《几何测度论导论》讲义笔记(1.1)-基本概念

外测度

回忆$X$上外测度$\mu$指$2^X\rightarrow [0,\infty]$的单调次可加函数, 满足$\mu(\varnothing)=0.$ 有时简称其为测度, 若不会导致混淆的话. 称集合$A\subset X$是$\mu$-可测的, 若$\,\forall\,S\subset X,$ $\mu(S)=\mu(S\setminus A)+\mu(S\cap A).$ 后者称为Caratheodory条件, 指集合$A$能够关于”$\mu$-可加地”切割其它集合.

注 1.1. 事实上由次可加性, 条件等价于$\mu(S)\ge\mu(S\setminus A)+\mu(S\cap A).$ 由此易证任意$\mu$-零测集都是$\mu$-可测的.

由于$S\setminus A=S\cap (X\setminus A),$ 易见$A$为$\mu$-可测的当且仅当$X\setminus A$也是. 设$A_1,A_2$是$\mu$-可测的, 那么$\,\forall\,S\subset X,$

$$\begin{aligned} \mu(S)=&\mu(S\setminus A_1)+\mu(S\cap A_1)& A_1\text{是$\mu$-可测的}\\ =&\mu((S\setminus A_1)\setminus A_2)+\mu((S\setminus A_1)\cap A_2)+\mu(S\cap A_1)& A_2\text{是$\mu$-可测的}\\ \ge&\mu(S\setminus (A_1\cup A_2))+\mu(S\cap((A_2\setminus A_1)\cup A_1))& \text{次可列可加性}\\ =&\mu(S\setminus(A_1\cup A_2))+\mu(S\cap(A_1\cup A_2)), \end{aligned}$$

从而$A_1\cup A_2$是$\mu$-可测的. 那么$X\setminus((X\setminus A_1)\cup (X\setminus A_2))=A_1\cap A_2$也是$\mu$-可测的, $A_2\setminus A_1=A_2\cap (X\setminus A_1)$也是. 归纳可知有限步并,交,补运算均保持$\mu$-可测的性质. Caratheodory条件给无交$\mu$-可测集赋予了可加性:

引理 1.2 (可列可加性). 若$A_1,\cdots$为无交$\mu$-可测集, 则$\,\forall\,S\subset X,$ $\mu(S\cap \bigcup_{i=1}^\infty A_i)=\sum_{i=1}^\infty\mu(S\cap A_i).$

证: 由次可列可加性, $\mu(S\cap \bigcup_{i=1}^\infty A_i)\le \sum_{i=1}^\infty\mu(S\cap A_i).$ 由Caratheodory条件, $\,\forall\,S\subset X,$ $\mu(S\cap (A_1\cup A_2))=\mu((S\cap (A_1\cup A_2))\setminus A_1)+\mu(S\cap(A_1\cup A_2)\cap A_1)= \mu(S\cap A_1)+\mu(S\cap A_2).$ 从而可归纳到有限个无交$\mu$-可测集的情况.

现有$\sum_{i=1}^n \mu(S\cap A_i)=\mu(S\cap\bigcup_{i=1}^n A_i)\le \mu(S\cap \bigcup_{i=1}^\infty A_i).$ 令$n\rightarrow\infty,$ 即得到另一个方向的不等式, 从而命题得证.

引理 1.3. 若$A_1,\cdots$为递增$\mu$-可测集, 则$\,\forall\,S\subset X,$ $\mu(S\cap \bigcup_{i=1}^\infty A_i)=\lim\limits_{i}\mu(S\cap A_i).$

证: 记$A_0=\varnothing,$ 由于$\bigcup_{i=1}^\infty A_i=\bigcup_{i=1}^\infty (A_i\setminus A_{i-1}),$ 由可列可加性,

$$\begin{aligned} \mu(S\cap \bigcup_{i=1}^\infty A_i)&=\mu(S\cap \bigcup_{i=1}^\infty (A_{i}\setminus A_{i-1}))\\ &= \sum_{i=1}^\infty \mu(S\cap (A_i\setminus A_{i-1}))\\ &=\lim\limits_n \sum_{i=1}^n \mu(S\cap (A_i\setminus A_{i-1}))\\ &=\lim\limits_{n}\mu(S\cap A_n), \end{aligned}$$

命题得证.

推论 1.4. 设$A_1,\cdots$为递减$\mu$-可测集, 若$\,\exists\,A_k$测度有限, 则$\,\forall\,S\subset X,$ $\mu(S\cap \bigcap_{i=1}^\infty A_i)=\lim\limits_{i}\mu(S\cap A_i).$

称$X$一系列子集的集合$\mathcal{S}$为$\sigma$-代数, 若:

(1) $\varnothing,X\in \mathcal{S};$

(2) $A\in \mathcal{S}\Rightarrow X\setminus A\in \mathcal{S};$

(3) $A_1,\cdots \in \mathcal{S}\Rightarrow \bigcup_{i=1}^\infty A_i\in \mathcal{S}.$

注 1.5. $\sigma$-代数对可列并与取补运算封闭, 易见它对可列交也是封闭的. 由于$\sigma$-代数的交仍为$\sigma$-代数, 可以找到包含一系列指定集合的最小的一个$\sigma$-代数. 特别地, 对拓扑空间$X,$ 包含所有开集的最小$\sigma$-代数称为$X$中的Borel代数, 其中集合称为Borel集.

引理 1.6. $\mu$-可测集全体构成$\sigma$-代数, 包含所有$X$的$\mu$-零测集.

证: 只需证若$A_1,\cdots$是$\mu$-可测的, 则$\bigcup_{i=1}^\infty A_i$也是. 记$B_i=\bigcup_{j=1}^i A_j$为递增$\mu$-可测集, 则$\,\forall\,S\subset X,$

$$\begin{aligned} \mu(S)&=\mu(S\setminus \bigcup_{i=1}^n A_i)+\mu(S\cap \bigcup_{i=1}^n A_i)\\ &\ge\mu(S\setminus \bigcup_{i=1}^\infty A_i)+\mu(S\cap B_n). \end{aligned}$$

令$n\rightarrow \infty,$ 即有

$$\begin{aligned} \mu(S)&\ge \mu(S\setminus \bigcup_{i=1}^\infty A_i)+ \mu(S\cap \bigcup_{n=1}^\infty B_n)\\ &= \mu(S\setminus \bigcup_{i=1}^\infty A_i)+\mu(S\cap \bigcup_{i=1}^\infty A_i), \end{aligned}$$

从而命题得证.

Borel正则测度

称外测度$\mu$是正则的, 若$\,\forall\,A\subset X,$ 存在$\mu$-可测集$B\supset A$满足$\mu(B)=\mu(A).$ 此时先前关于$A_i$为单调列的性质无需再假设$A_i$是$\mu$-可测的.

当$X$为拓扑空间时, 外测度$\mu$是Borel正则的若所有Borel集都是$\mu$-可测的, 且$\,\forall\,A\subset X,$ 存在Borel集$B\supset A$满足$\mu(B)=\mu(A).$

需要强调的是$\mu(B)=\mu(A)$不意味着$\mu(B\setminus A)=0,$ 除非$A$是$\mu-$可测的且$\mu(A)<\infty.$

注 1.7. 回忆Borel测度$\mu_0$为Borel集全体到$[0,\infty]$上的映射, 满足$\mu_0(\varnothing)=0$与可列可加性(从而蕴含单调性). 此时$\,\forall\,A\subset X,$ $\mu(A):=\inf\limits_{B\text{ Borel, }B\supset A}\mu_0(B)$便定义了一个Borel正则外测度, 在Borel集全体上与$\mu_0$一致. 反过来一个Borel正则外测度在Borel集全体上的限制即为一个Borel测度.

给定任意子集$Y\subset X$与外测度$\mu,$ 可以定义一个新的外测度$\mu |_Y,$ $\mu|_Y(Z):=\mu(Y\cap Z),$ $Z\subset X.$ 任意$\mu$-可测集都是$\mu|_Y$-可测的, 即使$Y$本身不见得$\mu$-可测. 当$Y$是$\mu$-可测的且$\mu(Y)<\infty$时, 若$\mu$是Borel正则的, 则$\mu|_Y$也是. 关于这一点只需利用$\mu$的Borel正则性取合适的$B_1\supset Y,$ $B_2\supset B_1\setminus Y,$ $B_3\supset A\cap Y.$ 那么$A\subset (X\setminus B_1)\cup B_2\cup B_3$且两者$\mu|_Y$测度相同.

当$X$为度量空间时, 对$A,B\subset X,$ 记$d(A,B)=\inf\{d(a,b):a\in A,b\in B\}.$ $\inf\varnothing:=+\infty.$

定理 1.8 (Caratheodory判别法). 设$(X,d)$为度量空间, $\mu$为外测度. 若$\mu(A\cup B)=\mu(A)+\mu(B),$ $\,\forall\,A,B\subset X$满足$d(A,B)>0,$ 则所有Borel集都是$\mu$-可测的.

证: 只需证明所有闭集均是$\mu$-可测的. $\,\forall\,$闭集$C\subset X,$ 取$C_j=\{x\in X:d(x,C)\le \frac{1}{j}\}.$ 那么$\,\forall\,S\subset X,$ $d(S\setminus C_j,S\cap C)>0.$ 从而有$$\mu(S)\ge \mu(S\setminus C_j)+\mu(S\cap C),\quad \,\forall\,j>0.$$

令$j\rightarrow \infty,$ 由于$S\setminus C_j=S\cap (X\setminus C_j),$ $X\setminus C_j$单增, 从而$\lim\limits_{j\rightarrow \infty}\mu(S\setminus C_j)=\mu(S\setminus \bigcap_{j}C_j)=\mu(S\setminus C).$ 最后一个等号是因为$C$为闭集, 故所有与$C$距离为零的点都在$C$中. 因此$$\mu(S)\ge \mu(S\setminus C)+\mu(S\cap C),$$从而任意闭集$C$是可测的, 命题得证.

定义 1.9. 称Borel正则测度$\mu$是开$\sigma$-有限的, 若$X=\bigcup_j V_j,$ $V_j$为开集且测度有限.

显然当$X$为可分度量空间, $\mu$局部有限时, 它是开$\sigma$-有限的.

定理 1.10. 设拓扑空间$X$中每个闭集均是可列个开集之交, 设$\mu$是开$\sigma$-有限的Borel正则测度, 那么$\,\forall\,A\subset X,$ $$\mu(A)=\inf_{\text{开集}U\supset A}\mu(U).$$ 若$A$是$\mu$-可测的, 那么还有$$\mu(A)=\sup_{\text{闭集}C\subset A}\mu(C).$$

思路: 分全空间测度有限/无限的情况讨论. 取使结论成立的集合全体, 证其包含$\sigma$-代数, 其中又包含所有开集, 进而包含整个Borel代数. 最后由正则性即得结论.

证: 首先假设$\mu(X)<\infty,$ 此时$(1)\Rightarrow (2).$ 由测度是Borel正则的, 只需证明$A$为Borel集的情况. 取$\mathcal{A}:=\{\text{Borel集}A\subset X:\text{(1)成立}\},$ 那么$\mathcal{A}$自然包含所有开集.

取$A_1,\cdots\in \mathcal{A}.$ $\,\forall\,\varepsilon>0,$ $\,\exists\,$开$U_1,\cdots,$ 满足$U_j\supset A_j,$ 且$\mu(U_j\setminus A_j)\le 2^{-j}\varepsilon.$ 那么$$\mu(\bigcup_j U_j \setminus \bigcup_j A_j),\mu(\bigcap_j U_j\setminus \bigcap_j A_j)\le \sum_j\mu(U_j\setminus A_j)\le \varepsilon,$$ 从而$\bigcup_j A_j,\bigcap_j A_j\in \mathcal{A},$ 集合对可列并, 可列交封闭. 由于任意闭集是开集可列交, 全体闭集均在$\mathcal{A}$中.

取$\widetilde{\mathcal{A} }:=\{A\in \mathcal{A}:X\setminus A\in \mathcal{A}\},$ 那么容易验证$\widetilde{\mathcal{A} }$为包含全体开集的$\sigma$-代数, 从而$\mathcal{A}$包含(即为)Borel集全体, $\mu(X)$有限的情况得证.

对$\mu(X)=\infty$的情况, 只需利用$X$的$\sigma$-有限性, 取$X=\bigcup_{j} V_j,$ 每个$V_j$测度有限. 此时$\mu|_{V_{j} }(X)<\infty,$ 应用前面的结论, $\,\forall\,\varepsilon>0,$ 有开集$U_j\supset A$满足$\mu(U_j\cap V_j\setminus A)\le2^{-j}\varepsilon,$ 从而$\mu(\bigcup_{j}(U_j\cap V_j)\setminus A)\le \varepsilon.$ 由于$\bigcup_j (U_j\cap V_j)$为含$A$开集, (1)成立.

对(2), 类似地$\,\forall\,\varepsilon>0,$ 有闭集$C_j\subset A$满足$\mu(A\cap V_j\setminus C_j)\le 2^{-j}\varepsilon.$ 那么$$\mu(A\setminus \bigcup_j C_j)=\mu( A\cap \bigcup_jV_j\setminus \bigcup_jC_j)\le \sum_j \mu(A\cap V_j\setminus C_j) \le \varepsilon.$$ 从而当$\mu(A)=\infty$时, $\mu(\bigcup_{j=1}^NC_j)\rightarrow \infty;$ $\mu(A)<\infty$时, 对足够大的$N$有$\mu(A\setminus \bigcup_{j=1}^N C_j)\le 2\varepsilon.$ 因此(2)得证.

注 1.11. 注意若$X$是$\sigma$-紧的Hausdorff空间, 即全空间为可列个紧集之并. 那么第二个性质可推出$$\mu(A)=\sup_{\text{紧集}K\subset A}\mu(K).$$ 这是因为在上述假设下, 任意闭集为一列递增紧集之并.

由上述定理可证明Lusin定理:

定理 1.12 (Lusin). 令$\mu$为Borel正则外测度, 拓扑空间$X$中所有闭集均为可列开集之交(如$X$取度量空间即可). 令$A$为$\mu$-可测集且测度有限, $f:A\rightarrow \mathbb{R}$为可测函数. 那么$\,\forall\,\varepsilon>0,$ 存在$X$中闭集$C\subset A,$ 满足$\mu(A\setminus C)<\varepsilon,$ 且$f|_C$连续.

证: 取$A_{ij}=f^{-1}[\frac{j-1}{i},\frac{j}{i}),$ $i\in \mathbb{Z}_+,$ $j\in \mathbb{Z}.$ 那么$\mu$-可测的$A_{ij}$两两无交, 测度有限, 且有$\bigcup_{j}A_{ij}=A$对任意$i$成立. 由于$\mu|_{A}$是Borel正则的, 应用上述定理, 可找到$X$中闭集$C_{ij}\subset A_{ij},$ $\mu(A_{ij}\setminus C_{ij})=\mu|_A(A_{ij}\setminus C_{ij})<2^{-i-|j|-2}\varepsilon.$ 进一步$\mu(A\setminus \bigcup_j C_{ij})< 2^{-i}\varepsilon,$ 因此对每个$i,$ 有$J(i)\in \mathbb{Z}_+$使得$\mu(A\setminus \bigcup_{|j|\le J(i)}C_{ij})\le 2^{-i}\varepsilon.$ 取$C=\bigcap_{i}\bigcup_{|j|\le j(i)}C_{ij}$为闭集, 那么$\mu(A\setminus C)\le \varepsilon.$

最后, 定义$g_i:\bigcup_{|j|\le J(i)}C_{ij}\rightarrow \mathbb{R},$ $g_i|_{C_{ij} }(x)=\frac{j-1}{i}.$ 由于固定$i,$ $C_{ij}$彼此无交, 因此$g_i|_C$连续. 由于$\,\forall\,x\in C,$ $0\le f(x)-g_i(x)\le \frac{1}{i},$ 因此$g_{i}|_C\rightrightarrows f|_C,$ 从而$f|_C$连续.

文章最后更新于 2021-10-09 13:26:19

Outer Measure

Recall that the external measure $\mu$ on $X$ refers to the monotonic subadditive function of $2^X\rightarrow [0,\infty]$, Satisfies $\mu(\varnothing)=0.$ It is sometimes simply called a measure, if this does not cause confusion. A set $A\subset X$ is said to be $\mu$-measurable if $\,\forall\,S\subset X,$ $\mu(S)=\mu(S\setminus A)+\mu(S\cap A).$ The latter is called the Caratheodory condition, It means that the set $A$ can cut other sets additively with respect to $\mu$.

Note 1.1. In fact, due to subadditivity, the condition is equivalent to $\mu(S)\ge\mu(S\setminus A)+\mu(S\cap A).$, so it is easy to prove that any $\mu$-zero test set is $\mu$-measurable.

Due to $S\setminus A=S\cap (X\setminus A),$ It is easy to see that $A$ is $\mu$-measurable if and only if $X\setminus A$ is also. Assume $A_1,A_2$ is $\mu$-measurable, then $\,\forall\,S\subset X,$

Thus $A_1\cup A_2$ is $\mu$-measurable. Then $X\setminus((X\setminus A_1)\cup (X\setminus A_2))=A_1\cap A_2$ is also $\mu$-measurable, $A_2\setminus A_1=A_2\cap (X\setminus A_1)$ Same thing. It can be concluded by induction that the finite-step union, intersection, and complement operations all maintain the $\mu$-measurable property. The Caratheodory condition confers additivity to unintersecting $\mu$-measurable sets:

Lemma 1.2 (Additivity is possible). If $A_1,\cdots$ is a disjoint $\mu$-measurable set, then $\,\forall\,S\subset X,$ $\mu(S\cap \bigcup_{i=1}^\infty A_i)=\sum_{i=1}^\infty\mu(S\cap A_i).$

Certificate: By subsumable additivity, $\mu(S\cap \bigcup_{i=1}^\infty A_i)\le \sum_{i=1}^\infty\mu(S\cap A_i).$ Conditioned by Caratheodory, $\,\forall\,S\subset X,$ $\mu(S\cap (A_1\cup A_2))=\mu((S\cap (A_1\cup A_2))\setminus A_1)+\mu(S\cap(A_1\cup A_2)\cap A_1)= \mu(S\cap A_1)+\mu(S\cap A_2).$ This can be summarized into the case of a finite number of disjoint $\mu$-measurable sets.

Existing$\sum_{i=1}^n \mu(S\cap A_i)=\mu(S\cap\bigcup_{i=1}^n A_i)\le \mu(S\cap \bigcup_{i=1}^\infty A_i).$ Let $n\rightarrow\infty,$ get the inequality in the other direction, so the proposition is proved.

Lemma 1.3. If $A_1,\cdots$ is an increasing $\mu$-measurable set, then $\,\forall\,S\subset X,$ $\mu(S\cap \bigcup_{i=1}^\infty A_i)=\lim\limits_{i}\mu(S\cap A_i).$

Certificate: Remember$A_0=\varnothing,$ Due to $\bigcup_{i=1}^\infty A_i=\bigcup_{i=1}^\infty (A_i\setminus A_{i-1}),$ By listability and additivity,

The proposition is proved.

Corollary 1.4. Let $A_1,\cdots$ be a decreasing $\mu$-measurable set. If $\,\exists\,A_k$ has limited measure, then $\,\forall\,S\subset X,$ $\mu(S\cap \bigcap_{i=1}^\infty A_i)=\lim\limits_{i}\mu(S\cap A_i).$

The set $\mathcal{S}$ of a series of subsets of $X$ is called $\sigma$-algebra, if:

(1) $\varnothing,X\in \mathcal{S};$

(2) $A\in \mathcal{S}\Rightarrow X\setminus A\in \mathcal{S};$

(3) $A_1,\cdots \in \mathcal{S}\Rightarrow \bigcup_{i=1}^\infty A_i\in \mathcal{S}.$

Note 1.5. The $\sigma$-algebra is closed to listable union and complement operations, and it is easy to see that it is also closed to listable intersections. Since the intersection of $\sigma$-algebras is still a $\sigma$-algebra, the smallest $\sigma$-algebra containing a series of specified sets can be found. In particular, the smallest $\sigma$-algebra containing all open sets for the topological space $X,$ is called the Borel algebra in $X$, and the set is called a Borel set.

Lemma 1.6. $\mu$-Measurable set as a whole constitutes $\sigma$-algebra, including $\mu$-zero measurable set of all $X$.

Certificate: Just prove that if $A_1,\cdots$ is $\mu$-measurable, Then $\bigcup_{i=1}^\infty A_i$ is also. Note $B_i=\bigcup_{j=1}^i A_j$ as increasing $\mu$-measurable set, Then $\,\forall\,S\subset X,$

$$\begin{aligned} \mu(S)&=\mu(S\setminus \bigcup_{i=1}^n A_i)+\mu(S\cap \bigcup_{i=1}^n A_i)\\ &\ge\mu(S\setminus \bigcup_{i=1}^\infty A_i)+\mu(S\cap B_n). \end{aligned}$$

Let $n\rightarrow \infty,$ be

Thus the proposition is proved.

Borel regular measure

The external measure $\mu$ is said to be regular, if$\,\forall\,A\subset X,$ There exists $\mu$-measurable set $B\supset A$ satisfying $\mu(B)=\mu(A).$ At this point, the previous property that $A_i$ is a monotonic sequence no longer needs to be assumed that $A_i$ is $\mu$-measurable.

When $X$ is a topological space, The outer measure $\mu$ is Borel regular If all Borel sets are $\mu$-measurable, And $\,\forall\,A\subset X,$ exists a Borel set $B\supset A$ that satisfies $\mu(B)=\mu(A).$

It should be emphasized that $\mu(B)=\mu(A)$ does not mean $\mu(B\setminus A)=0,$ Unless $A$ is measurable by $\mu-$ and $\mu(A)<\infty.$

Note 1.7. Recall that the Borel measure $\mu_0$ is the mapping of the entire Borel set to $[0,\infty]$, satisfying $\mu_0(\varnothing)=0$ and listable additivity (thus implying monotonicity). At this time, $\,\forall\,A\subset X,$ $\mu(A):=\inf\limits_{B\text{ Borel, }B\supset A}\mu_0(B)$ defines a Borel regular external measure, which is consistent with $\mu_0$ on the entire Borel set. In turn, the restriction of a Borel regular external measure on the entire Borel set is a Borel measure.

Given any subset $Y\subset X$ and external measure $\mu,$, a new external measure $\mu |_Y,$ can be defined $\mu|_Y(Z):=\mu(Y\cap Z),$ $Z\subset X.$ Any $\mu$-measurable set is $\mu|_Y$-measurable, even if $Y$ itself is not necessarily $\mu$-measurable. When $Y$ is $\mu$-measurable and $\mu(Y)<\infty$, if $\mu$ is Borel regular, Then $\mu|_Y$ is also. For this point, just use the Borel regularity of $\mu$ to choose the appropriate $B_1\supset Y,$ $B_2\supset B_1\setminus Y,$ $B_3\supset A\cap Y.$ Then $A\subset (X\setminus B_1)\cup B_2\cup B_3$ and both $\mu|_Y$ have the same measure.

When $X$ is a metric space, for $A,B\subset X,$ Note $d(A,B)=\inf\{d(a,b):a\in A,b\in B\}.$ $\inf\varnothing:=+\infty.$

Theorem 1.8 (Caratheodory discrimination method). Let $(X,d)$ be the metric space and $\mu$ be the external measure. If $\mu(A\cup B)=\mu(A)+\mu(B),$ $\,\forall\,A,B\subset X$ satisfies $d(A,B)>0,$, then all Borel sets are $\mu$-measurable.

Certificate: Just prove that all closed sets are $\mu$-measurable. $\,\forall\,$ Closed sets $C\subset X,$ Take $C_j=\{x\in X:d(x,C)\le \frac{1}{j}\}.$ then $\,\forall\,S\subset X,$ $d(S\setminus C_j,S\cap C)>0.$ Thus there is $$\mu(S)\ge \mu(S\setminus C_j)+\mu(S\cap C),\quad \,\forall\,j>0.$$

Let $j\rightarrow \infty,$ due to $S\setminus C_j=S\cap (X\setminus C_j),$ $X\setminus C_j$Single increase, Thus $\lim\limits_{j\rightarrow \infty}\mu(S\setminus C_j)=\mu(S\setminus \bigcap_{j}C_j)=\mu(S\setminus C).$ The last equal sign is because $C$ is a closed set, so all points with a zero distance from $C$ are in $C$. Therefore $$\mu(S)\ge \mu(S\setminus C)+\mu(S\cap C),$$ and thus any closed set $C$ is measurable, The proposition is proved.

Definition 1.9. The Borel regular measure $\mu$ is said to be open $\sigma$ -Limited , if $X=\bigcup_j V_j,$ $V_j$ is an open set and the measure is limited.

Obviously, when $X$ is a separable metric space and $\mu$ is locally finite, it is open $\sigma$-finite.

Theorem 1.10. Assume that every closed set in the topological space $X$ is the intersection of enumerable open sets, and $\mu$ is an open $\sigma$-finite Borel regular measure, then $\,\forall\,A\subset X,$ $$\mu(A)=\inf_{\text{开集}U\supset A}\mu(U).$$ If $A$ is $\mu$-measurable, then there is also $$\mu(A)=\sup_{\text{闭集}C\subset A}\mu(C).$$

Idea: Discuss the case where the measure of the whole space is finite/infinite. Take the total set that makes the conclusion true, Prove that it contains the $\sigma$-algebra, which in turn contains all open sets, and in turn contains the entire Borel algebra. Finally, the conclusion is drawn from regularity.

Certificate: First assume $\mu(X)<\infty,$ at this time $(1)\Rightarrow (2).$ Since the measure is Borel regular, we only need to prove that $A$ is a Borel set. Take $\mathcal{A}:=\{\text{Borel集}A\subset X:\text{(1)成立}\},$ Then $\mathcal{A}$ naturally contains all open sets.

Take $A_1,\cdots\in \mathcal{A}.$ $\,\forall\,\varepsilon>0,$ $\,\exists\,$ Open $U_1,\cdots,$ Meet $U_j\supset A_j,$ And $\mu(U_j\setminus A_j)\le 2^{-j}\varepsilon.$ Then $$\mu(\bigcup_j U_j \setminus \bigcup_j A_j),\mu(\bigcap_j U_j\setminus \bigcap_j A_j)\le \sum_j\mu(U_j\setminus A_j)\le \varepsilon,$$ Thus $\bigcup_j A_j,\bigcap_j A_j\in \mathcal{A},$ sets can be merged, Intersectable and closed. Since any closed set is an open set and can be intersected, all closed sets are in $\mathcal{A}$.

Take $\widetilde{\mathcal{A} }:=\{A\in \mathcal{A}:X\setminus A\in \mathcal{A}\},$ Then it is easy to verify that $\widetilde{\mathcal{A} }$ is an $\sigma$-algebra containing all open sets, Therefore, $\mathcal{A}$ contains (is) the entire Borel set, and $\mu(X)$ is proved in limited cases.

For the case of $\mu(X)=\infty$, just use the $\sigma$-finiteness of $X$, Take $X=\bigcup_{j} V_j,$ and each $V_j$ measure is limited. At this time $\mu|_{V_{j} }(X)<\infty,$ Applying the previous conclusion, $\,\forall\,\varepsilon>0,$ There is an open set $U_j\supset A$ that satisfies $\mu(U_j\cap V_j\setminus A)\le2^{-j}\varepsilon,$ Thus $\mu(\bigcup_{j}(U_j\cap V_j)\setminus A)\le \varepsilon.$ Since $\bigcup_j (U_j\cap V_j)$ is an open set containing $A$, (1) holds.

For (2), similarly $\,\forall\,\varepsilon>0,$ There is a closed set $C_j\subset A$ satisfying $\mu(A\cap V_j\setminus C_j)\le 2^{-j}\varepsilon.$ Then $$\mu(A\setminus \bigcup_j C_j)=\mu( A\cap \bigcup_jV_j\setminus \bigcup_jC_j)\le \sum_j \mu(A\cap V_j\setminus C_j) \le \varepsilon.$$ So when $\mu(A)=\infty$, $\mu(\bigcup_{j=1}^NC_j)\rightarrow \infty;$ When $\mu(A)<\infty$, For large enough $N$ there is $\mu(A\setminus \bigcup_{j=1}^N C_j)\le 2\varepsilon.$ Therefore (2) is proved.

Note 1.11. Note that if $X$ is a $\sigma$-compact Hausdorff space, that is, the whole space is the union of a list of compact sets. Then the second property can be derived from $$\mu(A)=\sup_{\text{紧集}K\subset A}\mu(K).$$. This is because under the above assumption, any closed set is the union of a list of increasing compact sets.

From the above theorem, Lusin’s theorem can be proved:

Theorem 1.12 (Lusin). Let $\mu$ be the Borel canonical outer measure, and all closed sets in the topological space $X$ are intersections of listable open sets (for example, $X$ can take the metric space). Let $A$ be $\mu$ - a measurable set with limited measure, and $f:A\rightarrow \mathbb{R}$ be a measurable function. Then $\,\forall\,\varepsilon>0,$ exists in $X$ and the closed set $C\subset A,$ satisfies $\mu(A\setminus C)<\varepsilon,$ and $f|_C$ is continuous.

Certificate: Take $A_{ij}=f^{-1}[\frac{j-1}{i},\frac{j}{i}),$ $i\in \mathbb{Z}_+,$ $j\in \mathbb{Z}.$ Then $\mu$-measurable $A_{ij}$ have no intersection, and the measure is limited. And $\bigcup_{j}A_{ij}=A$ holds for any $i$. Since $\mu|_{A}$ is Borel regular, Applying the above theorem, we can find the closed set $C_{ij}\subset A_{ij},$ in $X$ $\mu(A_{ij}\setminus C_{ij})=\mu|_A(A_{ij}\setminus C_{ij})<2^{-i-|j|-2}\varepsilon.$ Further$\mu(A\setminus \bigcup_j C_{ij})< 2^{-i}\varepsilon,$ Therefore for each $i,$ Having $J(i)\in \mathbb{Z}_+$ makes $\mu(A\setminus \bigcup_{|j|\le J(i)}C_{ij})\le 2^{-i}\varepsilon.$ Take $C=\bigcap_{i}\bigcup_{|j|\le j(i)}C_{ij}$ as a closed set, Then $\mu(A\setminus C)\le \varepsilon.$

Finally, define $g_i:\bigcup_{|j|\le J(i)}C_{ij}\rightarrow \mathbb{R},$ $g_i|_{C_{ij} }(x)=\frac{j-1}{i}.$ Since $i,$ and $C_{ij}$ are fixed and have no intersection with each other, Therefore $g_i|_C$ is continuous. Since $\,\forall\,x\in C,$ $0\le f(x)-g_i(x)\le \frac{1}{i},$ Therefore $g_{i}|_C\rightrightarrows f|_C,$ Thus $f|_C$ is continuous.

The article was last updated on 2021-10-09 13:26:19