measurable sets form a sigma algebra

Properties we want for this idea

$m (E)$ is defined for all $E \subset R$
if $I$ is an interval then $m (E) = ℓ (I)$ the "length" of $I$
If ${E_{n}}$ is a (countable) collection of disjoint subsets of $E$ such that $E = ⋃_{n} E_{n}$ , then we want $m (⋃_{n} E_{n}) = m (E)$
Translation invariance. ie, if $E \subset R$ and $x \in R$ , then $m (x + E) = m ({x + y | y \in E}) = m (E)$

Proof

We already know that $M$ 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 $E = ⋃_{n} E_{n}$ . Then since the other half is always satisfied via monotonicity, we just need to show $m^{*} (A \cap E^{c}) + m^{*} (A \cap E) \leq m^{*} (A)$

Let $N \in N$ . Since $M$ is an algebra, the union $⋃_{n = 1}^{N} E_{n} \in M$ so we have

m^{*} (A) = m^{*} (A \cap [⋃_{n = 1}^{N} E_{n}]) + m^{*} (A \cap {[⋃_{n = 1}^{N} E_{n}]}^{c})

And since $⋃_{n = 1}^{N} E_{n} \subset E$ , we have $E^{c} \subset {[⋃_{n = 1}^{N} E_{n}]}^{c}$ . Thus we have

\begin{aligned} m^{*} (A) & = m^{*} (A \cap [⋃_{n = 1}^{N} E_{n}]) + m^{*} (A \cap {[⋃_{n = 1}^{N} E_{n}]}^{c}) \\ \geq m^{*} (A \cap [⋃_{n = 1}^{N} E_{n}]) + m^{*} (A \cap E^{c}) \end{aligned}

And since the measure of finite disjoint measurable sets is the sum of the measures, we get

\begin{aligned} 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^{*} (⋃_{n} A \cap E_{n}) + m^{*} (A \cap E^{c}) \\ = m^{*} (A \cap E) + m^{*} (A \cap E^{c}) \end{aligned}

Where $(*)$ is from countable subadditivity.

\begin{matrix} ◼ \end{matrix}

References

See Also

Mentions

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