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Monotonicity and nonnegative homogeneity of the nonnegative integral

Statement

Let $(X, A, μ)$ be an arbitrary measure space. Let $f, g : X \to [0, + \infty]$ be measurable and let $c \geq 0$ .

If $f \leq g$ , then $\int f d μ \leq \int g d μ$ .
If $c > 0$ , then $\int c f d μ = c \int f d μ$ . For $c = 0$ , the integral of the zero function is $0$ . Neither clause forms the undefined extended-real product $0 \cdot (+ \infty)$ .

Facts & Assumptions

Given: An arbitrary measure space $(X, A, μ)$ ; nonnegative measurable functions $f, g$ and a scalar $c \geq 0$ .

[L1]

The nonnegative integral is the supremum of simple minorants (The nonnegative Lebesgue integral).

[L2]

On simple functions, the nonnegative and simple integrals agree (The nonnegative integral agrees with the simple integral on simple functions).

[L3]

The simple integral is homogeneous for positive scalars and has zero integral on the zero simple function (The simple integral is monotone, homogeneous, and additive).

Proof

technique · direct

1.1L1given

If $f \leq g$ , every simple minorant of $f$ is also a simple minorant of $g$ . Taking suprema in [L1] gives $\int f d μ \leq \int g d μ$ .

1.2L1L2L3

The zero function has just one nonnegative simple minorant: itself. [L1, L2, L3] Its simple integral is $0$ by [L3], so [L1] gives integral $0$ for the zero function, even on a space of infinite measure.

2.1L1givenalgebraL2L3∎

For $c > 0$ , multiplication by $c$ bijects simple minorants of $f$ with those of $c f$ . The inverse divides by $c$ and preserves nonnegativity and simplicity. By [L2] and [L3], the corresponding simple integrals differ by the factor $c$ . Multiplication by a positive finite real commutes with the supremum in $[0, + \infty]$ , including when that supremum is infinite. Hence $\int c f d μ = c \int f d μ$ . Together with steps 1.1 and 1.2, this proves both clauses.

Depends on

Dependency tree · 2 levels

Sources

John K. Hunter, Measure Theory Notes, Proposition 4.5