Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree

Statement

Let f,gL1(μ). Then the following are equivalent:

  1. f=g almost everywhere;
  2. for every measurable A, Afdμ=Agdμ.

For integrable real or complex h, the notation in condition 2 means Ahdμ:=hχAdμ; the product is integrable because hχAh.

Facts & Assumptions

Given: Integrable functions f,gL1(μ).

[L1]

The integral over a null set vanishes for nonnegative integrands (A nonnegative integral over a null set vanishes).

[L2]

The Lebesgue integral is linear on L1(μ) (The Lebesgue integral is linear on L1(μ)).

[L3]

A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere).

[L4]

Real and imaginary parts of an integrable complex function are integrable (Integrable real and complex functions, and their integrals).

Proof

technique · direct
1.1

Assume f=g almost everywhere, with exceptional null set N. Then for [L1, L2, L4, given] every measurable A, the real and imaginary parts of (fg)χA are supported on N, so [L1] and [L2] give A(fg)dμ=0, hence Afdμ=Agdμ.

1.2

Assume instead that Afdμ=Agdμ for every measurable [L2, L3, L4, given] A. Apply this to the real part u:=Re(fg) on the set A+:={u>0} and to u on A:={u<0}. In each case the corresponding nonnegative integral is 0, so [L3] gives u=0 almost everywhere. The same argument for v:=Im(fg) shows v=0 almost everywhere. Hence f=g almost everywhere.

2.1

Step 1.1 proves (1)(2) and step 1.2 proves (2)(1).

step 1.1step 1.2

Depends on

Used by

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Sources