Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-27
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Equality almost everywhere with a measurable function can fail on an incomplete space

Statement refuted

That equality almost everywhere with a measurable function forces measurability even on an incomplete measure space.

Facts & Assumptions

Given: The Axiom of Choice and the Borel measure space (R,B(R),λ∣B(R)), the Cantor set C, and the homeomorphism ψ(x)=x+c(x) from [0,1] onto [0,2].

Counterexample

technique · direct
1.1L1

Let f=0 and g=1E. The set C is Borel null, so g=f on [L1] R∖C and hence g=f almost everywhere.

2.1step 1.1L1∎

But [step 1.1, L1] g−1((1/2,∞))=E, and [L1] says E is not Borel. Thus g is not measurable for the Borel sigma-algebra, even though it agrees almost everywhere with the measurable function f.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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