measurable sets form a sigma algebra

[[concept]]

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Theorem

The collection of measurable sets $\mathcal{M}$ is a sigma-algebra

NOTE

We want measurable sets to form a sigma algebra based on the desirable properties for measure:

Properties we want for this idea

1. $m(E)$ is defined for all $E\subset \mathbb{R}$

2. if $I$ is an interval then $m(E)=\ell (I)$ the “length” of $I$

3. If $\{E_n\}$ is a (countable) collection of disjoint subsets of $E$ such that $E={\bigcup }_n{E}_n$, then we want $m\left({\bigcup }_n{E}_n\right)=m(E)$

4. Translation invariance. ie, if $E\subset \mathbb{R}$ and $x\in \mathbb{R}$, then $m(x+E)=m(\{x+y|y\in E\})=m(E)$

In particular, in order to get (3), we need to be able to define measure on any countable disjoint union.

Proof

We already know that $\mathcal{M}$ is an algebra, and we know since algebras have closure under finite disjoint countable unions that we just need to check countable disjoint unions are measurable.

So let $\{E_n\}$ be a countable collection of disjoint measurable sets with $E={\bigcup }_n{E}_n$. Then since the other half is always satisfied via monotonicity, we just need to show $m^{\ast }(A\cap E^c)+m^{\ast }(A\cap E)\le m^{\ast }(A)$

Let $N\in \mathbb{N}$. Since $\mathcal{M}$ is an algebra, the union ${\bigcup }_{n=1}^N{E}_n\in \mathcal{M}$ so we have $m^{\ast }(A)=m^{\ast }\left(A\cap \left[{\bigcup }_{n=1}^N{E}_n\right]\right)+m^{\ast }\left(A\cap {\left[{\bigcup }_{n=1}^N{E}_n\right]}^c\right)$ And since ${\bigcup }_{n=1}^N{E}_n\subset E$, we have $E^c\subset {\left[{\bigcup }_{n=1}^N{E}_n\right]}^c$. Thus we have

m^*(A) &= m^*\left( A \cap \left[ \bigcup_{n=1}^N E_{n} \right] \right)+m^*\left( A \cap \left[ \bigcup_{n=1}^N E_{n}\right]^c \right) \\ &\geq m^*\left( A \cap \left[ \bigcup_{n=1}^N E_{n} \right] \right) + m^*(A \cap E^c) \end{align}$$ And since the [[Concept Wiki/measure of finite disjoint measurable sets is the sum of the measures]], we get $$\begin{align} m^*(A) &\geq \sum_{n=1}^N m^*(A \cap E_{n}) + m^*(A \cap E^c) \\ (N\to \infty )&\geq \sum_{n=1}^\infty m^*(A \cap E_{n}) + m^*(A \cap E^c) \\ (*) &\geq m^*\left( \bigcup_{n} A \cap E_{n} \right) + m^*(A \cap E^c) \\ &= m^*(A \cap E) + m^*(A \cap E^c) \end{align}$$ Where $(*)$ is from [[Concept Wiki/outer measure has countable subadditivity\|countable subadditivity]]. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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