MATH 597: Measure Theory

这是个 trivial proof. 但是它具有一定理解上的启发.
我们对 $𝜎$-algebra 有一个直观理解: 如果我们想把一些集合做成一个 $𝜎$-algebra, 那么首先我们把它们的补集放进这个 $𝜎$-algebra 里, 其次我们把这些集合的 up to countable 的任意组合的并集也放进这个 $𝜎$-algebra 里.
因而即便我们把一些 $𝜎$-algebra 给 intersect 起来, 其中每个集合的补集和这些集合的 up to ctbl 的任意组合的并集也在这个 intersection 里.
这是个重要的直观理解. 我们想到, 如果我们要把一个 sigma-algebra 里的一部分去掉，并保持它仍然是一个 sigma-algebra，那么我们得把这些集合的补集, 以及能够 ctbly union 成这些集合的小集合也去掉, 并对这些小集合也 recursively 进行这个操作.

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trivial.

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我们 recall: 所有的 countable 以及 second countable 的 topological space 都具有 Lindelöf property: 任意 open covering 都存在一个 countable 的 subcovering.
Lindelöf property 的一个推论就是, 在具有 Lindelöf property 的 metric space 或者 second countable 的 space 中, 任意 open set 都可以写成 countable 个 open balls 的 union.
我们在 elementary 的 real analysis 中已经学过, $[𝑎,𝑏)={\cap }_{𝑛\ge 1}(𝑎-1/𝑛,𝑏)$, 以其作为例子, 这些 intervals 彼此之间都可以相互转换.

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显然, ctbl additive implies finite additive.

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而后使用 finite additive 可得. 这是一个 direct corollary of countable additivity.

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显然, ctbl additive implies finite additive.

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而后使用 finite additive 可得. 这是一个 direct corollary of countable additivity.

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• 1. Since $\varnothing \in \mathrm{𝒜︀}$, $\varnothing \cap 𝑌=\varnothing$, we have $\varnothing \in \mathrm{𝒜︀}{|}_{𝑌}$

  2. Let $𝐹\in \mathrm{𝒜︀}{|}_{𝑌}$, we must have $𝐸\in \mathrm{𝒜︀}$ s.t. $𝐸\cap 𝑌=𝐹$. Since $𝐸\in \mathrm{𝒜︀}$, we have $𝑋\backslash 𝐸\in \mathrm{𝒜︀}$, so $𝑋\backslash 𝐸\cap 𝑌\in \mathrm{𝒜︀}{|}_{𝑌}$. Since $𝐸\cap 𝑌=𝐹$ and $𝑌=(𝐸\cap 𝑌)\bigsqcup ((𝑋\backslash 𝐸)\cap 𝑌)$, it implies $(𝑋\backslash 𝐸)\cap 𝑌=𝑌\backslash 𝐹$, therefore $𝑌\backslash 𝐹\in \mathrm{𝒜︀}{|}_{𝑌}$.

  3. Let ${𝐹}_1,{𝐹}_2,\cdots$ be a sequence of subsets in $\mathrm{𝒜︀}{|}_{𝑌}$. Then for each $𝑖\in \mathbb{N}$, we have ${𝐹}_{𝑖}={𝐸}_{𝑖}\cap 𝑌$ for some ${𝐸}_{𝑖}\in \mathrm{𝒜︀}$. Then ${\bigcup }_{𝑖=1}^{\infty }{𝐹}_{𝑖}={\bigcup }_{𝑖=1}^{\infty }({𝐸}_{𝑖}\cap 𝑌)=({\bigcup }_{𝑖=1}^{\infty }{𝐸}_{𝑖})\cap 𝑌\in \mathrm{𝒜︀}{|}_{𝑌}$ since ${\bigcup }_{𝑖=1}^{\infty }{𝐸}_{𝑖}\in \mathrm{𝒜︀}$.

• Let $\mathrm{ℬ︀}$ be a $𝜎$-algebra on $𝑌$.

  prove that there exists a $𝜎$-algebra $\mathrm{𝒜︀}$ on $𝑋$ such that $\mathrm{𝒜︀}{|}_{𝑌}=\mathrm{ℬ︀}$. Consider let

  $$\mathrm{𝒜︀}≔\left\{𝐸\subset 𝑋\mid 𝐸\cap 𝑌\in \mathrm{ℬ︀}\right\}$$

  Then

  $$\mathrm{𝒜︀}{|}_{𝑌}=\left\{𝐸\cap 𝑌\mid 𝐸,𝑌\subset 𝑋,𝐸\cap 𝑌\in \mathrm{ℬ︀}\right\}=\mathrm{ℬ︀}$$

  We then prove that this is a $𝜎$-algebra on $𝑋$.
  

  1. $\varnothing \cap 𝑌=\varnothing$ so $\varnothing \in \mathrm{𝒜︀}$.

  2. Closed under complement: Let $𝐸\in \mathrm{𝒜︀}$, we have $𝐸\cap 𝑌\in \mathrm{ℬ︀}$, so $𝑌\backslash (𝐸\cap 𝑌)=𝑌\backslash 𝐸\in \mathrm{ℬ︀}$.
     Then $(𝑋\backslash 𝐸)\cap 𝑌=𝑌\backslash 𝐸\in \mathrm{ℬ︀}$, so $𝑋\backslash 𝐸\in \mathrm{𝒜︀}$.

  3. Closed under countable union: Let ${𝐸}_1,{𝐸}_2,\cdots$ be a sequence in $\mathrm{𝒜︀}$, then ${𝐸}_{𝑛}\cap 𝑌\in \mathrm{ℬ︀}$. for each $𝑛$. Hence

     $$\left(\bigcup _{𝑛=1}^{\infty }{𝐸}_{𝑛}\right)\cap 𝑌=\bigcup _{𝑛=1}^{\infty }({𝐸}_{𝑛}\cap 𝑌)\in \mathrm{ℬ︀},$$

     since $\mathrm{ℬ︀}$ is a $𝜎$-algebra on $𝑌$. Therefore, ${\bigcup }_{𝑛=1}^{\infty }{𝐸}_{𝑛}\in \mathrm{𝒜︀}$.

• This is not unique.
  Counterexample:

  $$𝑋=\left\{0,1,2\right\},𝑌=\left\{0\right\}\subset 𝑋$$

  Consider

  $${𝐴}_1≔\mathrm{𝒫︀}(𝑋),{𝐴}_2≔\left\{\varnothing ,\left\{0\right\},\left\{1,2\right\},𝑋\right\}$$

  are valid $𝜎$-algebra on $𝑋$.
  Then we have ${𝐴}_1{|}_{𝑌}={𝐴}_2{|}_{𝑌}=\left\{\varnothing ,\left\{0\right\}\right\}$, while ${𝐴}_1$ is different from ${𝐴}_2$.

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Fix $𝑡\in {ℝ}^{𝑛}$. Define

We want to show that $\mathrm{𝒜︀}=\mathrm{ℬ︀}({ℝ}^{𝑛})$. We first show that $\mathrm{𝒜︀}$ is a $𝜎$-algebra.

1. $\varnothing \in \mathrm{𝒜︀}$ since $𝑡+\varnothing =\varnothing \in \mathrm{ℬ︀}({ℝ}^{𝑛})$.

2. $\mathrm{𝒜︀}$ is closed under complement: Let $𝐵\in \mathrm{𝒜︀}$, then $𝑡+𝐵\in \mathrm{ℬ︀}({ℝ}^{𝑛})$. The complement $(𝑡+𝐵{)}^{𝑐}$ is also in $\mathrm{ℬ︀}({ℝ}^{𝑛})$. Observe

Since $𝑡+𝐵$ is Borel, its complement is Borel, hence $𝑡+{𝐵}^{𝑐}$ is Borel, so ${𝐵}^{𝑐}\in \mathrm{𝒜︀}$.

3. $\mathrm{𝒜︀}$ is closed under countable unions: Let ${𝐵}_{𝑘}\in \mathrm{𝒜︀}$ for $𝑘=1,2,\dots$, then $𝑡+{𝐵}_{𝑘}\in \mathrm{ℬ︀}({ℝ}^{𝑛})$. Thus

Hence ${\bigcup }_{𝑘=1}^{\infty }{𝐵}_{𝑘}\in \mathrm{𝒜︀}$. These three properties show that $\mathrm{𝒜︀}$ is a $𝜎$-algebra.
Since $𝑡+𝑈$ is open if $𝑈$ is open in ${ℝ}^{𝑛}$, $\mathrm{𝒜︀}$ contains all open sets. Since $\mathrm{ℬ︀}({ℝ}^{𝑛})$ is the smallest $𝜎$-algebra containing all open sets in ${ℝ}^{𝑛}$, we have:$\mathrm{ℬ︀}({ℝ}^{𝑛})\subseteq \mathrm{𝒜︀}$ Hence suppose $𝐴\in \mathrm{ℬ︀}({ℝ}^{𝑛})$, then $𝐴\in \mathrm{𝒜︀}$, so $𝑡+𝐴\in \mathrm{ℬ︀}({ℝ}^{𝑛})$. This completes the proof of translation invariance.

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Fix $𝜆\in ℝ$. Case 1: $𝜆=0$, then $𝜆𝐴=\left\{0\right\}$ if $𝐴\ne \varnothing$, and $𝜆𝐴=\varnothing$ otherwise. Both $\left\{0\right\}$(closed set) and $\varnothing$ is Borel set.

Case 2: $𝜆\ne 0$. We define

We want to show that $\mathrm{𝒜︀}=\mathrm{ℬ︀}({ℝ}^{𝑛})$. We first show that $\mathrm{𝒜︀}$ is a $𝜎$-algebra.

1. $\varnothing \in \mathrm{𝒜︀}$ since $𝜆\varnothing =\varnothing$.

2. $\mathrm{𝒜︀}$ is closed under complement: Let $𝐵\in \mathrm{𝒜︀}$, then $𝜆𝐵\in \mathrm{ℬ︀}({ℝ}^{𝑛})$, then $(𝜆𝐵{)}^{𝑐}$ is also in $\mathrm{ℬ︀}({ℝ}^{𝑛})$. Observe $(𝜆𝐵{)}^{𝑐}=𝜆{𝐵}^{𝑐}$, so $𝜆{𝐵}^{𝑐}\in \mathrm{ℬ︀}({ℝ}^{𝑛})$, therefore ${𝐵}^{𝑐}\in \mathrm{𝒜︀}$. 3. $\mathrm{𝒜︀}$ is closed under countable unions: Let ${𝐵}_{𝑘}\in \mathrm{𝒜︀}$ for $𝑘=1,2,\dots$, then $𝜆{𝐵}_{𝑘}\in \mathrm{ℬ︀}({ℝ}^{𝑛})$. Thus

Hence ${\bigcup }_{𝑘=1}^{\infty }{𝐵}_{𝑘}\in \mathrm{𝒜︀}$. These three properties show that $\mathrm{𝒜︀}$ is a $𝜎$-algebra.
Since $𝜆\ne 0$, $𝜆𝑈$ is open iff $𝑈$ is open in ${ℝ}^{𝑛}$, thus $\mathrm{𝒜︀}$ contains all open sets, so $\mathrm{ℬ︀}({ℝ}^{𝑛})\subseteq \mathrm{𝒜︀}$,

Hence if $𝐴\in \mathrm{ℬ︀}({ℝ}^{𝑛})$, we have $𝐴\in \mathrm{𝒜︀}$, therefore $𝜆𝐴\in \mathrm{ℬ︀}({ℝ}^{𝑛})$. This completes the proof of translation invariance.

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Define：

Then clearly

Hence it suffices to show that $𝐵$ and $𝐶$ are Borel sets, since intersection of two Borel sets is a Borel set. And thus it suffices to show that ${𝐵}^{𝑐}$ and $𝐶$ are Borel sets. Note

, so the proof for ${𝐵}^{𝑐}$ and $𝐶$ are about the same. We now show ${𝐵}^{𝑐}$ is a Borel set: We define

where each ${𝑑}_{𝑖}$ is one of the 16 hexadecimal digits $\left\{0,1,2,\dots ,9,𝐴,𝐵,𝐶,𝐷,𝐸,𝐹\right\}$. Then the set contains all real numbers between $\frac{{𝑑}_1{𝑑}_2\cdots {𝑑}_{𝑛}}{{16}^{𝑛}}$ and $\frac{{𝑑}_1{𝑑}_2\cdots {𝑑}_{𝑛}+1}{{16}^{𝑛}}$, so actually it is an interval:

Since it is an interval, it is a Borel set on $[0,1]$. And we define:

Then we have

So it suffices to prove that each ${𝐷}_{𝑁}$ is Borel set, since a countable union of Borel sets is Borel set.

Claim : any ${𝐷}_{𝑁}$ is a Borel set. To prove this, we fix an $𝑁$ and define for each $𝑛\ge 𝑁$

Then we have

Thus each ${𝐸}_{𝑛}$ is a Borel set since it is a finite union of Borel set, which shows that ${𝐷}_{𝑁}$ is Borel set, since

This finishes the proof that ${𝐵}^{𝑐}$ is a Borel set, and by a similar argument, $𝐶$ is a Borel set, and thus $𝐴=𝐵\cap 𝐶$ is a Borel set.

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Let

And we define

Since a Annuli defined by this $(𝑎,𝑏,𝑟,𝑅)$ is unique, this is a well-defined function; and since every admissible annulis can be defined by an element of ${\mathbb{Q}}^4$, this map is surjective. Therefore $\text{card}(𝐴)\le \text{card}({\mathbb{Q}}^4)$, so $𝐴$ is countable.

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Claim 1: every open set in ${ℝ}^2$ is a countable union of open balls, each centered at some $𝑞\in {\mathbb{Q}}^2$.
Proof for Claim 1:
Let $𝑈$ be an open set in ${ℝ}^2$. Define

By definition, every point in $𝑈$ have an open ball centered at it that is completely contained in $𝑈$, so we pick such ball ${𝐵}_{{𝑟}_{𝑥}}(𝑥)$ for each $𝑥\in 𝑈$. Since ${\mathbb{Q}}^2$ is dense in ${ℝ}^2$, for each $𝑥\in 𝑈$ and each corresponding ${𝑟}_{𝑥}$, we can find a rational point ${𝑞}_{𝑥}\in {\mathbb{Q}}^2$ such that $|{𝑞}_{𝑥}-𝑥|<\frac{{𝑟}_{𝑥}}{3}$. (Or more generally, as small as we wish.)

Let ${𝑟}_{{𝑞}_{𝑥}}>0$ be chosen so that ${𝑟}_{{𝑞}_{𝑥}}=\frac{{𝑟}_{𝑥}}{3},$ Then observe that $𝑥\in 𝐵({𝑞}_{𝑥},{𝑟}_{{𝑞}_{𝑥}})$

which follows from the triangle inequality.

For each $𝑞\in {\mathbb{Q}}_{𝑈}$, we define:

Now we have:

This is because for each each $𝑥\in 𝑈$, $𝑥\in {𝐵}_{{𝑟}_{{𝑞}_{𝑥}}}({𝑞}_{𝑥})\subset {𝐵}_{{𝑟}_{{𝑞}_{𝑥},𝑠𝑢𝑝}}({𝑞}_{𝑥})$

And we also have the other direction:

since every ${𝐵}_{{𝑟}_{𝑞}}(𝑞)$ is guaranteed to be the subset of some ball around some $𝑥\in 𝑈$. All togethe we have

This finishes the proof of claim 1.

Claim 2: every open ball centered at some $𝑞\in {\mathbb{Q}}^2$ is a countable union of admissible annulises with the same center, together with another admissible annulis whose center is also rational. Proof for Claim 2: Let $𝑞=(𝑎,𝑏)\in {\mathbb{Q}}^2$.
We have

-1, 这里写的略有问题, 因为 $𝑅$ 不一定是 rational 的, 不过我们可以用 density of $\mathbb{Q}$ in $ℝ$ 来写. It remains to cover the center. Let ${𝑞}^{\prime }≔({𝑎}^{\prime },{𝑏}^{\prime })\in {\mathbb{Q}}^2$ such that $𝑅/6<|{𝑞}^{\prime }-𝑞|<𝑅/3$, ${𝑟}^{\prime }≔𝑅/6$ and ${𝑅}^{\prime }≔𝑅/2$ . Then the annuli $𝐴({𝑎}^{\prime },{𝑏}^{\prime },{𝑟}^{\prime },{𝑅}^{\prime })$ defined by the four parameters is contained in the $𝐵(𝑞,𝑅)$ and it covers $\left\{𝑞\right\}$. Therefore

This finishes the proof of Claim 2.
Combining Claim 1 and Claim 2, we can conclude that every open subset of ${ℝ}^2$ is a countable union of admissible annuli.

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As defined,

Let

Every admissible annuli is open in ${ℝ}^2$, so

and since $\mathrm{ℬ︀}({ℝ}^2)$ is a $𝜎$-algebra, we have

by the proposition proved in class. And by (2), any open set is a countable union of admissible annulis, therefore every open set is in $<𝐴>$ since any countable union of sets in a $𝜎$-algebra is still in the set. So

This finishes the proof that

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这一部分的 proof 以及 remark 在 hw2. 这里, $\overline{𝑀}$ 称为 completion of $\mathrm{ℳ︀}$ with respect to $𝜇$, 以及 $\overline{𝜇}$ 称为 completion of $𝜇$.

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1. 取所有 ${𝐸}_{𝑗}=⌀$, 得到 ${𝜇}^{\ast }(⌀)=0$

2. monotonicity 显然, 因为如果 $𝐴\subseteq 𝐵$, 那么 $𝐴$ 取 inf 的这个集合是包含于 $𝐵$ 的, 因而取到的 inf 是小于等于的.

3. 证明 ctbl subadditivity, 我们使用经典的 $𝜖/2^{𝑖}$ argument. 这个 statement 直观上是显然的, 因为对一个 seq of sets, 每一个里面都有一个 seq of covering, 那么这个 seq of seq of covering 总体也是这个 seq union 的一个 covering. 不过我们不能这么说, 因为这里有一个 inf 操作的换序. 所以我们令 $𝜖>0$, 对于每个 ${𝐴}_{𝑖}$ 的 covering $({𝐸}_{𝑖,𝑘}{)}_{𝑘\in \mathbb{N}}$, 我们令 ${\sum }_{𝑘}𝜌({𝐸}_{𝑖,𝑘})\le {𝜇}^{\ast }({𝐴}_{𝑖})+𝜖/2^{𝑖}$, 最后可以得到 ${𝜇}^{\ast }({\bigcup }_{𝑖}{𝐴}_{𝑖})\le {\sum }_{𝑖}{𝜇}^{\ast }({𝐴}_{𝑖})$. 由于 $𝜖$ arbitrary, 得证.

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我们首先证明这个 $\mathrm{ℳ︀}$ 是一个 $𝜎$-algebra

1. $⌀\in \mathrm{ℳ︀}$ by def.

2. $\mathrm{ℳ︀}$ closed under complement, by def of ${𝜇}^{\ast }$-measurablity. (它对于 complement 是对称的.)

3. 为证明 $\mathrm{ℳ︀}$ closed under countable union, 我们首先 prove it for two sets. 假设 $𝐴,𝐵\in \mathrm{ℳ︀}$, 且 disjoint. Let $𝐸\subseteq 𝑋$. 我们已知

   $${𝜇}^{\ast }(𝐸)={𝜇}^{\ast }(𝐸\cap 𝐴)+{𝜇}^{\ast }(𝐸\cap {𝐴}^{𝑐})$$

   我们 WTS: ${𝜇}^{\ast }(𝐸)={𝜇}^{\ast }(𝐸\cap (𝐴\cup 𝐵))+{𝜇}^{\ast }(𝐸\cap (𝐴\cup 𝐵{)}^{𝑐})$
   我们对于 $𝐸\cap 𝐴$, $𝐸\cap {𝐴}^{𝑐}$ 可以得到:

   $${𝜇}^{\ast }(𝐸\cap 𝐴)={𝜇}^{\ast }(𝐸\cap 𝐴\cap 𝐵)+{𝜇}^{\ast }(𝐸\cap 𝐴\cap {𝐵}^{𝑐})$$

By $𝐴\cup 𝐵=(𝐴\backslash 𝐵)\bigsqcup (𝐴\cap 𝐵)\bigsqcup (𝐵\backslash 𝐴)$, 可以得到:

结合以上四个 equations 可以得到

又 $\le$ by countable subadditivity 成立, 我们得证 closed under two union (从而 inductively closed under any finite union, $\mathrm{ℳ︀}$ 因而是一个 algebra).

(Continuing the proof:) 现在我们再把这个 closed under finite union 推广到 closed under countable union, 以映证 $\mathrm{ℳ︀}$ 是一个 $𝜎$-algebra. 注意到 STS (suffices to show): $\mathrm{ℳ︀}$ closed under countable disjoint union. 因为任意不 disjoint 的两个集合都可以拆分成三个 disjoint 的集合.
我们令 $({𝐴}_{𝑖})$ 为一个 $\mathrm{ℳ︀}$ 中的 disjoint sequence, 并定义 ${𝐵}_{𝑛}≔{\bigcup }_{𝑖=1}^{𝑛}{𝐴}_{𝑖}$, 我们由上一步的结论知道, ${𝐵}_{𝑛}\in \mathrm{ℳ︀}$ for all $𝑛$. Define $𝐵≔{\bigcup }_{𝑖=1}^{\infty }{𝐴}_{𝑖}$, Let $𝐸\subseteq 𝑋$, WTS: ${𝜇}^{\ast }(𝐸)={𝜇}^{\ast }(𝐸\cap 𝐵)+{𝜇}^{\ast }(𝐸\cap {𝐵}^{𝑐})$.
考虑 ${𝜇}^{\ast }(𝐸\cap {𝐵}_{𝑛})={𝜇}^{\ast }(𝐸\cap {𝐵}_{𝑛}\cap {𝐴}_{𝑛})+{𝜇}^{\ast }(𝐸\cap {𝐵}_{𝑛}\cap {𝐴}_{𝑛}^{𝑐})={𝜇}^{\ast }(𝐸\cap {𝐴}_{𝑛})+{𝜇}^{\ast }(𝐸\cap {𝐵}_{𝑛-1})$, 因为 inductively 可得到:

从而：

by monotonicity (${𝜇}^{\ast }(𝐸\cap {𝐵}_{𝑛}^{𝑐})\ge {𝜇}^{\ast }(𝐸\cap {𝐵}^{𝑐})$), 这里是一个 infinite sum, 并且 true for every $𝑛$, 因而可以推广到 infinity, 得到

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这个 outer measure 的 construction directly follows from Theorem 2.5.
Proof that ${𝜇}^{\ast }$ restricted to ${\mathrm{𝒜︀}}_0$ is ${𝜇}_0$: 令 $𝐸\in {\mathrm{𝒜︀}}_0$, 假设 $𝐸\subseteq {\bigcup }_{𝑖=1}^{\infty }{𝐴}_{𝑖}$, 我们令 ${𝐵}_{𝑛}≔𝐸\cap ({𝐴}_{𝑛}\backslash {\bigcup }_{𝑖=1}^{𝑛-1}{𝐴}_{𝑖})$, 即把 covering intersecting $𝐸$ 变成 disjoint covering $({𝐵}_{𝑛})$, 从而由 ${𝜇}_0$ 的 ctbl disjoint additivity 可得, 这一个新 covering 的 measure sum ${\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐵}_{𝑖})≔{𝜇}_0(𝐸)$. 并且由于 ${\mathrm{𝒜︀}}_0$ 是一个 algebra, 这些 ${𝐵}_{𝑛}$ 也在 ${\mathrm{𝒜︀}}_0$ 里面, 从而它满足 monotonicty, then ${𝜇}_0(𝐸)={\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐵}_{𝑖})\le {\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐴}_{𝑖})$
Proof that every set in ${\mathrm{𝒜︀}}_0$ is ${𝜇}^{\ast }$-measurable: Fix $𝐴\in {\mathrm{𝒜︀}}_0$, 我们取任意 $𝐸\subseteq 𝑋$. Let $𝜖>0$, by def of the outer measure, 存在一个 seq ${\left\{{𝐵}_{𝑖}\right\}}_{𝑖=1}^{\infty }\subseteq {\mathrm{𝒜︀}}_0$, 使得 $𝐸\subseteq {\bigcup }_{𝑖=1}^{\infty }{𝐵}_{𝑖}$ 并且 ${\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐵}_{𝑖})\le {𝜇}^{\ast }(𝐸)+𝜖$. 有 disjoint additivity of ${𝜇}_0$ 可得, ${\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐵}_{𝑖})={\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐵}_{𝑖}\cap 𝐴)+{\sum }_{𝑖=1}^{\infty }{𝜇}_0({𝐵}_{𝑖}\cap {𝐴}^{𝑐})$. 从而 ${𝜇}^{\ast }(𝐸)\ge {𝜇}^{\ast }(𝐸\cap 𝐴)+{𝜇}^{\ast }(𝐸\cap {𝐴}^{𝑐})$, 得证. (实际上这是个 trivial argument, 通过$𝜖$ argument 来严格证明.)

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Proof of $(𝑋,{\mathrm{𝒜︀}}_0,{𝜇}_0)$ extends to $(𝑋,\mathrm{ℳ︀},𝜇)$:
这个 Statement directly follows from Theorem 2.6(Carathéodory’s Theorem) 以及上一个 proposition Proposition 2.1.
. 我们首先用 ${𝜇}_0$ induce 出 ${𝜇}^{\ast }$, 再 restrict ${𝜇}^{\ast }$ to ${\mathrm{ℳ︀}}^{\ast }≔\left\{\text{all}{𝜇}^{\ast }\text{-measurable sets}\right\}$, 得到一个 $𝜎$-algebra ${\mathrm{ℳ︀}}^{\ast }$.
注意此时: 由上一个 proposition Proposition 2.1 可得 ${\mathrm{ℳ︀}}_0$ 中所有集合都是 ${𝜇}^{\ast }$-measurable 的, thus ${𝑀}_0\subseteq {\mathrm{ℳ︀}}^{\ast }$, 由于 ${\mathrm{ℳ︀}}^{\ast }$ 是一个 $𝜎$-algebra, 由 Lemma 2.2 可得: $\mathrm{ℳ︀}≔<{\mathrm{ℳ︀}}_0>\subseteq {\mathrm{ℳ︀}}^{\ast }$.
. 由 Carathéodory’s Theorem 可以得到: ${𝜇}^{\ast }{|}_{{\mathrm{ℳ︀}}^{\ast }}$ 是一个 measure, 从而 $𝜇≔{𝜇}^{\ast }{|}_{\mathrm{ℳ︀}}$ 也是一个 measure(等于把 ${𝜇}^{\ast }{|}_{{\mathrm{ℳ︀}}^{\ast }}$ 限制在了一个更小的 sub-$𝜎$-algebra 上).
(Note: this is a trivial fact that if ${𝑀}^{\ast }$ is a $𝜎$-algebra and $𝑀\subset {𝑀}^{\ast }$is also a $𝜎$-algebra, then $𝜇{|}_{𝑀}$ is a measure if given that $𝜇$ is a $𝜎$-algebra on ${𝑀}^{\ast }$)

Proof of $𝜇$ being the largest extension of ${𝜇}_0$ to $\mathrm{ℳ︀}$: 假设 $𝜈$ 是一个 $\mathrm{ℳ︀}$ 上的 $𝜎$-algebra s.t. $𝜈{|}_{{\mathrm{ℳ︀}}_0}={𝜇}_0$.
Let $𝐸\subseteq \mathrm{ℳ︀}$. (WTS: $𝜈(𝐸)\le 𝜇(𝐸)$, 即$𝜈(𝐸)\le {𝜇}^{\ast }(𝐸)$ .)
由外测度 ${𝜇}^{\ast }$ 的定义, 对于任意 $𝜖>0$, 存在一列集合 ${\left\{{𝐴}_{𝑖}\right\}}_{𝑖=1}^{\infty }\subset {\mathrm{𝒜︀}}_0$ 满足

由于 $𝜈$ 在 ${\mathrm{𝒜︀}}_0$ 上和 ${𝜇}_0$ 一致，即

因此，

利用 $𝜈$ 的 additivity 和 monotoncity 得

由于 $𝜖$ arbitrary, 得到

(证明思路: 在 $\mathrm{ℳ︀}$ 上 $𝜇$ 就等于 ${𝜇}_0$ induce 的外测度, 对于其他的 extended measure, 其作用在一个集合上的测度一定小于等于任意的 ${\mathrm{ℳ︀}}_0$ covering 的 premeasure 和, 而我们可以通过控制这个 covering 的测度和与它的外测度的差距(since inf), 从而使得这个测度小于等它的外测度加一个无限小的 $𝜖$, 从而得证.)

Proof of $𝜇$ being the unique extension of ${𝜇}_0$ to $\mathrm{ℳ︀}$, provided that ${𝜇}_0$ is $𝜎$-finite:
(recall ${𝜇}_0$ is $𝜎$-finite 即 ${𝜇}_0(𝑋)<\infty$) It remains to show that $𝜈(𝐸)\ge {𝜇}^{\ast }(𝐸)$.

Continuing 上一个 proof, we have:

我们只要 controling ${𝜇}^{\ast }({\bigcup }_{𝑖=1}^{\infty }{𝐴}_{𝑖}\backslash 𝐸)={𝜇}^{\ast }({\bigcup }_{𝑖=1}^{\infty }{𝐴}_{𝑖})-{𝜇}^{\ast }(𝐸)=𝜖$ 逼近 0, 即可得到反向的不等式关系.
(证明思路: 我们证明了 $𝜈(𝐸)\le {𝜇}^{\ast }(𝐸)$ 之后, 注意到 covering set 和 $𝐸$ 之间的差集的 $𝜈$-measure 自然也小于等于这个差集的 ${𝜇}^{\ast }$-measure, which can approximate 0.)

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略.(嘻嘻)

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