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The indefinite integral of a nonnegative measurable function is a measure

Statement

Let $(X, A, μ)$ be an arbitrary measure space. Let $f : X \to [0, + \infty]$ be measurable and define $ν_{f} (A) : = \int_{A} f d μ (A \in A) .$ Then $ν_{f}$ is a measure on $(X, A)$ .

Facts & Assumptions

Given: An arbitrary measure space $(X, A, μ)$ ; a nonnegative measurable function $f$ .

[L1]

The set function $A \mapsto \int_{A} f d μ$ is defined by $A \mapsto \int f χ_{A} d μ$ (Integral over a measurable subset).

[L2]

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

[L3]

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

[L4]

A measure must vanish at the empty set and be countably additive on disjoint measurable sequences (Measures on sigma-algebras).

Proof

technique · direct

1.1L1L2given

One has $ν_{f} (\emptyset) = 0$ . If $(A_{n})$ is a pairwise disjoint sequence, put $B_{n} : = ⋃_{k < n} A_{k}$ . Then $χ_{B_{n}} ↑ χ_{⋃_{k} A_{k}}$ , so $f χ_{B_{n}} ↑ f χ_{⋃_{k} A_{k}}$ , including where $f = + \infty$ under the convention $0 \cdot \infty = 0$ . By [L2], $ν_{f} (⋃_{k} A_{k}) = \lim_{n} ν_{f} (B_{n})$ .

2.1step 1.1L3algebra

Because the sets $A_{k}$ are disjoint, $f χ_{B_{n}} = \sum_{k < n} f χ_{A_{k}}$ . Repeated use of [L3] gives $ν_{f} (B_{n}) = \sum_{k < n} ν_{f} (A_{k})$ . Taking the increasing limit in step 1.1 proves countable additivity.

3.1step 1.1step 2.1L4∎

Steps 1.1 and 2.1 verify the two conditions in [L4], so $ν_{f}$ is a measure.

The indefinite integral of a nonnegative measurable function is a measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources