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Integrating against a density agrees with integrating the product

Statement

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

Facts & Assumptions

Given: An arbitrary measure space $(X, A, μ)$ ; 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

1.1L1L2L5givenalgebra

Suppose first that $g = \sum_{j = 1}^{m} c_{j} χ_{E_{j}}$ is simple with pairwise disjoint measurable $E_{j}$ . By [L5] for the measure $f d μ$ and then [L1], $\int g d (f d μ) = \sum_{j} c_{j} (f d μ) (E_{j}) = \sum_{j} c_{j} \int_{E_{j}} f d μ$ . Also $g f = \sum_{j} c_{j} f χ_{E_{j}}$ has pairwise disjoint summand supports, so [L2] and [L5] give $\int g f d μ = \sum_{j} c_{j} \int_{E_{j}} f d μ$ . Hence $\int g d (f d μ) = \int g f d μ$ .

2.1step 1.1L3L4∎

For general measurable $g \geq 0$ , choose simple $g_{n} ↑ g$ by [L3]. Then $g_{n} f ↑ g f$ pointwise (using $0 \cdot \infty = 0$ ). Applying [L4] to both measures and step 1.1 to each $g_{n}$ yields $\int g d (f d μ) = \lim_{n} \int g_{n} d (f d μ) = \lim_{n} \int g_{n} f d μ = \int g f d μ$ .

Depends on

Dependency tree · 2 levels

Sources

Gerald B. Folland, Real Analysis, 2nd ed., §2.2