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Suppose that we defined $\mathcal{M}_{\mu^{*}}=\left\{B:B \ \ \mu^{*}-\text{measurable}\right\}$ as the set of all $\mu^{*}$ measurable sets.A set is measurable with respect to outter measure $\mu^{*}$ if for every $A\in\mathcal{P}(X)$ we have $\mu^{*}(A)=\mu^{*}(A\cap B)+\mu^{*}(A\cap B^{c})$ then $B\in\mathcal{M}_{\mu^{*}}$

Outter measure is defined as

$\mu^{*}:\mathcal{P}(X)\rightarrow (0,\infty]$

$A\subset B \Rightarrow \mu^{*}(A)\leq \mu^{*}(B)$

for every $(A_{n})_{n\mathbb{N}}$ , $\mu^{*}(\cup_{n=1}^{\infty}A_{n})\leq \sum_{n=1}^{\infty}\mu^{*}(A_{n})$

$\mu^{*}(\varnothing )=0$

I proved that $M_{\mu^{*}}$ is algebra and based on this result I found the following proposition:

An algebra is sigma-algebra if for every $(A_{n})_{n\mathbb{N}}$ (where $A_{n}$ disjoint pairwise) ,$\cup_{n=1}^{\infty}A_{n}$ belongs in algebra.

It seems rational that we want to have the infinite countable union of $A_{n}$ in algebra because that's the definition of sigma algebra.But I'm confused on how to prove this proposition.

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Let $\mathcal A$ be an algebra that is closed under the formation of countable unions of sets in $\mathcal A$ that are mutually disjoint.

For $n=1,2,\dots$ let $B_n\in\mathcal A$.

Then $A_1:=B_1\in\mathcal A$ and for every $n\geq2$ also $A_n:=B_n\cap\left(\bigcup_{k=1}^{n-1}B_k\right)^{\complement}\in\mathcal A$.

This because $\mathcal A$ is closed under the formation of finite unions and complements.

Observe that the sets $A_n$ are mutually disjoint so that: $$\bigcup_{n=1}^{\infty}B_n=\bigcup_{n=1}^{\infty}A_n\in\mathcal A$$

So $\mathcal A$ is an algebra closed under the formation of countable unions, hence is a $\sigma$-algebra.

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