measure of finite disjoint measurable sets is the sum of the measures

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Theorem

Let $A \subset R$ and let $E_{1}, \dots, E_{n}$ be disjoint and measurable. Then

m^{*} (A \cap [⋃_{k = 1}^{n} E_{k}]) = \sum_{k = 1}^{n} m^{*} (A \cap E_{k})

Proof

Via induction. Trivially true for $n = 1$ . Suppose that the equality is true for $n = m$ . Now, suppose we have pairwise disjoint measurable sets $E_{1}, \dots, E_{m + 1}$ and $A \subset R$ . Since $E_{k} \cap E_{m + 1} = \emptyset$ for all $k$ , we have (since $E_{m + 1}$ is measurable ) that

\begin{aligned} A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1} & = A \cap E_{m + 1} \\ and A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1}^{c} & = A \cap [⋃_{k = 1}^{m} E_{k}] \\ ⟹ m^{*} (A \cap [⋃_{k = 1}^{m + 1} E_{k}]) & = m^{*} (A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1}) + m^{*} (A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1}^{c}) \\ = m^{*} (A \cap E_{m + 1}) + m^{*} (A \cap [⋃_{k = 1}^{m} E_{k}]) \\ = m^{*} (A \cap E_{m + 1}) + \sum_{k = 1}^{m} m^{*} (A \cap E_{k}) \end{aligned}

Where the last inequality is due to the induction hypothesis. Thus the result follows via induction.

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File	Last Modified
Functional Analysis Lecture 8	2025-07-14
measurable sets form a sigma algebra	2025-07-08
measure of union of nested sets converges to measure of limiting set	2025-07-08
measure satisfies countable additivity	2025-07-08

File

Last Modified

Functional Analysis Lecture 8

2025-07-14

measurable sets form a sigma algebra

2025-07-08

measure of union of nested sets converges to measure of limiting set

2025-07-08

measure satisfies countable additivity

2025-07-08

measure of finite disjoint measurable sets is the sum of the measures

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