Session-authored (Fable 5 assisted)

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.

✓ 25 results · all verified · 13 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 12 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Modes of Convergence Egorov and Lusin

1 · Prerequisites

2 · Summary

This page fixes the core dictionary of measure-theoretic convergence modes used later in the library: almost-everywhere convergence, convergence in measure, almost uniform convergence, and convergence in $L^{1}$ . It then proves the exact finite-measure bridges among them, culminating in Riesz's subsequence theorem, Egorov's theorem, Lusin's theorem, and the finite-measure form of Vitali's convergence theorem.

The page is deliberately honest about where extra hypotheses are spent. Finite total measure is what turns almost-everywhere convergence into convergence in measure and what makes Egorov work; uniform integrability is what recovers $L^{1}$ from convergence in measure; tightness is what restores that recovery on sigma-finite spaces. The false statements and the closing implication-table remark point every failed arrow to a concrete witness on the companion examples page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Convergence almost everywhere relative to a measure

Definition

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. The sequence $(f_{n})$ converges to $f$ $μ$ -almost everywhere when there is a measurable $μ$ -null set $N$ such that for every $x \in X ∖ N$ the real sequence $(f_{n} (x))$ converges to $f (x)$ .

Equivalently, $(f_{n})$ converges to $f$ $μ$ -almost everywhere when the pointwise convergence statement $f_{n} (x) \to f (x)$ holds $μ$ -almost everywhere in the sense of Measure-null sets and almost-everywhere statements relative to a measure.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

Convergence in measure

Definition

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. The sequence $(f_{n})$ converges to $f$ in measure when for every real $ε > 0$ ,

$μ ({x \in X : ∣ f_{n} (x) - f (x) ∣ > ε}) \to 0.$

Thus convergence in measure asks only that, for each fixed threshold $ε$ , the measure of the bad set ${∣ f_{n} - f ∣ > ε}$ vanish as $n \to \infty$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

Cauchy sequences in measure

Definition

Let $(X, A, μ)$ be a measure space and let $f_{n} : X \to R$ be measurable for every $n \in N$ . The sequence $(f_{n})$ is Cauchy in measure when for every real $ε > 0$ , $μ ({x \in X : ∣ f_{n} (x) - f_{m} (x) ∣ > ε}) \to 0$ as $m, n \to \infty$ .

Equivalently: for every $ε > 0$ and every $η > 0$ there is $N \in N$ such that $m, n \geq N ⟹ μ ({∣ f_{n} - f_{m} ∣ > ε}) < η .$

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Almost uniform convergence

Definition

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. The sequence $(f_{n})$ converges to $f$ almost uniformly when for every real $ε > 0$ there is a measurable set $E \in A$ with $μ (E) < ε$ such that $(f_{n})$ converges uniformly to $f$ on $X ∖ E$ .

In expanded quantifiers, this means that for every $ε > 0$ there is a measurable $E$ with $μ (E) < ε$ such that $(\forall η > 0) (\exists N \in N) (\forall n \geq N) (\forall x \in X ∖ E) ∣ f_{n} (x) - f (x) ∣ < η .$

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

Convergence in L^1(mu)

Definition

Let $(X, A, μ)$ be a measure space and let $f_{n}, f \in L^{1} (μ)$ (The class $L^{1} (μ)$ of integrable functions). The sequence $(f_{n})$ converges to $f$ in $L^{1} (μ)$ when

$\int ∣ f_{n} - f ∣ d μ \to 0.$

On this page $L^{1} (μ)$ is still the class of integrable representatives rather than the quotient by almost-everywhere equality, so the expression above is read on the functions themselves.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Convergence in measure determines the limit almost everywhere

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f, g : X \to R$ be measurable. If $f_{n} \to f$ in measure and $f_{n} \to g$ in measure, then $f = g$ $μ$ -almost everywhere.

Facts & Assumptions

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

[L1]

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

[L2]

A property holds $μ$ -almost everywhere when its exceptional set is contained in a measurable $μ$ -null set. (Measure-null sets and almost-everywhere statements relative to a measure)

[L3]

For measurable $(E_{k})$ one has $μ (⋃_{k} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k})$ , and in particular $μ (E_{0} \cup E_{1}) \leq μ (E_{0}) + μ (E_{1})$ . (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

For $m \geq 1$ put $E_{m} : = {∣ f - g ∣ > 1 / m}$ , and for $n \in N$ put $A_{n, m} : = {∣ f_{n} - f ∣ > 1 / (2 m)}$ and $B_{n, m} : = {∣ f_{n} - g ∣ > 1 / (2 m)}$ . If $x \in E_{m}$ and $x \notin A_{n, m} \cup B_{n, m}$ , then $∣ f (x) - g (x) ∣ \leq ∣ f (x) - f_{n} (x) ∣ + ∣ f_{n} (x) - g (x) ∣ \leq 1 / m$ , a contradiction. So $E_{m} \subseteq A_{n, m} \cup B_{n, m}$ for every $n, m$ . [given, L1, algebra] 2.1 Fix $m \geq 1$ and let $η > 0$ . By [L1] choose $n$ so large that $μ (A_{n, m}) < η / 2$ and $μ (B_{n, m}) < η / 2$ . Then step 1.1 and [L3] give $μ (E_{m}) \leq μ (A_{n, m}) + μ (B_{n, m}) < η$ . Since $η$ was arbitrary, $μ (E_{m}) = 0$ . [step 1.1, L1, L3] 3.1 If $f (x) \neq g (x)$ , then $∣ f (x) - g (x) ∣ > 1 / m$ for some $m \geq 1$ , so ${f \neq g} = ⋃_{m = 1}^{\infty} E_{m}$ . Step 2.1 makes every $E_{m}$ null, hence [L3] gives $μ ({f \neq g}) = 0$ . By [L2], $f = g$ $μ$ -almost everywhere. ∎

step 2.1L2L3

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Convergence in L^1(mu) implies convergence in measure

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable with $f_{n}, f \in L^{1} (μ)$ . If $f_{n} \to f$ in $L^{1} (μ)$ , then $f_{n} \to f$ in measure.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable real-valued integrable functions $f_{n}, f \in L^{1} (μ)$ , and convergence of $(f_{n})$ to $f$ in $L^{1} (μ)$ .

[L1]

Convergence in $L^{1} (μ)$ means $\int ∣ f_{n} - f ∣ d μ \to 0$ . (Convergence in L^1(mu))

[L2]

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

[L3]

For a nonnegative measurable function $h$ and a real $t > 0$ , $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ . (Chebyshev-Markov inequality for the integral)

Proof

technique · direct

1.1

Fix $ε > 0$ . Applying [L3] to $h : = ∣ f_{n} - f ∣$ and $t : = ε$ gives $μ ({∣ f_{n} - f ∣ \geq ε}) \leq ε^{- 1} \int ∣ f_{n} - f ∣ d μ$ for every $n$ .

L1L3algebra

2.1

By [L1], the right-hand side in step 1.1 tends to $0$ as $n \to \infty$ . Hence $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ for every $ε > 0$ , which is exactly [L2].

step 1.1L1L2∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

FALSE: convergence in L^1(mu) forces almost-everywhere convergence

Statement refuted

convergence in $L^{1} (μ)$ forces almost-everywhere convergence.

Facts & Assumptions

Given: Lebesgue measure on $[0, 1]$ and the dyadic typewriter sequence $f_{n}$ defined by $f_{0} : = 0$ and $f_{2^{k} + j} : = χ_{I_{k, j}} for k \geq 0, 0 \leq j < 2^{k},$ where $I_{k, j} = [j 2^{- k}, (j + 1) 2^{- k})$ for $j < 2^{k} - 1$ and $I_{k, 2^{k} - 1} = [1 - 2^{- k}, 1]$ .

[L1]

Convergence in $L^{1} (μ)$ means $\int ∣ f_{n} - f ∣ d μ \to 0$ . (Convergence in L^1(mu))

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

Refutation

technique · direct

1.1

If $2^{k} \leq n < 2^{k + 1}$ , then $f_{n}$ is the indicator of an interval of length $2^{- k}$ , so $\int_{0}^{1} ∣ f_{n} ∣ d λ = 2^{- k} \to 0.$ Thus $f_{n} \to 0$ in $L^{1} ([0, 1])$ by [L1].

givenL1algebra

1.2

Fix $x \in [0, 1]$ . For each $k \geq 1$ there is exactly one $j_{k} \in {0, \dots, 2^{k} - 1}$ with $x \in I_{k, j_{k}}$ , so $f_{2^{k} + j_{k}} (x) = 1$ . Because the same generation contains other dyadic intervals as well, there are also infinitely many indices $n$ with $f_{n} (x) = 0$ . So $(f_{n} (x))$ does not converge for any $x \in [0, 1]$ .

given

2.1

Step 1.1 gives convergence in $L^{1}$ , while step 1.2 shows failure of pointwise convergence at every point and hence failure of [L2]. This refutes the claim.

step 1.1step 1.2L2∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Almost uniform convergence implies almost-everywhere convergence and convergence in measure

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ almost uniformly, then $f_{n} \to f$ $μ$ -almost everywhere and $f_{n} \to f$ in measure.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable functions $f_{n}, f : X \to R$ , and almost-uniform convergence of $(f_{n})$ to $f$ .

[L1]

Almost-uniform convergence means that for every $ε > 0$ there is a measurable $E$ with $μ (E) < ε$ such that $f_{n} \to f$ uniformly on $X ∖ E$ . (Almost uniform convergence)

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L3]

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

[L4]

If $A \subseteq B$ are measurable, then $μ (A) \leq μ (B)$ . (Measures are monotone)

Proof

technique · direct

1.1

For each $m \geq 1$ , [L1] gives a measurable set $E_{m}$ with $μ (E_{m}) < 1 / m$ such that $f_{n} \to f$ uniformly on $X ∖ E_{m}$ . Let $N : = ⋂_{m = 1}^{\infty} E_{m}$ . Since $N \subseteq E_{m}$ for every $m$ , [L4] gives $μ (N) \leq 1 / m$ for every $m$ , hence $μ (N) = 0$ . If $x \in X ∖ N$ , then $x \notin E_{m}$ for some $m$ , and uniform convergence on $X ∖ E_{m}$ implies $f_{n} (x) \to f (x)$ . Therefore $f_{n} \to f$ almost everywhere by [L2].

L1L2L4

1.2

Fix $ε > 0$ and $η > 0$ . By [L1] choose a measurable set $E$ with $μ (E) < η$ such that $f_{n} \to f$ uniformly on $X ∖ E$ . Then there is $N$ such that for $n \geq N$ and $x \in X ∖ E$ one has $∣ f_{n} (x) - f (x) ∣ \leq ε$ , so ${∣ f_{n} - f ∣ > ε} \subseteq E$ . Hence $μ ({∣ f_{n} - f ∣ > ε}) \leq μ (E) < η$ for $n \geq N$ . Since $η$ was arbitrary, [L3] follows.

L1L3L4

2.1

Steps 1.1 and 1.2 prove the two asserted conclusions.

step 1.1step 1.2∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

On a finite measure space, almost-everywhere convergence implies convergence in measure

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ , and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ $μ$ -almost everywhere, then $f_{n} \to f$ in measure.

Facts & Assumptions

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

[L1]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L2]

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

[L3]

If $(E_{n})$ is a decreasing sequence of measurable sets and one $E_{n_{0}}$ has finite measure, then $μ (⋂_{n} E_{n}) = \inf_{n} μ (E_{n})$ . (Continuity from above when one set has finite measure)

[L4]

If $A \subseteq B$ are measurable, then $μ (A) \leq μ (B)$ . (Measures are monotone)

Proof

technique · direct

1.1

Fix $ε > 0$ , and let $N$ be a measurable null set outside which $f_{n} (x) \to f (x)$ . For $r \geq 0$ put $E_{r} : = ⋃_{n \geq r} {∣ f_{n} - f ∣ > ε}$ . Then $(E_{r})$ is a decreasing sequence of measurable sets, each contained in $X$ , and $⋂_{r = 0}^{\infty} E_{r} \subseteq N$ because outside $N$ only finitely many indices can satisfy $∣ f_{n} (x) - f (x) ∣ > ε$ .

givenL1

2.1

Because $μ (X) < + \infty$ , [L3] applies to $(E_{r})$ . The intersection in step 1.1 is null, so $μ (E_{r}) ↓ μ ⁣ (⋂_{r = 0}^{\infty} E_{r}) = 0.$ For each $r$ one has ${∣ f_{r} - f ∣ > ε} \subseteq E_{r}$ , hence by [L4] $μ ({∣ f_{r} - f ∣ > ε}) \leq μ (E_{r}) \to 0$ . This is exactly [L2].

step 1.1L2L3L4

3.1

Since $ε > 0$ was arbitrary, the sequence converges in measure.

step 2.1L2∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

FALSE: almost-everywhere convergence implies convergence in measure on every measure space

Statement refuted

almost-everywhere convergence implies convergence in measure on every measure space.

Facts & Assumptions

Given: Lebesgue measure on $R$ and the sequence $f_{n} : = χ_{[n, n + 1]}$ .

[L1]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L2]

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

Refutation

technique · direct

1.1

Fix $x \in R$ . If $n > x + 1$ , then $x \notin [n, n + 1]$ , so $f_{n} (x) = 0$ . Thus $f_{n} (x) \to 0$ for every $x \in R$ , hence almost everywhere by [L1].

givenL1

1.2

For every $n$ one has ${∣ f_{n} - 0 ∣ > 1 / 2} = [n, n + 1]$ , whose Lebesgue measure is $1$ . So the bad-set measures do not tend to $0$ , and [L2] fails.

givenL2

2.1

This single sequence satisfies the premise and violates the conclusion, so the claim is false.

step 1.1step 1.2∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

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

1.1

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 $d (f, h) \leq d (f, g) + d (g, h)$ .

L3algebra

1.2

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].

L1L2L3algebra

1.3

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 $d (f_{n}, f) \to 0$ .

L1L2L3algebra

1.4

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 $d (f, g) = 0$ .

2.1

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.

step 1.1step 1.4algebra

3.1

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.

step 1.2step 1.3step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Riesz's subsequence theorem for convergence in measure

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ in measure, then there is a strictly increasing sequence $(n_{k})_{k \geq 0}$ of natural numbers such that $f_{n_{k}} \to f$ $μ$ -almost everywhere.

The construction below uses the least admissible index at each stage, so no choice principle is spent.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable functions $f_{n}, f : X \to R$ , and convergence in measure of $(f_{n})$ to $f$ .

[L1]

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

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L3]

For measurable $(E_{k})$ one has $μ (⋃_{k} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k})$ . (Finite and countable subadditivity of measures)

[L4]

If $(E_{r})$ is a decreasing sequence of measurable sets and one $E_{r_{0}}$ has finite measure, then $μ (⋂_{r} E_{r}) = \inf_{r} μ (E_{r})$ . (Continuity from above when one set has finite measure)

Proof

technique · direct

1.1

For each $k \geq 0$ , [L1] applied with $ε : = 2^{- (k + 1)}$ yields an index after which $μ ({∣ f_{n} - f ∣ > 2^{- (k + 1)}}) < 2^{- (k + 1)}$ . Define $n_{0}$ to be the least admissible index for $k = 0$ , and recursively define $n_{k + 1}$ to be the least admissible index larger than $n_{k}$ . Then $(n_{k})_{k \geq 0}$ is strictly increasing and $μ ({∣ f_{n_{k}} - f ∣ > 2^{- (k + 1)}}) < 2^{- (k + 1)} (k \geq 0) .$

L1choose

2.1

Put $A_{k} : = {∣ f_{n_{k}} - f ∣ > 2^{- (k + 1)}}$ and $E_{r} : = ⋃_{k \geq r} A_{k}$ . Then $(E_{r})$ is a decreasing sequence of measurable sets, and step 1.1 together with [L3] gives $μ (E_{r}) \leq \sum_{k = r}^{\infty} μ (A_{k}) \leq \sum_{k = r}^{\infty} 2^{- (k + 1)} = 2^{- r} .$ In particular $μ (E_{0}) \leq 1 < + \infty$ .

step 1.1L3algebra

3.1

Let $N : = ⋂_{r = 0}^{\infty} E_{r}$ . By [L4] and step 2.1, $μ (N) = \lim_{r \to \infty} μ (E_{r}) = 0.$ If $x \in X ∖ N$ , then $x \notin E_{r}$ for some $r$ , hence $x \notin A_{k}$ for every $k \geq r$ . Therefore $∣ f_{n_{k}} (x) - f (x) ∣ \leq 2^{- (k + 1)}$ for all $k \geq r$ , so $f_{n_{k}} (x) \to f (x)$ . By [L2], $f_{n_{k}} \to f$ almost everywhere.

step 2.1L2L4

4.1

The subsequence constructed in step 1.1 has the required almost-everywhere limit.

step 3.1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Convergence in L^1(mu) has an almost-everywhere convergent subsequence

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable with $f_{n}, f \in L^{1} (μ)$ . If $f_{n} \to f$ in $L^{1} (μ)$ , then some subsequence of $(f_{n})$ converges to $f$ $μ$ -almost everywhere.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable real-valued integrable functions $f_{n}, f \in L^{1} (μ)$ , and convergence of $(f_{n})$ to $f$ in $L^{1} (μ)$ .

[L1]

Convergence in $L^{1} (μ)$ implies convergence in measure. (Convergence in L^1(mu) implies convergence in measure)

[L2]

Convergence in measure has a subsequence converging almost everywhere to the same limit. (Riesz's subsequence theorem for convergence in measure)

Proof

technique · direct

1.1

By [L1], the sequence $(f_{n})$ converges to $f$ in measure.

2.1

Apply [L2] to step 1.1. The resulting subsequence converges to $f$ almost everywhere.

step 1.1L2∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-28Open item page →

FALSE: convergence in measure implies almost-everywhere convergence

Statement refuted

convergence in measure implies almost-everywhere convergence.

Facts & Assumptions

[L1]

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

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

Refutation

technique · direct

1.1

If $2^{k} \leq n < 2^{k + 1}$ , then $f_{n}$ is the indicator of an interval of length $2^{- k}$ . Hence for every $ε \in (0, 1)$ , $λ ({∣ f_{n} - 0 ∣ > ε}) = 2^{- k} \to 0,$ so $f_{n} \to 0$ in measure by [L1].

givenL1algebra

2.1

Fix $x \in [0, 1]$ . In each dyadic generation there is exactly one interval containing $x$ , so $f_{n} (x) = 1$ infinitely often; the same generation also contains intervals missing $x$ , so $f_{n} (x) = 0$ infinitely often. Therefore $(f_{n} (x))$ has no limit for any $x \in [0, 1]$ , and [L2] fails.

step 1.1L2

3.1

This refutes the claim.

step 1.1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Cauchy sequences in measure converge in measure

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n} : X \to R$ be measurable. If $(f_{n})$ is Cauchy in measure, then there is a measurable $f : X \to R$ such that $f_{n} \to f$ in measure.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and a measurable sequence $f_{n} : X \to R$ that is Cauchy in measure.

[L1]

Cauchy in measure means that for every real $ε > 0$ and every $η > 0$ there is $N$ such that $m, n \geq N \Rightarrow μ ({∣ f_{n} - f_{m} ∣ > ε}) < η$ . (Cauchy sequences in measure)

[L2]

For measurable $(E_{k})$ one has $μ (⋃_{k} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k})$ . (Finite and countable subadditivity of measures)

[L3]

[L4]

A uniformly Cauchy sequence of real-valued functions on a set converges uniformly to some real-valued function on that set. (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy)

[L5]

For a measurable set $E$ , the indicator $1_{E}$ is measurable. (An indicator function is measurable exactly when its set is measurable)

[L6]

Products of measurable real-valued functions are measurable. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[L7]

Pointwise limits of measurable functions are measurable. (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable)

[L8]

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

Proof

technique · direct

1.1

For each $k \geq 0$ , [L1] with $ε = η = 2^{- (k + 1)}$ gives an index $N_{k}$ such that $m, n \geq N_{k} ⟹ μ ({∣ f_{n} - f_{m} ∣ > 2^{- (k + 1)}}) < 2^{- (k + 1)} .$ Choose a strictly increasing sequence $(n_{k})_{k \geq 0}$ with $n_{k} \geq N_{k}$ for every $k$ . Then $n \geq n_{k} ⟹ μ ({∣ f_{n} - f_{n_{k}} ∣ > 2^{- (k + 1)}}) < 2^{- (k + 1)} .$

L1choose

2.1

For $k \geq 0$ put $A_{k} : = {∣ f_{n_{k + 1}} - f_{n_{k}} ∣ > 2^{- (k + 1)}}$ , $E_{r} : = ⋃_{k \geq r} A_{k}$ , and $G_{r} : = X ∖ E_{r}$ . Step 1.1 gives $μ (A_{k}) < 2^{- (k + 1)}$ , so $μ (E_{r}) \leq \sum_{k = r}^{\infty} 2^{- (k + 1)} = 2^{- r} .$ Thus $μ (E_{0}) \leq 1 < + \infty$ , the sets $E_{r}$ decrease with $r$ , and [L3] gives $μ ⁣ (⋂_{r = 0}^{\infty} E_{r}) = 0.$ Call this null set $N$ .

step 1.1L2L3algebra

3.1

Fix $r \geq 0$ . If $x \in G_{r}$ and $m > ℓ \geq r$ , then $x \notin A_{j}$ for every $j \geq r$ , so $∣ f_{n_{j + 1}} (x) - f_{n_{j}} (x) ∣ \leq 2^{- (j + 1)}$ for $j = ℓ, \dots, m - 1$ . Therefore $∣ f_{n_{m}} (x) - f_{n_{ℓ}} (x) ∣ \leq \sum_{j = ℓ}^{m - 1} 2^{- (j + 1)} \leq 2^{- ℓ} .$ Hence the tail $(f_{n_{k}} ∣_{G_{r}})_{k \geq r}$ is uniformly Cauchy on $G_{r}$ , so [L4] gives a function $g_{r} : G_{r} \to R$ with $f_{n_{k}} ∣_{G_{r}} \to g_{r}$ uniformly on $G_{r}$ .

step 2.1L4algebra

4.1

By [L5], $1_{G_{r}}$ is measurable. Since each $f_{n_{k}}$ is measurable, [L6] makes $u_{k, r} : = f_{n_{k}} 1_{G_{r}}$ measurable on $X$ . For $x \in G_{r}$ one has $u_{k, r} (x) = f_{n_{k}} (x) \to g_{r} (x)$ , while for $x \notin G_{r}$ all $u_{k, r} (x) = 0$ . So [L7] gives a measurable function $h_{r} : X \to R$ such that $h_{r} = g_{r}$ on $G_{r}$ and $h_{r} = 0$ on $E_{r}$ . The sets $G_{r}$ increase and cover $X ∖ N$ , and on overlaps the limits agree, so $(h_{r} (x))_{r \geq 0}$ stabilizes for every $x$ . Define $f (x) : = \lim_{r \to \infty} h_{r} (x) .$ By [L7] again, $f$ is measurable.

step 3.1L5L6L7construct

5.1

Fix $k \geq 0$ and $n \geq n_{k}$ . If $x \in G_{k}$ , then step 4.1 makes $f (x) = g_{k} (x)$ , so letting $m \to \infty$ in step 3.1 with $ℓ = k$ gives $∣ f_{n_{k}} (x) - f (x) ∣ \leq 2^{- k}$ . Hence on $G_{k}$ , $∣ f_{n} (x) - f (x) ∣ \leq ∣ f_{n} (x) - f_{n_{k}} (x) ∣ + 2^{- k} .$ Therefore ${∣ f_{n} - f ∣ > 3 \cdot 2^{- (k + 1)}} \subseteq {∣ f_{n} - f_{n_{k}} ∣ > 2^{- (k + 1)}} \cup E_{k} .$ Step 1.1 makes the first set have measure below $2^{- (k + 1)}$ , and step 2.1 gives $μ (E_{k}) \leq 2^{- k}$ . So $μ ({∣ f_{n} - f ∣ > 3 \cdot 2^{- (k + 1)}}) < 3 \cdot 2^{- (k + 1)} (n \geq n_{k}) .$

step 1.1step 2.1step 3.1step 4.1algebra

6.1

Given $ε, η > 0$ , choose $k$ with $3 \cdot 2^{- (k + 1)} < \min {ε, η}$ . Then for $n \geq n_{k}$ , $μ ({∣ f_{n} - f ∣ > ε}) \leq μ ({∣ f_{n} - f ∣ > 3 \cdot 2^{- (k + 1)}}) < η .$ This is exactly [L8].

step 5.1L8choose

7.1

The measurable function $f$ from step 4.1 is the limit of $(f_{n})$ in measure.

step 4.1step 6.1L8∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Egorov's theorem

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ , and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ $μ$ -almost everywhere, then $f_{n} \to f$ almost uniformly.

The finite-measure hypothesis is used exactly at the continuity-from-above step below.

Facts & Assumptions

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

[L1]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L2]

Almost-uniform convergence means that for every $ε > 0$ there is a measurable $E$ with $μ (E) < ε$ such that $f_{n} \to f$ uniformly on $X ∖ E$ . (Almost uniform convergence)

[L3]

[L4]

For measurable $(E_{k})$ one has $μ (⋃_{k} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k})$ . (Finite and countable subadditivity of measures)

[L5]

If $A \subseteq B$ are measurable, then $μ (A) \leq μ (B)$ . (Measures are monotone)

Proof

technique · direct

1.1

Let $ε > 0$ , and let $N$ be a measurable null set outside which $f_{n} (x) \to f (x)$ . For $m, k \geq 1$ put $E_{k, m} : = ⋃_{j \geq k} {∣ f_{j} - f ∣ > 1 / m} .$ For fixed $m$ the sets $E_{k, m}$ decrease with $k$ , each lies in $X$ , and $⋂_{k = 1}^{\infty} E_{k, m} \subseteq N$ because outside $N$ only finitely many $j$ satisfy $∣ f_{j} (x) - f (x) ∣ > 1 / m$ . Therefore [L3] and [L5] give $μ (E_{k, m}) ↓ μ ⁣ (⋂_{k = 1}^{\infty} E_{k, m}) = 0.$ So for each $m \geq 1$ there is a least index $k (m)$ with $μ (E_{k (m), m}) < ε 2^{- m}$ .

L1L3L5choose

2.1

Put $E : = ⋃_{m = 1}^{\infty} E_{k (m), m}$ . Then [L4] and step 1.1 give $μ (E) \leq \sum_{m = 1}^{\infty} μ (E_{k (m), m}) < \sum_{m = 1}^{\infty} ε 2^{- m} = ε .$ If $x \in X ∖ E$ , then for every $m \geq 1$ and every $j \geq k (m)$ one has $∣ f_{j} (x) - f (x) ∣ \leq 1 / m$ . Hence for any $η > 0$ one may choose $m$ with $1 / m < η$ and then $j \geq k (m)$ gives $∣ f_{j} (x) - f (x) ∣ < η$ . So $f_{n} \to f$ uniformly on $X ∖ E$ .

step 1.1L4algebra

3.1

Since $ε > 0$ was arbitrary, step 2.1 is exactly [L2]. Therefore $f_{n} \to f$ almost uniformly.

step 2.1L2∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

FALSE: Egorov's theorem holds on every measure space

Statement refuted

Egorov's theorem holds on every measure space.

Facts & Assumptions

Given: Lebesgue measure on $R$ and the sequence $f_{n} : = χ_{[n, n + 1]}$ .

[L2]

Almost-uniform convergence means that for every $ε > 0$ there is a measurable $E$ with $μ (E) < ε$ such that $f_{n}$ converges uniformly on $X ∖ E$ . (Almost uniform convergence)

Refutation

technique · direct

1.1

For each fixed $x \in R$ one has $f_{n} (x) = 0$ for all $n > x + 1$ , so $f_{n} (x) \to 0$ pointwise on $R$ .

given

2.1

Let $E \subseteq R$ be measurable with $λ (E) < + \infty$ . If $[n, n + 1] \subseteq E$ for infinitely many $n$ , then $λ (E) \geq \sum 1 = + \infty$ , impossible. Hence infinitely many indices $n$ satisfy $[n, n + 1] \subseteq̸ E$ , so for each such $n$ there is $x_{n} \in [n, n + 1] ∖ E$ with $f_{n} (x_{n}) = 1$ . Therefore $\sup_{x \in R ∖ E} ∣ f_{n} (x) - 0 ∣ \geq 1$ for infinitely many $n$ , and the convergence cannot be uniform on $R ∖ E$ . Thus [L2] fails.

step 1.1L2algebra

3.1

So pointwise almost-everywhere convergence does not force almost-uniform convergence on this infinite-measure space. The claim is false.

step 1.1step 2.1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

On a finite measure space, convergence in measure has an almost-uniformly convergent subsequence

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ , and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ in measure, then some subsequence of $(f_{n})$ converges to $f$ almost uniformly.

Facts & Assumptions

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

[L1]

Convergence in measure has a subsequence converging almost everywhere to the same limit. (Riesz's subsequence theorem for convergence in measure)

[L2]

On a finite measure space, almost-everywhere convergence implies almost-uniform convergence. (Egorov's theorem)

Proof

technique · direct

1.1

By [L1], there is a subsequence $(f_{n_{k}})$ converging to $f$ almost everywhere.

2.1

Apply [L2] to the subsequence from step 1.1. It converges to $f$ almost uniformly.

step 1.1L2∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Assuming countable choice, simple functions are continuous on a large closed core

Statement

Assume the Axiom of Countable Choice.

Let $n \geq 1$ , let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < + \infty$ , and let $s : E \to R$ be a simple measurable function. Then for every $ε > 0$ there is a closed set $F \subseteq E$ such that $λ_{n} (E ∖ F) < ε$ and $s ∣_{F}$ is continuous.

Facts & Assumptions

Given: The Axiom of Countable Choice, a Lebesgue measurable set $E \subseteq R^{n}$ of finite measure, a simple measurable function $s : E \to R$ , and a real $ε > 0$ .

[L1]

If the distinct values of $s$ are $c_{1}, \dots, c_{m}$ and $E_{j} : = s^{- 1} ({c_{j}})$ , then the sets $E_{j}$ are measurable, pairwise disjoint, their union is $E$ , and $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ . (A simple function and its canonical representation)

[L2]

Assuming countable choice, every Lebesgue measurable subset of $R^{n}$ has compact subsets of arbitrarily close measure from inside. (Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets)

[L3]

For measurable $(A_{k})$ one has $μ (⋃_{k} A_{k}) \leq \sum_{k = 0}^{\infty} μ (A_{k})$ . (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

Let $c_{1}, \dots, c_{m}$ be the distinct values of $s$ , and let $E_{j} : = s^{- 1} ({c_{j}})$ . By [L1], the sets $E_{j}$ are measurable, pairwise disjoint, and cover $E$ . For each $j$ choose a compact set $K_{j} \subseteq E_{j}$ with $λ_{n} (E_{j} ∖ K_{j}) < ε / m .$ This is possible by [L2].

L1L2choose

2.1

Put $F : = ⋃_{j = 1}^{m} K_{j}$ . Each $K_{j}$ is closed in $R^{n}$ , so the finite union $F$ is closed and lies in $E$ . Also $E ∖ F = ⋃_{j = 1}^{m} (E_{j} ∖ K_{j}),$ so [L3] and step 1.1 give $λ_{n} (E ∖ F) \leq \sum_{j = 1}^{m} λ_{n} (E_{j} ∖ K_{j}) < ε .$

step 1.1L3algebra

3.1

Fix $x \in F$ . By step 1.1, $x$ lies in exactly one $K_{j}$ . In the subspace $F$ , the set $K_{j} = F ∖ ⋃_{i \neq j} K_{i}$ is open because the other $K_{i}$ are closed and finite in number. On that neighbourhood $K_{j}$ , the restriction $s ∣_{F}$ is constant with value $c_{j}$ . So $s ∣_{F}$ is locally constant at every point of $F$ , hence continuous.

step 1.1step 2.1

4.1

The closed set $F$ from step 2.1 has $λ_{n} (E ∖ F) < ε$ , and step 3.1 makes $s ∣_{F}$ continuous.

step 2.1step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Assuming countable choice, simple approximants to a measurable function can be made uniformly convergent on a large closed set

Statement

Assume the Axiom of Countable Choice.

Let $n \geq 1$ , let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < + \infty$ , and let $f : E \to R$ be bounded and measurable. Then for every $ε > 0$ there are a closed set $F \subseteq E$ and simple functions $s_{m} : E \to R$ such that

$λ_{n} (E ∖ F) < ε$ ;
each restriction $s_{m} ∣_{F}$ is continuous;
$s_{m} \to f$ uniformly on $F$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, a bounded measurable function $f : E \to R$ on a finite Lebesgue-measure set $E \subseteq R^{n}$ , and a real $ε > 0$ .

[L1]

Every measurable function admits simple approximations $s_{m}$ with $∣ s_{m} ∣ \leq ∣ f ∣$ for every $m$ and $s_{m} (x) \to f (x)$ for every $x$ . (Every measurable function admits simple approximations dominated by its absolute value)

[L2]

On a finite measure space, almost-everywhere convergence implies almost-uniform convergence. (Egorov's theorem)

[L3]

[L4]

Every simple measurable function on a finite-measure Lebesgue set is continuous on a large closed core. (Assuming countable choice, simple functions are continuous on a large closed core)

[L5]

For measurable $(A_{k})$ one has $μ (⋃_{k} A_{k}) \leq \sum_{k = 0}^{\infty} μ (A_{k})$ . (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

By [L1], choose simple functions $s_{m} : E \to R$ with $∣ s_{m} ∣ \leq ∣ f ∣$ and $s_{m} (x) \to f (x)$ for every $x \in E$ . Since $λ_{n} (E) < + \infty$ , [L2] gives a measurable set $G \subseteq E$ with $λ_{n} (E ∖ G) < ε / 2$ such that $s_{m} \to f$ uniformly on $G$ .

L1L2choose

1.2

By [L3], choose a compact set $K \subseteq G$ with $λ_{n} (G ∖ K) < ε / 4$ . For each $m \geq 1$ , apply [L4] to $s_{m}$ with tolerance $ε 2^{- m - 3}$ , obtaining a closed set $F_{m} \subseteq E$ such that $λ_{n} (E ∖ F_{m}) < ε 2^{- m - 3}$ and $s_{m} ∣_{F_{m}}$ is continuous.

L3L4choose

2.1

Put $F : = K \cap ⋂_{m = 1}^{\infty} F_{m}$ . Then $F$ is closed, $F \subseteq K \subseteq G \subseteq E$ , and $E ∖ F \subseteq (E ∖ G) \cup (G ∖ K) \cup ⋃_{m = 1}^{\infty} (E ∖ F_{m}) .$ So [L5] together with steps 1.1 and 1.2 gives $λ_{n} (E ∖ F) < \frac{ε}{2} + \frac{ε}{4} + \sum_{m = 1}^{\infty} ε 2^{- m - 3} < ε .$ Because $F \subseteq G$ , the convergence $s_{m} \to f$ remains uniform on $F$ . And because $F \subseteq F_{m}$ , each $s_{m} ∣_{F}$ is continuous as a restriction of $s_{m} ∣_{F_{m}}$ .

step 1.1step 1.2L5algebra

3.1

The closed set $F$ and the simple approximants $s_{m}$ satisfy all three assertions.

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n

Statement

Assume the Axiom of Countable Choice.

Let $n \geq 1$ , let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < + \infty$ , and let $f : E \to R$ be measurable. Then for every $ε > 0$ there exist a real $M > 0$ and a closed set $F \subseteq E \cap f^{- 1} ([- M, M])$ such that $λ_{n} (E ∖ F) < ε$ and $f ∣_{F}$ is continuous. In particular, $f ∣_{F}$ is bounded.

Facts & Assumptions

Given: The Axiom of Countable Choice, a measurable function $f : E \to R$ on a Lebesgue measurable set $E \subseteq R^{n}$ with $λ_{n} (E) < + \infty$ , and a real $ε > 0$ .

[L1]

Measurability means preimages of Borel sets are measurable. (A measurable function between measurable spaces)

[L2]

Measures are continuous from below on increasing measurable sets. (Continuity from below for measures)

[L3]

Assuming countable choice, for bounded measurable functions on finite-measure Lebesgue sets there is a large closed set on which simple approximants are continuous and converge uniformly to the function. (Assuming countable choice, simple approximants to a measurable function can be made uniformly convergent on a large closed set)

[L4]

A uniform limit of continuous functions is continuous. (A uniform limit of continuous functions is continuous, so $C (X, Y)$ is closed in $Y^{X}$ under the uniform metric)

[L5]

For measurable $(A_{k})$ one has $μ (⋃_{k} A_{k}) \leq \sum_{k = 0}^{\infty} μ (A_{k})$ . (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

For $m \geq 1$ , put $E_{m} : = E \cap f^{- 1} ([- m, m])$ . Because $[- m, m]$ is a Borel subset of $R$ , [L1] makes every $E_{m}$ measurable. The sets $E_{m}$ increase with $m$ , and their union is all of $E$ because $f$ is real-valued. So [L2] gives $λ_{n} (E) = \sup_{m \geq 1} λ_{n} (E_{m}) .$ Choose $M \geq 1$ with $λ_{n} (E ∖ E_{M}) < ε / 2$ .

L1L2choose

2.1

On $E_{M}$ one has $∣ f ∣ \leq M$ , so $f ∣_{E_{M}}$ is bounded. Apply [L3] to the bounded measurable function $f ∣_{E_{M}}$ with tolerance $ε / 2$ . This gives a closed set $F \subseteq E_{M}$ and simple functions $s_{m} : E_{M} \to R$ such that $λ_{n} (E_{M} ∖ F) < ε / 2,$ each $s_{m} ∣_{F}$ is continuous, and $s_{m} \to f ∣_{E_{M}}$ uniformly on $F$ . By [L4], $f ∣_{F}$ is continuous.

step 1.1L3L4

3.1

Because $F \subseteq E_{M} \subseteq E$ , one has $E ∖ F = (E ∖ E_{M}) \cup (E_{M} ∖ F) .$ Hence [L5] together with steps 1.1 and 2.1 gives $λ_{n} (E ∖ F) < ε / 2 + ε / 2 = ε .$ The set $F$ also lies in $E \cap f^{- 1} ([- M, M])$ by step 1.1.

step 1.1step 2.1L5algebra

4.1

The set $F$ and the bound $M$ satisfy the theorem.

step 2.1step 3.1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Assuming countable choice and dependent choice, a measurable function on a finite-measure subset of R^n agrees there, off a small set, with a continuous function on R^n

Statement

Assume the Axiom of Countable Choice and the Axiom of Dependent Choice. Let $n \geq 1$ , let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < + \infty$ , and let $f : E \to R$ be measurable. Then for every $ε > 0$ there are a continuous function $g : R^{n} \to R$ and a closed set $F \subseteq E$ such that $λ_{n} (E ∖ F) < ε and g (x) = f (x) for every x \in F .$

In particular, $λ_{n} ({x \in E : g (x) \neq f (x)}) < ε .$

Facts & Assumptions

Given: The Axiom of Countable Choice, dependent choice, a measurable function $f : E \to R$ on a finite-measure Lebesgue set $E \subseteq R^{n}$ , and a real $ε > 0$ .

[L1]

Assuming countable choice, Lusin's theorem gives a real $M > 0$ and a closed set $F \subseteq E \cap f^{- 1} ([- M, M])$ such that $λ_{n} (E ∖ F) < ε$ and $f ∣_{F}$ is continuous. (Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n)

[L2]

Assuming dependent choice, every continuous map from a closed subspace of a normal space into a closed interval $[a, b]$ extends continuously to the whole space. (Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a, b]$ extends continuously to the whole space, and this property characterises normality)

[L3]

For $n \geq 1$ , $R^{n}$ carries its Euclidean metric. ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it)

[L4]

Every metric space is normal. (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal)

Proof

technique · direct

1.1

By [L1], choose $M > 0$ and a closed set $F \subseteq E \cap f^{- 1} ([- M, M])$ with $λ_{n} (E ∖ F) < ε$ such that $f ∣_{F} : F \to [- M, M]$ is continuous.

L1choose

2.1

By [L3], $R^{n}$ is a metric space, so [L4] makes it normal. Apply [L2] to the closed subspace $F \subseteq R^{n}$ and to the interval $[- M, M]$ . This gives a continuous function $g : R^{n} \to [- M, M]$ with $g ∣_{F} = f ∣_{F}$ .

step 1.1L2L3L4

3.1

Since $g = f$ on $F$ , one has ${x \in E : g (x) \neq f (x)} \subseteq E ∖ F$ . Therefore $λ_{n} ({x \in E : g (x) \neq f (x)}) < ε$ . The function $g$ and the closed set $F$ satisfy the corollary.

step 1.1step 2.1algebra∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Assuming countable choice, on a bounded measurable set, Lusin's closed core can be chosen compact

Statement

Assume the Axiom of Countable Choice.

Let $n \geq 1$ , let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < + \infty$ , and suppose $E$ is bounded. Let $f : E \to R$ be measurable. Then for every $ε > 0$ there is a compact set $K \subseteq E$ such that $λ_{n} (E ∖ K) < ε$ and $f ∣_{K}$ is continuous.

Facts & Assumptions

Given: The Axiom of Countable Choice, a bounded Lebesgue measurable set $E \subseteq R^{n}$ of finite measure, a measurable function $f : E \to R$ , and a real $ε > 0$ .

[L1]

Assuming countable choice, Lusin's theorem gives a closed set $F \subseteq E$ with $λ_{n} (E ∖ F) < ε$ such that $f ∣_{F}$ is continuous. (Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n)

[L2]

In $R^{n}$ , a subset is compact if and only if it is closed and bounded. (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line)

Proof

technique · direct

1.1

By [L1], choose a closed set $F \subseteq E$ with $λ_{n} (E ∖ F) < ε$ such that $f ∣_{F}$ is continuous.

L1choose

2.1

Because $F \subseteq E$ and $E$ is bounded, the set $F$ is bounded. Since $F$ is also closed in $R^{n}$ , [L2] makes $F$ compact.

step 1.1L2

3.1

Taking $K : = F$ proves the claim.

step 2.1∎

False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-28Open item page →

FALSE: assuming countable choice, Lusin's theorem says measurable functions are continuous off a null set

Statement refuted

Assume the Axiom of Countable Choice.

Lusin's theorem says measurable functions are continuous off a null set.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Dirichlet function $D : = 1_{Q \cap [0, 1]} : [0, 1] \to R$ .

[L1]

Assuming countable choice, Lusin's theorem provides large closed sets on which a measurable real-valued function is continuous. (Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n)

[L2]

Both the rationals and the irrationals are dense in $R$ . (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable)

Refutation

technique · direct

1.1

The function $D$ is measurable and, by [L1], for every $ε > 0$ there is a closed set $F \subseteq [0, 1]$ with $λ ([0, 1] ∖ F) < ε$ such that $D ∣_{F}$ is continuous.

1.2

Fix $x \in [0, 1]$ . By [L2], every neighbourhood of $x$ contains both a rational point $q$ and an irrational point $u$ . Then $D (q) = 1$ and $D (u) = 0$ , so $D$ is not continuous at $x$ . Thus $D$ is nowhere continuous on $[0, 1]$ .

2.1

Step 1.1 is exactly Lusin's conclusion, while step 1.2 shows that no null set deletion can make $D$ continuous at the remaining points as a function on $[0, 1]$ . So the stated reading of Lusin's theorem is false.

step 1.1step 1.2∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

A uniformly integrable family

Definition

Let $(X, A, μ)$ be a measure space. A family $F \subseteq L^{1} (μ)$ of integrable real-valued functions is uniformly integrable when $\sup_{f \in F} \int_{{∣ f ∣ > M}} ∣ f ∣ d μ \to 0 as M \to \infty .$

Equivalently, for every $ε > 0$ there is $M > 0$ such that $f \in F ⟹ \int_{{∣ f ∣ > M}} ∣ f ∣ d μ < ε .$

This page adopts the tail-integral definition. On finite measure spaces it is equivalent to $L^{1}$ -boundedness plus uniform absolute continuity, proved later on this page.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

On a finite measure space, uniform integrability is equivalent to L^1-boundedness plus uniform absolute continuity

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ , and let $F \subseteq L^{1} (μ)$ be a family of integrable real-valued functions. Then the following are equivalent:

$F$ is uniformly integrable;
$\sup_{f \in F} \int ∣ f ∣ d μ < + \infty$ and for every $ε > 0$ there is $δ > 0$ such that $μ (E) < δ \Rightarrow \sup_{f \in F} \int_{E} ∣ f ∣ d μ < ε$ .

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ and a family $F \subseteq L^{1} (μ)$ .

[L1]

Uniform integrability means that for every $ε > 0$ there is $M > 0$ such that $f \in F \Rightarrow \int_{{∣ f ∣ > M}} ∣ f ∣ d μ < ε$ . (A uniformly integrable family)

Proof

technique · direct

1.1

Assume $F$ is uniformly integrable. Applying [L1] with $ε : = 1$ gives $M > 0$ such that $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ < 1$ for every $f \in F$ . Then $\int ∣ f ∣ d μ \leq \int_{{∣ f ∣ \leq M}} ∣ f ∣ d μ + \int_{{∣ f ∣ > M}} ∣ f ∣ d μ \leq M μ (X) + 1,$ so $F$ is $L^{1}$ -bounded.

L1algebra

1.2

Conversely, assume $F$ is $L^{1}$ -bounded by some constant $C$ , and assume the stated uniform absolute continuity. Let $ε > 0$ , and choose $δ > 0$ such that $μ (E) < δ \Rightarrow \sup_{f \in F} \int_{E} ∣ f ∣ d μ < ε$ . Choose $M > C / δ$ . For $f \in F$ , put $E_{f} : = {∣ f ∣ > M}$ . Then $M μ (E_{f}) \leq \int ∣ f ∣ d μ \leq C$ , so $μ (E_{f}) < δ$ and hence $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ = \int_{E_{f}} ∣ f ∣ d μ < ε .$ Since this bound is uniform in $f$ , [L1] holds.

L1algebra

2.1

Still under step 1.1, let $ε > 0$ . Use [L1] with $ε / 2$ to choose $M > 0$ such that $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ < ε / 2$ for every $f \in F$ , and put $δ : = ε / (2 M + 1)$ . If $μ (E) < δ$ , then for every $f \in F$ , $\int_{E} ∣ f ∣ d μ \leq \int_{E \cap {∣ f ∣ \leq M}} ∣ f ∣ d μ + \int_{{∣ f ∣ > M}} ∣ f ∣ d μ \leq M μ (E) + ε / 2 < ε .$ So $F$ has the stated uniform absolute continuity.

step 1.1L1algebra

3.1

Steps 1.1 and 2.1 prove that uniform integrability implies clause 2, and step 1.2 proves the converse implication.

step 1.1step 2.1step 1.2∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Dominated families are uniformly integrable

Statement

Let $(X, A, μ)$ be a measure space. If $F \subseteq L^{1} (μ)$ and there is a nonnegative function $g \in L^{1} (μ)$ such that $∣ f ∣ \leq g$ almost everywhere for every $f \in F$ , then $F$ is uniformly integrable.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , a family $F \subseteq L^{1} (μ)$ , and a nonnegative integrable function $g$ with $∣ f ∣ \leq g$ almost everywhere for every $f \in F$ .

[L1]

If $g \in L^{1} (μ)$ and $ε > 0$ , then there is $δ > 0$ such that $μ (E) < δ \Rightarrow \int_{E} g d μ < ε$ . (Absolute continuity of the integral)

[L2]

If $h : X \to [0, + \infty]$ is measurable and $t > 0$ , then $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ . (Chebyshev-Markov inequality for the integral)

[L3]

If two integrable functions are equal almost everywhere, then their integrals over every measurable set agree. (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree)

[L4]

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)

[L5]

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

Proof

technique · direct

1.1

Let $ε > 0$ . Use [L1] for $g$ to choose $δ > 0$ such that $μ (E) < δ$ implies $\int_{E} g d μ < ε$ . Since $g \in L^{1} (μ)$ , choose $M > δ^{- 1} \int g d μ$ . Then [L2] applied to $h : = g$ gives $μ ({g \geq M}) \leq M^{- 1} \int g d μ < δ,$ and so [L1] yields $\int_{{g \geq M}} g d μ < ε .$

L1L2choosealgebra

2.1

Fix $f \in F$ , and choose a measurable null set $N_{f}$ such that $∣ f ∣ \leq g$ on $X ∖ N_{f}$ . Put $u_{f} : = ∣ f ∣ χ_{X ∖ N_{f}} .$ Then $u_{f}$ is integrable, $u_{f} = ∣ f ∣$ almost everywhere, and $u_{f} χ_{{∣ f ∣ > M}} \leq g χ_{{g \geq M}}$ pointwise. Therefore [L3], [L4], and [L5] give $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ = \int_{{∣ f ∣ > M}} u_{f} d μ \leq \int_{{g \geq M}} g d μ < ε .$ Since the same $M$ works for every $f \in F$ , this is exactly uniform integrability.

step 1.1L3L4L5construct

3.1

The family $F$ is uniformly integrable.

step 2.1L1∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

FALSE: uniform integrability implies domination by one integrable function

Statement refuted

uniform integrability implies domination by one integrable function.

Facts & Assumptions

Given: On $[0, 1]$ with Lebesgue measure, the pairwise disjoint intervals $I_{k} : = [a_{k}, a_{k} + \frac{1}{2 k^{2}})$ , where $a_{k} : = \sum_{j = 1}^{k - 1} \frac{1}{2 j^{2}}$ , and the functions $f_{k} : = k χ_{I_{k}}$ .

[L1]

Uniform integrability means that for every $ε > 0$ there is $M > 0$ such that $\int_{{∣ f_{k} ∣ > M}} ∣ f_{k} ∣ d μ < ε$ for every $k$ . (A uniformly integrable family)

Refutation

technique · direct

1.1

The intervals $I_{k}$ are pairwise disjoint and lie in $[0, 1]$ because $\sum_{k = 1}^{\infty} \frac{1}{2 k^{2}} < 1$ . Also $\int ∣ f_{k} ∣ d λ = k λ (I_{k}) = \frac{1}{2 k} .$

L1algebra

2.1

If $M > 0$ and $k \leq M$ , then ${∣ f_{k} ∣ > M} = \emptyset$ ; if $k > M$ , then ${∣ f_{k} ∣ > M} = I_{k}$ and $\int_{{∣ f_{k} ∣ > M}} ∣ f_{k} ∣ d λ = \frac{1}{2 k} \leq \frac{1}{2 ⌊ M ⌋ + 2} .$ Therefore the family $(f_{k})$ is uniformly integrable by [L1].

step 1.1L1algebra

2.2

If an integrable function $g$ satisfied $g \geq f_{k}$ almost everywhere for every $k$ , then $∣ g ∣ \geq g \geq k$ almost everywhere on $I_{k}$ . Since the intervals are pairwise disjoint, $\int ∣ g ∣ d λ \geq \sum_{k = 1}^{\infty} \int_{I_{k}} ∣ g ∣ d λ \geq \sum_{k = 1}^{\infty} k λ (I_{k}) = \sum_{k = 1}^{\infty} \frac{1}{2 k} = + \infty,$ contradicting integrability of $g$ . Hence no single integrable majorant exists.

step 1.1algebra

3.1

This uniformly integrable family is not dominated by any integrable function, so the claim is false.

step 2.1step 2.2∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

A tight family of integrable functions

Definition

Let $(X, A, μ)$ be a measure space. A family $F \subseteq L^{1} (μ)$ of integrable real-valued functions is tight when for every $ε > 0$ there is a measurable set $E \in A$ with $μ (E) < + \infty$ such that

$\sup_{f \in F} \int_{X ∖ E} ∣ f ∣ d μ < ε .$

On a finite measure space every family is automatically tight by taking $E = X$ , so tightness matters only on infinite spaces.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Vitali convergence theorem on finite and sigma-finite measure spaces

Statement

Let $(X, A, μ)$ be a measure space, let $f_{n} : X \to R$ be measurable with $f_{n} \in L^{1} (μ)$ for every $n$ , and let $f : X \to R$ be measurable.

If $μ (X) < + \infty$ , then $f_{n} \to f$ in $L^{1} (μ)$ if and only if $f_{n} \to f$ in measure and the family ${f_{n} : n \in N}$ is uniformly integrable.
If $μ$ is sigma-finite and $f_{n} \to f$ in measure, and the family ${f_{n} : n \in N}$ is uniformly integrable and tight, then $f \in L^{1} (μ)$ and $f_{n} \to f$ in $L^{1} (μ)$ .

Facts & Assumptions

Given: Integrable real-valued functions $f_{n} \in L^{1} (μ)$ and a measurable $f : X \to R$ .

[L1]

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

[L2]

Convergence in $L^{1} (μ)$ means $\int ∣ f_{n} - f ∣ d μ \to 0$ . (Convergence in L^1(mu))

[L3]

On a finite measure space, uniform integrability is equivalent to $L^{1}$ -boundedness plus uniform absolute continuity. (On a finite measure space, uniform integrability is equivalent to L^1-boundedness plus uniform absolute continuity)

[L4]

Convergence in $L^{1} (μ)$ implies convergence in measure. (Convergence in L^1(mu) implies convergence in measure)

[L5]

Convergence in measure has a subsequence converging almost everywhere. (Riesz's subsequence theorem for convergence in measure)

[L6]

If $g \in L^{1} (μ)$ and $ε > 0$ , then there is $δ > 0$ such that $μ (E) < δ \Rightarrow \int_{E} ∣ g ∣ d μ < ε$ . (Absolute continuity of the integral)

[L7]

For nonnegative measurable functions $u_{n}$ , $\int {lim inf}_{n} u_{n} d μ \leq {lim inf}_{n} \int u_{n} d μ$ . (Fatou's lemma)

[L8]

A family is tight when for every $ε > 0$ there is a measurable set $E$ of finite measure such that $\sup_{n} \int_{X ∖ E} ∣ f_{n} ∣ d μ < ε$ . (A tight family of integrable functions)

Proof

technique · direct

1.1

Assume $μ (X) < + \infty$ and $f_{n} \to f$ in $L^{1} (μ)$ . By [L4], the sequence converges to $f$ in measure. To prove uniform integrability, it suffices by [L3] to show $L^{1}$ -boundedness and uniform absolute continuity. For $L^{1}$ -boundedness, choose $N$ so that $\int ∣ f_{n} - f ∣ d μ < 1$ for $n \geq N$ . Then $\int ∣ f_{n} ∣ d μ \leq \int ∣ f ∣ d μ + 1$ for $n \geq N$ , and the finitely many initial terms may be absorbed into one larger bound.

L2L3L4algebra

1.2

Still on a finite measure space, assume $f_{n} \to f$ in measure and the family ${f_{n}}$ is uniformly integrable. By [L3], the family is $L^{1}$ -bounded and uniformly absolutely continuous on small sets. By [L5], some subsequence $f_{n_{k}}$ converges to $f$ almost everywhere. Fatou's lemma [L7] then gives $\int ∣ f ∣ d μ \leq {lim inf}_{k \to \infty} \int ∣ f_{n_{k}} ∣ d μ < + \infty,$ so $f \in L^{1} (μ)$ .

L3L5L7algebra

2.1

Still under step 1.1, let $ε > 0$ . Choose $N$ so that $\int ∣ f_{n} - f ∣ d μ < ε / 2$ for $n \geq N$ . Apply [L6] to $f$ and to the finitely many functions $f_{0}, \dots, f_{N - 1}$ , and let $δ$ be the minimum of the resulting positive numbers. Then $μ (E) < δ$ implies $\int_{E} ∣ f_{n} ∣ d μ < ε$ for every $n < N$ , while for $n \geq N$ one has $\int_{E} ∣ f_{n} ∣ d μ \leq \int_{E} ∣ f ∣ d μ + \int ∣ f_{n} - f ∣ d μ < ε .$ So [L3] yields uniform integrability.

L2L3L6choosealgebra

2.2

Let $ε > 0$ . Choose $δ > 0$ such that $μ (E) < δ$ implies both $\sup_{n} \int_{E} ∣ f_{n} ∣ d μ < ε / 3$ and $\int_{E} ∣ f ∣ d μ < ε / 3$ ; the first uses step 1.2 together with [L3], and the second uses [L6]. Choose $η > 0$ with $η μ (X) < ε / 3$ . Since $f_{n} \to f$ in measure, there is $N$ such that $μ (B_{n}) < δ$ for $n \geq N$ , where $B_{n} : = {∣ f_{n} - f ∣ > η}$ . Then for $n \geq N$ , $\int ∣ f_{n} - f ∣ d μ \leq \int_{B_{n}} ∣ f_{n} ∣ d μ + \int_{B_{n}} ∣ f ∣ d μ + \int_{X ∖ B_{n}} η d μ < ε .$ So $f_{n} \to f$ in $L^{1} (μ)$ .

step 1.2L1L2L3L6choosealgebra

3.1

Assume now that $μ$ is sigma-finite, that $f_{n} \to f$ in measure, and that the family ${f_{n}}$ is uniformly integrable and tight. Let $ε > 0$ . By [L8], choose a measurable set $E$ with $μ (E) < + \infty$ and $\sup_{n} \int_{X ∖ E} ∣ f_{n} ∣ d μ < ε / 4.$ Because $f_{n} \to f$ in measure on $X$ , the restricted sequence converges in measure on $E$ as well, and uniform integrability persists on $E$ . Therefore step 2.2 applied on the finite measure space $E$ gives $f ∣_{E} \in L^{1} (μ)$ and $\int_{E} ∣ f_{n} - f ∣ d μ \to 0$ . By [L5], after passing to a subsequence if necessary we may assume $f_{n_{k}} \to f$ almost everywhere on $X$ . Applying [L7] to the nonnegative functions $∣ f_{n_{k}} ∣ 1_{X ∖ E}$ gives $\int_{X ∖ E} ∣ f ∣ d μ \leq {lim inf}_{k \to \infty} \int_{X ∖ E} ∣ f_{n_{k}} ∣ d μ \leq ε / 4.$ Hence $f \in L^{1} (μ)$ . For all large $n$ one has $\int_{E} ∣ f_{n} - f ∣ d μ < ε / 2$ , and then $\int ∣ f_{n} - f ∣ d μ \leq \int_{E} ∣ f_{n} - f ∣ d μ + \int_{X ∖ E} ∣ f_{n} ∣ d μ + \int_{X ∖ E} ∣ f ∣ d μ < ε .$ So $f_{n} \to f$ in $L^{1} (μ)$ .

step 2.2L5L7L8choosealgebra

4.1

Step 1.1 proves the forward finite implication, steps 1.2 and 2.2 prove the reverse finite implication, and step 3.1 proves the sigma-finite tight form.

step 1.1step 2.1step 1.2step 2.2step 3.1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Dominated convergence is a Vitali corollary

Statement

Let $(X, A, μ)$ be a sigma-finite measure space. Let $f_{n} : X \to R$ be measurable, let $g \in L^{1} (μ)$ be nonnegative, and suppose $∣ f_{n} ∣ \leq g$ almost everywhere for every $n$ and $f_{n} \to f$ almost everywhere. Then $f \in L^{1} (μ)$ and $f_{n} \to f$ in $L^{1} (μ)$ .

Facts & Assumptions

Given: A sigma-finite measure space $(X, A, μ)$ , measurable functions $f_{n}, f : X \to R$ , and a nonnegative integrable function $g$ with $∣ f_{n} ∣ \leq g$ almost everywhere and $f_{n} \to f$ almost everywhere.

[L1]

A dominated family is uniformly integrable. (Dominated families are uniformly integrable)

[L2]

On a sigma-finite measure space, convergence in measure together with uniform integrability and tightness implies convergence in $L^{1}$ . (Vitali convergence theorem on finite and sigma-finite measure spaces)

[L3]

On a finite measure space, almost-everywhere convergence implies convergence in measure. (On a finite measure space, almost-everywhere convergence implies convergence in measure)

[L4]

If $h : X \to [0, + \infty]$ is measurable and $t > 0$ , then $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ . (Chebyshev-Markov inequality for the integral)

[L5]

For an increasing sequence of nonnegative measurable functions, the integrals increase to the integral of the limit. (Monotone convergence for the integral)

[L6]

[L7]

Sigma-finiteness means that $X = ⋃_{m} X_{m}$ for some measurable sets $X_{m}$ of finite measure. (Finite, sigma-finite, and semifinite measures)

[L8]

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)

[L9]

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

[L10]

Countable unions of null sets are null. (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

Let $N_{*}$ be a null set outside which $f_{n} (x) \to f (x)$ , and for each $n \in N$ let $N_{n}$ be a null set outside which $∣ f_{n} ∣ \leq g$ . Put $N : = N_{*} \cup ⋃_{n = 0}^{\infty} N_{n} .$ By [L10], the set $N$ is null. On $X ∖ N$ one has $f_{n} (x) \to f (x)$ for every $x$ and $∣ f_{n} (x) ∣ \leq g (x)$ for every $n$ , so also $∣ f (x) ∣ \leq g (x)$ there.

L10construct

1.2

By [L1], the family ${f_{n} : n \in N}$ is uniformly integrable.

2.1

By [L7], choose measurable sets $Y_{m}$ of finite measure with $X = ⋃_{m} Y_{m}$ , and put $X_{m} : = ⋃_{j = 0}^{m} Y_{j}$ . Then $X_{m}$ is measurable of finite measure, $X_{m} ↑ X$ , and hence $g χ_{X_{m}} ↑ g$ . Fact [L5] gives $\int_{X_{m}} g d μ ↑ \int g d μ$ , so $\int_{X ∖ X_{m}} g d μ \to 0$ . Fix $ε > 0$ and choose $m$ with $\int_{X ∖ X_{m}} g d μ < ε$ . For each $n$ , put $u_{n, m} : = ∣ f_{n} ∣ χ_{(X ∖ X_{m}) ∖ N}$ . Then $u_{n, m} = ∣ f_{n} ∣ χ_{X ∖ X_{m}}$ almost everywhere and $u_{n, m} \leq g χ_{X ∖ X_{m}}$ pointwise, so [L6], [L8], and [L9] give $\int_{X ∖ X_{m}} ∣ f_{n} ∣ d μ = \int_{X ∖ X_{m}} u_{n, m} d μ \leq \int_{X ∖ X_{m}} g d μ < ε .$ So the family ${f_{n}}$ is tight.

step 1.1L5L6L7L8L9algebra

2.2

Let $ε, η > 0$ . Choose $m$ so that $\int_{X ∖ X_{m}} g d μ < ε η / 8$ . On the finite measure space $X_{m}$ , step 1.1 and [L3] make $f_{n} \to f$ in measure. Define $h_{n} : = ∣ f_{n} - f ∣ χ_{(X ∖ X_{m}) ∖ N}$ . Then ${h_{n} > ε} = {∣ f_{n} - f ∣ > ε} \cap ((X ∖ X_{m}) ∖ N),$ and $h_{n} \leq 2 g χ_{X ∖ X_{m}}$ pointwise. Since $N$ is null, $μ ({∣ f_{n} - f ∣ > ε} \cap (X ∖ X_{m})) = μ ({h_{n} > ε}) .$ Therefore [L4], [L8], and [L9] give $μ ({∣ f_{n} - f ∣ > ε} \cap (X ∖ X_{m})) \leq ε^{- 1} \int_{X ∖ X_{m}} h_{n} d μ \leq \frac{2}{ε} \int_{X ∖ X_{m}} g d μ < η / 4.$ Combining this with convergence in measure on $X_{m}$ shows $f_{n} \to f$ in measure on all of $X$ .

step 1.1L3L4L8L9choosealgebra

3.1

Step 1.2 gives uniform integrability, step 2.1 gives tightness, and step 2.2 gives convergence in measure. Applying [L2] yields $f \in L^{1} (μ)$ and $f_{n} \to f$ in $L^{1} (μ)$ .

step 1.2step 2.1step 2.2L2∎

RemarkRemark: AI-adaptedProof: Not applicableaudited 2026-08-28Open item page →

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

On a finite measure space, the page establishes the following implication picture for the four main convergence modes it develops.

from	to	status	reason or witness
almost uniform	almost everywhere	proved	Almost uniform convergence implies almost-everywhere convergence and convergence in measure
almost uniform	in measure	proved	Almost uniform convergence implies almost-everywhere convergence and convergence in measure
almost everywhere	in measure	proved	On a finite measure space, almost-everywhere convergence implies convergence in measure
in $L^{1}$	in measure	proved	Convergence in L^1(mu) implies convergence in measure
in measure + uniform integrability	in $L^{1}$	proved	Vitali convergence theorem on finite and sigma-finite measure spaces
in measure	almost everywhere	false	the typewriter sequence on $[0, 1]$
in $L^{1}$	almost everywhere	false	the same typewriter sequence
in measure	in $L^{1}$	false	the spikes $f_{0} = 0$ and $f_{n} = n χ_{(0, 1 / n)}$ for $n \geq 1$
almost everywhere	in $L^{1}$	false	the same spike sequence

Two companion observations matter just as much as the table.

Convergence in measure on a finite measure space is exactly convergence in the truncated metric of On a finite measure space, the truncated L^1 metric metrises convergence in measure.
Even when convergence in measure does not imply almost-everywhere convergence of the whole sequence, Riesz gives an almost-everywhere convergent subsequence and, on finite measure spaces, Egorov upgrades that subsequence to almost uniform convergence (On a finite measure space, convergence in measure has an almost-uniformly convergent subsequence).

The companion examples page still spells out the same concrete witnesses: the typewriter sequence, the translated unit intervals on $R$ , the spike family $f_{0} = 0$ , $f_{n} = n χ_{(0, 1 / n)}$ for $n \geq 1$ , the explicit Egorov core for $x^{k}$ , the Dirichlet-function Lusin core, and the uniformly integrable but non-dominated disjoint-spike family.

5 · Examples, counterexamples and false statements

None yet.