TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Continuity from above when one set has finite measure

Statement

Let $(E_{n})_{n \in N}$ be a decreasing sequence of measurable sets for a measure $μ$ . If $μ (E_{n_{0}}) < + \infty$ for some $n_{0}$ , then

$μ (⋂_{n \in N} E_{n}) = \inf_{n \in N} μ (E_{n}) .$

Facts & Assumptions

Given: Measurable sets $E_{0} \supseteq E_{1} \supseteq \dots$ , an index $n_{0}$ with $μ (E_{n_{0}}) < + \infty$ , and $E = ⋂_{n} E_{n}$ .

[L1]

For increasing measurable $A_{n}$ , $μ (⋃_{n} A_{n}) = \sup_{n} μ (A_{n})$ (Continuity from below for measures).

[L2]

If $A \subseteq B$ and $μ (A) < + \infty$ , then $μ (B) = μ (A) + μ (B ∖ A)$ , with real subtraction valid when $μ (B) < + \infty$ (Measure of a set difference when the smaller set has finite measure).

[L3]

Every subset of $\overline{R}$ has an infimum 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

1.1given

For $k \in N$ put $H_{k} : = E_{n_{0}} ∖ E_{n_{0} + k}$ . Then $(H_{k})$ increases and $⋃_{k} H_{k} = E_{n_{0}} ∖ E$ .

1.2givenL2

Every $E_{n_{0} + k}$ and $E$ has finite measure by inclusion in $E_{n_{0}}$ , and [L2] gives $μ (H_{k}) = μ (E_{n_{0}}) - μ (E_{n_{0} + k})$ and $μ (E_{n_{0}} ∖ E) = μ (E_{n_{0}}) - μ (E)$ .

2.1step 1.1step 1.2L1L3algebra

Apply continuity from below to $(H_{k})$ and substitute step 1.2: taking the supremum of the left differences is the same as subtracting the infimum of the decreasing finite values, so cancellation of the finite number $μ (E_{n_{0}})$ yields $μ (E) = \inf_{k} μ (E_{n_{0} + k})$ .

3.1step 2.1L3∎

A decreasing sequence has the same infimum as any of its tails, so step 2.1 gives $μ (E) = \inf_{n} μ (E_{n})$ ; this includes $E = \emptyset$ , $μ (E) = 0$ , and a sequence that is constant from $n_{0}$ onward.

Depends on

Dependency tree · 2 levels

Sources

S. Axler, Measure, Integration & Real Analysis, Theorem 2.60