Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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The measure of a set liminf is at most the liminf of the measures

Statement

For every sequence (Ek) of measurable sets,

μ(lim infkEk)lim infkμ(Ek),

where the numerical liminf is taken in [0,+] as in Limit superior and limit inferior of a nonnegative extended-real sequence.

Facts & Assumptions

Given: A measure μ and a sequence (Ek) of measurable sets.

[L1]

For increasing measurable sets, the measure of the union is the supremum of their measures (Continuity from below for measures).

[L2]

Measures are monotone under inclusion (Measures are monotone).

[L3]

The set liminf is NkNEk (Limit superior and limit inferior of a sequence of sets).

[L4]

For a nonnegative extended sequence (ak), lim infkak=supNinfkNak (Limit superior and limit inferior of a nonnegative extended-real sequence), and all these bounds exist (Every subset of R has a least upper bound and a greatest lower bound in R, agreeing with the real supremum and infimum on nonempty sets bounded in R).

Proof

technique · direct
1.1

Put FN:=kNEk. Then FNFN+1 and NFN=lim infkEk.

givenL3
1.2

For every N and every kN, FNEk, so μ(FN)μ(Ek) and hence μ(FN)infkNμ(Ek).

givenL2L4
2.1

Continuity from below and step 1.2 give μ(lim infkEk)=supNμ(FN)supNinfkNμ(Ek)=lim infkμ(Ek).

step 1.1step 1.2L1L4

Depends on

Used by

Dependency tree · two levels

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Sources