TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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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

1.1

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

given

1.2

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 $μ (E_{n_{0}} ∖ E) = μ (E_{n_{0}}) - μ (E)$ .

givenL2

2.1

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 $μ (E) = \inf_{k} μ (E_{n_{0} + k})$ .

step 1.1step 1.2L1L3algebra

3.1

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.

step 2.1L3∎

Depends on

Used by

Dependency tree · two levels

14 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 2.60 (standard reference, not scraped)