (A_j)$

Recall that we can extend $\mu$ to $\nu :\mathcal{A}(\mathcal{S})⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$, then it is equal to prove that $\nu (A)={\sum }_{j⩾1}\nu (A_j)$. To see that, let $A={\sum }_{j⩾1}{A}_j$ and it contain the finite union ${\sum }_{j=1}^n{A}_j\subset {\sum }_{j⩾1}{A}_j$, then $$\nu (A)(\nu isadditive)⩾\nu (j=1\sum nAj)=j=1\sum n\nu (Aj)$$Let $n$ be replaced by $\infty$, we obtain $\nu (A)⩾{\sum }_{j⩾1}\nu (A_j)$.

In the other hand, we assume that $A=(a,b]$ and $A={\sum }_{j⩾1}{A}_j$ where $A_j$$=$$(a_j,b_j]$. Our goal is the following inequality $$b-a⩽\sum _{k⩾}(b_j-a_j)$$Fix $ϵ>0$, we have $$[a+ϵ,b]\subset (a,b]=\sum _{j⩾1}(a_j,b_j]\subset \bigcup _{j⩾1}(a_j,b_j+ϵ/2^j)$$Since $[a+ϵ,b]$ is closed and bounded, then $[a+ϵ,b]$ is compact by the fact of *Heine-Borel* theorem.

Remark (*Heine-Borel* theorem)

$A\subset {\mathbb{R}}^n$ is closed and bounded $⟺$ $A$ is compact.

Therefore, $[a+ϵ,b]\subset {\bigcup }_{k=1}^m(a_{j,k},b_{j,k}+ϵ/2^{j,k})$, then $(a+ϵ,b]\subset {\bigcup }_{k=1}^m(a_{j,k},b_{j,k}+$$ϵ/2^{j,k}]$ and hence $$\nu ((a+ϵ,b])⩽\nu (k=1\bigcup m(aj,k,bj,k+ϵ/2j,k])⩽k=1\sum m\nu ((aj,k,bj,k+ϵ/2j,k])$$So we obtain $$b-a-ϵ⩽k=1\sum m(bj,k-aj,k+ϵ/2j,k)⩽j⩾1\sum (bj-aj+ϵ/2j)=j⩾1\sum (bj-aj)+ϵ$$Since $b-a⩽{\sum }_{j⩾1}(b_j-a_j)+2ϵ$ holds for any $ϵ>0$, then $$b-a⩽\sum _{j⩾1}(b_j-a_j)$$which is what we want to obtain.

Consider an increasing converges sequence ${\left\{E_n=(-n,n]\right\}}_{n⩾1}⟶\mathbb{R}$. For any $A$$\in$$\mathcal{S}$, $A\bigcap E_n$ is bounded.

Observation $\nu (A)={\lim }_{n\to +\infty }\nu (A\bigcap E_n)$.

$A\bigcap E_n={\sum }_{j⩾1}{A}_j\bigcap E_n$ since $A={\sum }_{j⩾1}{A}_j$ for some $A_j\in \mathcal{S}$. Then $\nu (A\bigcap$$E_n)=$${\sum }_{j⩾1}\nu (A_j\bigcap E_n)⩽$${\sum }_{j⩾1}\nu (A_j)$ holds for any $n$. So $$\nu (A)=\underset{n\to +\infty }{\lim }\nu (A\bigcap E_n)⩽\sum _{j⩾1}\nu (A_j)$$which is what we want to prove. Consequently, $\nu$ is $\sigma$-additive and hence $\mu$ is $\sigma$-additive.

Let $E_n=(-n,n]$, $\mathbb{R}={\bigcup }_{n⩾1}{E}_n$, $\nu (E_n)=2n<+\infty$, therefore $\mathbb{R}$ is $\sigma$-finite with respect to $\nu$. By the Caratheodory theorem we now know that the extension $\pi$ of $\nu$ on the $\sigma$-algebra which generated by the interval is unique.