TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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Continuity from below for measures

Statement

Let $(E_{n})_{n \in N}$ be an increasing sequence of measurable sets for a measure $μ$ , so $E_{n} \subseteq E_{n + 1}$ . Then

$μ (⋃_{n \in N} E_{n}) = \sup_{n \in N} μ (E_{n}) .$

No finiteness hypothesis is required.

Facts & Assumptions

Given: A measure $μ$ and measurable sets $E_{0} \subseteq E_{1} \subseteq \dots$ ; write $E = ⋃_{n} E_{n}$ .

[L1]

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

If $A \subseteq B$ are measurable and $μ (A) < + \infty$ , then $μ (B) = μ (A) + μ (B ∖ A)$ , with the corresponding finite and infinite cases stated explicitly (Measure of a set difference when the smaller set has finite measure).

[L3]

A nonnegative extended series is the supremum of its finite partial sums (Series in the nonnegative extended real line).

[L4]

Every subset of $\overline{R}$ has a supremum there (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

Proof

technique · direct

1.1

Define $D_{0} : = E_{0}$ and $D_{k + 1} : = E_{k + 1} ∖ E_{k}$ . The sets $D_{k}$ are measurable and pairwise disjoint, $E_{n} = ⋃_{k < n + 1} D_{k}$ , and $E = ⋃_{k} D_{k}$ .

given

2.1

Countable additivity gives $μ (E) = \sum_{k} μ (D_{k})$ .

step 1.1L1

2.2

If $μ (E_{n}) = + \infty$ for some $n$ , then $μ (E) = + \infty$ by step 1.1 and [L1], while $\sup_{k} μ (E_{k}) = + \infty$ because that value occurs.

step 1.1L1L4

2.3

If every $μ (E_{n})$ is finite, then [L2] gives $μ (D_{0}) = μ (E_{0})$ and $μ (D_{k + 1}) = μ (E_{k + 1}) - μ (E_{k})$ ; hence the first $n + 1$ terms telescope to $μ (E_{n})$ .

step 1.1L2algebra

3.1

In the finite-valued case, [L3], steps 2.1 and 2.3 give $μ (E) = \sup_{n} μ (E_{n})$ ; step 2.2 gives the same equality in the remaining case, including $E_{0} = \emptyset$ and a sequence that stabilizes.

step 2.1step 2.2step 2.3L3∎

Depends on

Used by

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Axler, Measure, Integration & Real Analysis, Theorem 2.59 (standard reference, not scraped)