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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.