PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

On a finite measure space, the truncated L^1 metric metrises convergence in measure

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ . For measurable $f, g : X \to R$ , define $d (f, g) : = \int \min {∣ f - g ∣, 1} d μ .$

Then:

$d (f, g) = 0$ exactly when $f = g$ $μ$ -almost everywhere, so $d$ descends to a metric on almost-everywhere equivalence classes of measurable functions.
For a sequence $(f_{n})$ of measurable functions and a measurable $f$ , $d (f_{n}, f) \to 0$ if and only if $f_{n} \to f$ in measure.

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ and measurable functions $f, g, h, f_{n} : X \to R$ .

[L1]

Convergence in measure means that for every real $ε > 0$ , $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ . (Convergence in measure)

[L2]

For a measurable set $E$ and a nonnegative measurable function $u$ , $\int_{E} u d μ = \int u χ_{E} d μ$ . (Integral over a measurable subset)

[L3]

The nonnegative integral is monotone and homogeneous. (Monotonicity and nonnegative homogeneity of the nonnegative integral)

[L4]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

Proof

technique · direct

1.1L3algebra

One has $d (f, f) = 0$ and $d (f, g) = d (g, f)$ by the pointwise identities $\min {∣ f - f ∣, 1} = 0$ and $\min {∣ f - g ∣, 1} = \min {∣ g - f ∣, 1}$ . Also $∣ f - h ∣ \leq ∣ f - g ∣ + ∣ g - h ∣$ implies $\min {∣ f - h ∣, 1} \leq \min {∣ f - g ∣, 1} + \min {∣ g - h ∣, 1},$ so [L3] gives $d (f, h) \leq d (f, g) + d (g, h)$ .

1.2L1L2L3algebra

Suppose $d (f_{n}, f) \to 0$ . Fix $ε > 0$ and put $A_{n} : = {∣ f_{n} - f ∣ > ε}$ . On $A_{n}$ one has $\min {∣ f_{n} - f ∣, 1} \geq \min {ε, 1}$ , so $\min {ε, 1} μ (A_{n}) \leq d (f_{n}, f) .$ Hence $μ (A_{n}) \to 0$ , which is exactly [L1].

1.3L1L2L3algebra

Conversely, assume $f_{n} \to f$ in measure. Let $ε \in (0, 1)$ and put $A_{n} : = {∣ f_{n} - f ∣ > ε}$ . Then $d (f_{n}, f) \leq \int_{X ∖ A_{n}} ε d μ + \int_{A_{n}} 1 d μ \leq ε μ (X) + μ (A_{n}) .$ By [L1], $μ (A_{n}) \to 0$ , so ${lim sup}_{n \to \infty} d (f_{n}, f) \leq ε μ (X)$ . Since $ε \in (0, 1)$ was arbitrary and $μ (X) < + \infty$ , it follows that $d (f_{n}, f) \to 0$ .

1.4L4

If $d (f, g) = 0$ , then [L4] applied to $\min {∣ f - g ∣, 1}$ shows $\min {∣ f - g ∣, 1} = 0$ almost everywhere, which is equivalent to $f = g$ almost everywhere. Conversely, if $f = g$ almost everywhere, then [L4] gives $d (f, g) = 0$ .

2.1step 1.1step 1.4algebra

Let $[f]$ denote the almost-everywhere class of $f$ . If $f \sim f^{'}$ and $g \sim g^{'}$ , then step 1.4 gives $d (f, f^{'}) = d (g, g^{'}) = 0$ , so step 1.1 yields $d (f, g) \leq d (f, f^{'}) + d (f^{'}, g^{'}) + d (g^{'}, g) = d (f^{'}, g^{'}) .$ The same argument with the pairs reversed gives $d (f^{'}, g^{'}) \leq d (f, g)$ , so $d (f, g) = d (f^{'}, g^{'})$ . Thus the value $d ([f], [g])$ is well defined and step 1.1 makes it a metric on the quotient.

3.1step 1.2step 1.3step 2.1∎

Step 2.1 identifies the quotient metric, and steps 1.2 and 1.3 prove that its convergence is exactly convergence in measure. So the truncated distance metrises convergence in measure on finite measure spaces.

Depends on

Used by

Implication table for the main modes of convergence on a finite measure space Remark

Dependency tree · two levels

11 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., Exercise 32 (standard reference, not scraped)