TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27

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The modulus of an integral is bounded by the integral of the modulus

Statement

If $f \in L^{1} (μ)$ , then $∣ \int f d μ ∣ \leq \int ∣ f ∣ d μ .$

Facts & Assumptions

Given: An integrable function $f$ .

[L1]

The integral is linear on $L^{1} (μ)$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[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.

[L4]

The nonnegative integral is monotone (Monotonicity and nonnegative homogeneity of the nonnegative integral).

Proof

technique · direct

1.1L1L3L4algebra

For real-valued $f$ , the functions $∣ f ∣ + f = 2 f^{+}$ and $∣ f ∣ - f = 2 f^{-}$ are nonnegative. Therefore [L1] and [L4] give $0 \leq \int (∣ f ∣ + f) d μ = \int ∣ f ∣ d μ + \int f d μ,$ $0 \leq \int (∣ f ∣ - f) d μ = \int ∣ f ∣ d μ - \int f d μ .$ So $- \int ∣ f ∣ d μ \leq \int f d μ \leq \int ∣ f ∣ d μ,$ and hence $∣ \int f d μ ∣ \leq \int ∣ f ∣ d μ$ .

2.1L1L2step 1.1L4algebra∎

For complex-valued $f$ , let $I : = \int f d μ$ . If $I = 0$ there is nothing to prove. Otherwise set $α : = \overline{I} / ∣ I ∣$ , so $∣ α ∣ = 1$ by [L2]. Then $∣ I ∣ = α I = \int α f d μ,$ and $∣ Re (α f) ∣ \leq ∣ α f ∣ = ∣ f ∣$ , so step 1.1 applies to the integrable real-valued function $Re (α f)$ . Taking real parts gives $∣ I ∣ = \int Re (α f) d μ \leq \int ∣ α f ∣ d μ = \int ∣ f ∣ d μ,$ because $Re z \leq ∣ z ∣$ for every complex $z$ .

The modulus of an integral is bounded by the integral of the modulus

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources