outer measure has countable subadditivity

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Proof

If there is any $n$ such that $m^{*} (A_{n}) = \infty$ (ie, if any of the $A_{n}$ are unbounded) or $\sum_{n} m^{*} (A_{n}) = \infty$ then the inequality is true.

So consider only when all $m^{*} (A_{n}) < \infty$ and $\sum_{n} m^{*} (A_{n}) < \infty$ .

So let ${I_{n_{k}}}_{k \in N}$ be a collection of open intervals with

\begin{aligned} A_{n} & \subset ⋃_{k \in N} I_{n_{k}} \\ \sum_{k = 1}^{\infty} ℓ (I_{n_{k}}) & < m^{*} (A_{n}) + \frac{ϵ}{2^{n}} \end{aligned}

Then we have

\begin{aligned} ⋃_{n \in N} A_{n} & \subset ⋃_{n \in N, k \in N} I_{n_{k}} \\ ⟹ m^{*} (⋃_{n} A_{n}) & \leq \sum_{n, k} ℓ (I_{n_{k}}) \\ = \sum_{n} \sum_{k} ℓ (I_{n_{k}}) \\ < \sum_{n} m^{*} (A_{n}) + \frac{ϵ}{2^{n}} \\ = [\sum_{n} m^{*} (A_{n})] + ϵ \end{aligned}

So letting $ϵ \to 0$ , we get the desired result.

\begin{matrix} ◼ \end{matrix}

File	Last Modified
Functional Analysis Lecture 8	2025-07-14
Functional Analysis Lecture 6	2025-07-08
measurable sets form a sigma algebra	2025-07-08
measure satisfies countable additivity	2025-07-08
we can always find an open set with outer measure slightly more	2025-07-01

File

Last Modified

Functional Analysis Lecture 8

2025-07-14

Functional Analysis Lecture 6

2025-07-08

measurable sets form a sigma algebra

2025-07-08

measure satisfies countable additivity

2025-07-08

we can always find an open set with outer measure slightly more

2025-07-01

outer measure has countable subadditivity

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