TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Integrating against a density agrees with integrating the product

Statement

Let $f, g : X \to [0, + \infty]$ be measurable. Then $\int g d (f d μ) = \int g f d μ .$

Facts & Assumptions

Given: Nonnegative measurable functions $f$ and $g$ .

[L1]

The density measure is defined by $(f d μ) (A) = \int_{A} f d μ$ (The measure with density $f$ relative to $μ$ ).

[L2]

The nonnegative integral is additive on measurable sets (Additivity of the nonnegative Lebesgue integral).

[L3]

Nonnegative measurable functions admit increasing simple approximations, and products with simple functions are measurable by finite sums of indicator products (Every nonnegative measurable function is the increasing limit of simple measurable functions, Closure properties of measurable functions used by the integral).

[L4]

Monotone convergence holds for the nonnegative integral (Monotone convergence for the integral).

[L5]

The nonnegative integral is homogeneous, and on simple functions it agrees with the simple integral for any measure. (Monotonicity and nonnegative homogeneity of the nonnegative integral, The nonnegative integral agrees with the simple integral on simple functions)

Proof

technique · direct

1.1

Suppose first that $g = \sum_{j = 1}^{m} c_{j} χ_{E_{j}}$ is a simple representation, so the sets $E_{j}$ are pairwise disjoint. Applying [L5] on the measure space $(X, A, f d μ)$ and then using [L1], one gets.

L1L2L5givenalgebra

$\int g d (f d μ) = \sum_{j = 1}^{m} c_{j} (f d μ) (E_{j}) = \sum_{j = 1}^{m} c_{j} \int_{E_{j}} f d μ .$

Also $g f = \sum_{j = 1}^{m} c_{j} f χ_{E_{j}}$ , and the summands have pairwise disjoint supports. Therefore [L2] and [L5] give

$\int g f d μ = \sum_{j = 1}^{m} \int c_{j} f χ_{E_{j}} d μ = \sum_{j = 1}^{m} c_{j} \int_{E_{j}} f d μ .$

Hence $\int g d (f d μ) = \int g f d μ$ .

2.1

For general measurable $g \geq 0$ , choose simple $g_{n} ↑ g$ by [L3]. Then $g_{n} f ↑ g f$ pointwise. Applying [L4] twice and step 1.1 to each $g_{n}$ yields.

step 1.1L3L4∎

$\int g d (f d μ) = \lim_{n} \int g_{n} d (f d μ) = \lim_{n} \int g_{n} f d μ = \int g f d μ .$

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Gerald B. Folland, Real Analysis, 2nd ed., §2.2 (standard reference, not scraped)