Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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The modulus of an integral is bounded by the integral of the modulus

Statement

If fL1(μ), then fdμfdμ.

Facts & Assumptions

Given: An integrable function f.

[L1]

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

[L2]

Real and imaginary parts, complex conjugation, and modulus are as in Real and imaginary parts, complex conjugation, and modulus.

[L3]

Real and complex integrability are defined in Integrable real and complex functions, and their integrals.

Proof

technique · direct
1.1

For real-valued f, the functions f+f=2f+ and ff=2f are nonnegative. Therefore [L1] and [L4] give 0(f+f)dμ=fdμ+fdμ, 0(ff)dμ=fdμfdμ. So fdμfdμfdμ, and hence fdμfdμ.

L1L3L4algebra
2.1

For complex-valued f, let I:=fdμ. If I=0 there is nothing to prove. Otherwise set α:=I/I, so α=1 by [L2]. Then I=αI=αfdμ, and Re(αf)αf=f, so step 1.1 applies to the integrable real-valued function Re(αf). Taking real parts gives I=Re(αf)dμαfdμ=fdμ, because Rezz for every complex z.

L1L2step 1.1L4algebra

Depends on

Used by

Dependency tree · two levels

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Sources