\begin{proof} We first begin with some definitions and lemmas:

Step 1: prove For $A_{1}\in\mathcal{M},A_{2}\subseteq\Omega$, such that $A_{1},A_{2}$ are disjoint $$\begin{align*} \mathbb{P}^{*}(A_{1}\cap A_{2})&= \mathbb{P}^{*}(A_{1}\cap(A_{1}\cup A_{2}))+\mathbb{P}^{*}(A_{1}^{c}\cap(A_{1}\cup A_{2}))\\ &= \mathbb{P}^{*}(A_{1})+\mathbb{P}^{*}(A_{2}) \end{align*}$$Where the first equality is done by applying the definition of and the second by disjoint property of both events. By induction, for $A_{1},\dots,A_{n}\in\mathcal{M}$ disjoint (by and finite superadditivity) $$\mathbb{P}^{*}(A_{1}\cup\dots \cup A_{n})=\sum_{i=1}^{n}\mathbb{P}^{*}(A_{i})$$Thus $$\mathbb{P}^{*}\left( \bigcup_{i\ge 1}A_{i} \right)\ge \mathbb{P}^{*}\left( \bigcup_{i=1}^{n}A_{i} \right)=\sum_{i=1}^{n}\mathbb{P}^{*}(A_{i})\quad\forall n\ge 1$$ $$\implies \mathbb{P}^{*}(\cup_{i}A_{i})\ge \sum_{i\ge 1} \mathbb{P}^{*}(A_{i})$$But, by we have that $$\mathbb{P}^{*}(\cup_{i}A_{i})=\sum_{i\ge 1}\mathbb{P}^{*}(A_{i})$$ Step 2: Show $\mathcal{M}$ is a σ-algebra where $$\mathcal{M}:=\{ A\subseteq\Omega:\mathbb{P}^{*}(E)=\mathbb{P}^{*}(E\cap A)+\mathbb{P}^{*}(E\cap A^{c})\quad\forall E\subseteq\Omega \}$$ a) Show $\mathcal{M}$ is a Algebra

1. $\emptyset,\Omega \in\mathcal{M}$ ✅
2. $A\in\mathcal{M}\implies A^{c}\in\mathcal{M}$ ✅
3. Let $A,B\in\mathcal{M}$, $E\subseteq\Omega$, then $$\begin{align*} &\mathbb{P}^{*}((A\cap B)\cap E)+\mathbb{P}^{*}((A\cap B)^{c}\cap E)\\ &= \mathbb{P}^{*}(A\cap B\cap E)+\mathbb{P}^{*}((A^{c}\cap B\cap E)\cup(A\cap B^{c}\cap E)\cup(A^{c}\cap B^{c}\cap E))\\ &\le \mathbb{P}^{*}(A\cap B\cap E)+\mathbb{P}^{*}(A^{c}\cap B\cap E)+\mathbb{P}^{*}(A\cap B^{c}\cap E)+\mathbb{P}^{c}(A^{c}\cap B^{c}\cap E)\\ &= \mathbb{P}^{*}(B\cap E)+\mathbb{P}^{*}(B^{c}\cap E)\\ &= \mathbb{P}^{*}(E) \end{align*}$$ where the last two arguments are because $A,B\in\mathcal{M}$. Since this is the the one side of the inequality we wanted to prove (the other is trivial) we have that $A\cap B\in\mathcal{M}$.

b) Lemmas… We now state some additional lemmas: >[!lemma|2.3.11] >Let $A_{1},A_{2},\dots \in\mathcal{M}$ be disjoint. For each $n\in\mathbb{N}$, let $B_{n}=\bigcup_{i=1}^{n}A_{i}$. Then $\forall n\in\mathbb{N},\forall E\subseteq\Omega$ we have $$\mathbb{P}^{*}(E\cap B_{n})=\sum_{i=1}^{n}\mathbb{P}^{*}(E\cap A_{i})$$

c) $\mathcal{M}$ is a σ-algebra: >[!lemma|2.3.14] >$\mathcal{M}$ is a σ-algebra

\begin{proof} Using we just use the fact that for any $A_{1},A_{2},\dots \in\mathcal{M}$ we have that $$\bigcup_{i}A_{i}=\underbrace{ A_{1}\cup(A_{2}\setminus A_{1})\cup(A_{3}\setminus(A_{1}\cup A_{2}))\cup\dots }_{ \text{disjoint} }\in\mathcal{M}$$ \end{proof}

Step 3: $\mathcal{J}\subseteq \mathcal{M}$:

\begin{proof} Let $A\in \mathcal{J}$. Since $\mathcal{J}$ is a semialgebra we can write $A^{c}=J_{1}\sqcup\dots\sqcup J_{k}$ for some disjoint $J_{1},\dots,J_{k}\in \mathcal{J}$. Also, for any $E\subseteq\Omega$ and $\epsilon>0$ by the definition of : we can find $A_{1},A_{2},\dots \in \mathcal{J}$ with $E\subseteq \bigcup_{n}A_{n}$ and $\sum_{n}\mathbb{P}(A_{n})\le\mathbb{P}^{*}(E)+\epsilon$. Then

$$\begin{align*} &\mathbb{P}^{*}(E\cap A)+\mathbb{P}^{*}(E\cap A^{c})\\ &\le \mathbb{P}^{*}\left( \left( \bigcup_{n}A_{n} \right)\cap A \right)+\mathbb{P}^{*}\left( \left( \bigcup_{n}A_{n} \right)\cap A^{c} \right)&\text{monotonicity}\\ &= \mathbb{P}^{*}\left( \bigcup_{n}(A_{n}\cap A) \right)+\mathbb{P}^{*}\left( \bigcup_{n}\bigcup_{i=1}^{k}(A_{n}\cap J_{i}) \right)&\text{definition of semialgebra}\\ &\le \sum_{n}\mathbb{P}^{*}(A_{n}\cap A) + \sum_{n}\sum_{i=1}^{k}\mathbb{P}^{*}(A_{n}\cap J_{i})&\text{subadditivity}\\ &= \sum_{n}\mathbb{P}(A_{n}\cap A)+\sum_{n}\sum_{i=1}^{k}\mathbb{P}(A_{n}\cap J_{i})&\text{since }\mathbb{P}^{*}|_{\mathcal{J}}=\mathbb{P}\\ &= \sum_{n} \left(\mathbb{P}(A_{n}\cap A)+\sum_{i=1}^{k}\mathbb{P}(A_{n}\cap J_{i})\right)\\ &\le \sum_{n}\mathbb{P}(A_{n})&\text{superadditivity then Carathéodory}\\ &\le \mathbb{P}^{*}(E)+\epsilon&\text{by assumption}. \end{align*}$$ This is true $\forall\epsilon>0$, hence since $\epsilon$ is arbitrary, $\mathbb{P}^{*}(E\cap A)+\mathbb{P}^{*}(E\cap A^{c})\le\mathbb{P}^{*}(E),\forall E\subseteq\Omega$. Since the other direction is trivial we have that the condition holds giving us $A\in \mathcal{M}$, since this is for any $A\in \mathcal{J}$ we have $\mathcal{J}\subseteq \mathcal{M}$. \end{proof} All these lemmas prove what we needed to prove ✅

\end{proof}

Extensions of the Extension Theorem