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

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Almost-everywhere monotone convergence

Statement

Let $f$ be measurable, let $0 \leq f_{1} \leq f_{2} \leq \dots$ be measurable, and suppose $f_{n} ↑ f$ almost everywhere. Then $\int f_{n} d μ ↑ \int f d μ .$

Facts & Assumptions

Given: A measurable nonnegative function $f$ and a nondecreasing sequence $(f_{n})$ of nonnegative measurable functions with $f_{n} ↑ f$ almost everywhere.

[L1]

Monotone convergence holds when the pointwise increase is everywhere (Monotone convergence for the integral).

[L2]

A nonnegative integral over a measurable null set is $0$ (A nonnegative integral over a null set vanishes).

[L3]

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

Proof

technique · direct

1.1

Let $N$ be a measurable null set outside which $f_{n} ↑ f$ , and define $g_{n} : = f_{n} χ_{X ∖ N}$ and $g : = f χ_{X ∖ N}$ .

L1givenconstruct

Then $g_{n} ↑ g$ everywhere, and $g$ is measurable because $f$ is. So [L1] gives $\int g_{n} d μ ↑ \int g d μ$ .

2.1

Each difference $f_{n} - g_{n} = f_{n} χ_{N}$ and $f - g = f χ_{N}$ is supported on the null set $N$ .

step 1.1L2L3

Therefore [L2] and [L3] give

$\int f_{n} d μ = \int g_{n} d μ, \int f d μ = \int g d μ .$

Substituting into step 1.1 yields the result. [step 1.1, L2, L3] ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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., Corollary 2.17 (standard reference, not scraped)