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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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On a complete measure space, equality almost everywhere preserves measurability

Statement

Let (X,A,μ) be a complete measure space, let f:XR be measurable, and let g:XR satisfy g=f almost everywhere. Then g is measurable.

Facts & Assumptions

Given: A complete measure space (X,A,μ), a measurable function f:XR, a function g:XR, and a measurable null set N such that f=g on XN.

[L1]

In a complete measure space, every subset of a measurable null set is measurable and null. (Null sets are closed under countable unions and, in a complete space, under arbitrary subsets)

[L2]

Threshold measurability characterizes measurable R-valued functions. (Threshold characterisations of real-valued and extended-real-valued measurability)

Proof

technique · direct
1.1

Fix a real a. On XN, the equality g=f gives

givenalgebra

{g>a}(XN)={f>a}(XN).

Therefore

{g>a}=({f>a}(XN))({g>a}N).

[given, algebra]

2.1

The set {f>a} is measurable by [L2]. The set {g>a}N is a [step 1.1, L1, L2] subset of the measurable null set N, so [L1] makes it measurable. Hence {g>a} is measurable for every real a.

step 1.1L1L2
3.1

By [L2], step 2.1 proves that g is measurable.

step 2.1L2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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