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✓ 35 results · all verified · 19 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 16 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Measures and Their Basic Properties

1 · Prerequisites

2 · Summary

Sigma-algebras provide the measurable domains, while generated sigma-algebras, pi-systems, lambda-systems and Dynkin's theorem provide the machinery used to prove uniqueness. Set liminf and limsup describe eventual and repeated membership. Together with the Borel sigma-algebra, these prerequisite structures support countable set operations without requiring the underlying set itself to be finite or countable.

Nonnegative extended sums lead to measures, probability measures, nullity, completeness and the finite, sigma-finite and semifinite conditions. Monotonicity and subadditivity yield continuity under monotone set limits, inclusion-exclusion, Borel-Cantelli and the set-limit inequalities. Completion and restriction enlarge or localize a measure, weighted Dirac sums classify atomic cases, semifinite parts separate finite-measure information, and finite uniqueness extends along a pi-system exhaustion.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Series in the nonnegative extended real line

Definition

Let $a = (a_{k})_{k \in N}$ take values in $[0, + \infty] \subseteq \overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). Its partial sums are the unique sequence $(s_{n})$ in $[0, + \infty]$ satisfying

$s_{0} = 0, s_{n + 1} = s_{n} + a_{n} .$

To apply The recursion theorem with a fixed successor function, use the state space $N \times [0, + \infty]$ and the self-map $T (n, s) = (n + 1, s + a_{n})$ , starting from $(0, 0)$ . Recursion gives a unique state sequence; induction makes its first coordinate $n$ , and its second coordinates are exactly the unique $(s_{n})$ satisfying the displayed recurrence. Addition of two nonnegative extended reals is always defined, including when either is $+ \infty$ . The sequence $(s_{n})$ is nondecreasing, and its nonnegative extended sum is

$\sum_{k = 0}^{\infty} a_{k} : = \sup_{n \in N} s_{n} \in [0, + \infty],$

whose existence follows from completeness of the extended real line (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ). More generally, for $m \in N$ ,

$\sum_{k \geq m} a_{k} : = \sum_{j = 0}^{\infty} a_{m + j} .$

Finite sums use the same recursion: $\sum_{k < n} a_{k} = s_{n}$ , so the empty sum at $n = 0$ is $0$ . A double sum such as $\sum_{i} \sum_{j} a_{i j}$ means that the inner nonnegative extended sum is formed first and the resulting nonnegative extended sequence is then summed.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Limit superior and limit inferior of a nonnegative extended-real sequence

Definition

Let $(a_{n})_{n \in N}$ be a sequence in $[0, + \infty] \subseteq \overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). For each $N \in N$ , completeness of $\overline{R}$ (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ) gives the tail bounds

$u_{N} : = \sup_{k \geq N} a_{k}, ℓ_{N} : = \inf_{k \geq N} a_{k} .$

The limit superior and limit inferior of $(a_{n})$ are

${lim sup}_{n \to \infty} a_{n} : = \inf_{N \in N} u_{N}, {lim inf}_{n \to \infty} a_{n} : = \sup_{N \in N} ℓ_{N} .$

All four suprema and infima exist in $\overline{R}$ , including when some terms are $+ \infty$ . This definition is therefore distinct from the real-sequence definition: it applies directly to sequences of measure values.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Tonelli's theorem for double series of nonnegative extended real numbers

Statement

For every double sequence $(a_{i j})_{i, j \in N}$ in $[0, + \infty]$ ,

$\sum_{i = 0}^{\infty} (\sum_{j = 0}^{\infty} a_{i j}) = \sup_{m, n \in N} \sum_{i < m} \sum_{j < n} a_{i j} = \sum_{j = 0}^{\infty} (\sum_{i = 0}^{\infty} a_{i j}),$

where every sum is the nonnegative extended sum of Series in the nonnegative extended real line. Thus the order of summation may be interchanged, even when the common value is $+ \infty$ .

Facts & Assumptions

Given: A double sequence $(a_{i j})_{i, j \in N}$ with $0 \leq a_{i j} \leq + \infty$ .

[L1]

For a nonnegative extended sequence, the partial sums start at the empty sum $0$ , increase, and the series is their supremum in $[0, + \infty]$ (Series in the nonnegative extended real line).

[L2]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there, with $\sup \emptyset = - \infty$ and $\inf \emptyset = + \infty$ (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

[L3]

A natural-number-indexed finite family of nonempty sets has a choice function in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1givenL1

Put $r_{i, n} : = \sum_{j < n} a_{i j}$ and $R_{m, n} : = \sum_{i < m} r_{i, n}$ . These finite sums exist for $m, n \in N$ , $R_{0, n} = R_{m, 0} = 0$ , and $R_{m, n}$ is nondecreasing in each index.

1.2L2

The set ${R_{m, n} : m, n \in N}$ is nonempty and bounded above by $+ \infty$ , so let $S : = \sup_{m, n} R_{m, n} \in [0, + \infty]$ .

2.1step 1.1L1L2L3choose

For fixed $m$ , $\sup_{n} R_{m, n} = \sum_{i < m} \sup_{n} r_{i, n}$ : the inequality $\leq$ follows from $r_{i, n} \leq \sup_{q} r_{i, q}$ ; for the reverse inequality, if one of the finitely many row suprema is $+ \infty$ then its partial sums make $R_{m, n}$ unbounded, while if they are all finite, for every $ε > 0$ finite choice selects for each $i < m$ an index $n_{i}$ with $r_{i, n_{i}} > \sup_{n} r_{i, n} - ε / m$ when $m > 0$ ; take the largest $n_{i}$ and use monotonicity. The case $m = 0$ is the empty equality $0 = 0$ .

3.1step 1.1step 1.2step 2.1L1L2

Repeating steps 1.1 and 2.1 with the two indices interchanged gives $\sum_{j} (\sum_{i} a_{i j}) = \sup_{n} \sup_{m} R_{m, n} = S$ .

3.2step 1.2step 2.1L1

By [L1] and step 2.1, $\sum_{i} (\sum_{j} a_{i j}) = \sup_{m} \sum_{i < m} \sup_{n} r_{i, n} = \sup_{m} \sup_{n} R_{m, n} = S$ .

4.1step 3.2step 3.1∎

Both iterated sums equal the supremum of the finite rectangular sums, so they equal one another; the argument includes zero rows, zero columns, infinite entries, and unbounded finite rectangles without subtraction or an undefined product.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Measures on sigma-algebras

Definition

Let $X$ be a set and let $A$ be a sigma-algebra on $X$ (Sigma-algebras). A measure on $(X, A)$ is a function

$μ : A ⟶ [0, + \infty] \subseteq \overline{R}$

such that:

$μ (\emptyset) = 0$ ;
for every pairwise disjoint sequence $(E_{k})_{k \in N}$ in $A$ , $μ (⋃_{k \in N} E_{k}) = \sum_{k = 0}^{\infty} μ (E_{k}),$ where the right side is the nonnegative extended sum of Series in the nonnegative extended real line.

The second condition is countable additivity. It begins at index $0$ and includes the case in which some term or the total sum is $+ \infty$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Measure spaces

Definition

A measure space is a triple $(X, A, μ)$ in which $X$ is a set, $A$ is a sigma-algebra on $X$ , and $μ$ is a measure on $(X, A)$ (Measures on sigma-algebras). The members of $A$ are the measurable sets of the space.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Finitely additive nonnegative set functions

Definition

Let $A$ be a sigma-algebra on $X$ (Sigma-algebras). A finitely additive nonnegative set function is a function $ϕ : A \to [0, + \infty]$ such that $ϕ (\emptyset) = 0$ and

$ϕ (A \cup B) = ϕ (A) + ϕ (B)$

whenever $A, B \in A$ are disjoint. By induction, the same equality holds for every finite pairwise disjoint family, with the empty union and empty sum both equal to $0$ . No countable-additivity condition is part of this definition.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Finite, sigma-finite, and semifinite measures

Definition

Let $(X, A, μ)$ be a measure space (Measure spaces).

The measure $μ$ is finite if $μ (X) < + \infty$ .
It is sigma-finite if there is a sequence $(E_{n})_{n \in N}$ in $A$ such that $X = ⋃_{n} E_{n}$ and $μ (E_{n}) < + \infty$ for every $n$ .
It is semifinite if, whenever $E \in A$ and $μ (E) > 0$ , there is $F \in A$ with $F \subseteq E$ and $0 < μ (F) < + \infty$ .

The last condition is automatic when $0 < μ (E) < + \infty$ , by taking $F = E$ ; its substantive case is $μ (E) = + \infty$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Probability measures and probability spaces

Definition

A probability measure on a measurable space $(X, A)$ is a measure $P$ with $P (X) = 1$ (Measures on sigma-algebras). The triple $(X, A, P)$ is a probability space (Measure spaces); $X$ is its sample space and the members of $A$ are its events.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Measure-null sets and almost-everywhere statements relative to a measure

Definition

In a measure space $(X, A, μ)$ (Measure spaces), a measurable set $N \in A$ is $μ$ -null if $μ (N) = 0$ .

A property $P (x)$ holds $μ$ -almost everywhere, or for $μ$ -almost every $x$ , if its exceptional set is contained in a measurable $μ$ -null set: there is $N \in A$ with $μ (N) = 0$ such that $P (x)$ holds for every $x \in X ∖ N$ . Both notions are relative to the named measure $μ$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Complete measure spaces

Definition

A measure space $(X, A, μ)$ is complete if every subset of every measurable $μ$ -null set is measurable: whenever $N \in A$ , $μ (N) = 0$ , and $S \subseteq N$ , one has $S \in A$ (Measure-null sets and almost-everywhere statements relative to a measure).

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Measures are monotone

Statement

Let $μ$ be a measure on $(X, A)$ . If $A, B \in A$ and $A \subseteq B$ , then $μ (A) \leq μ (B)$ .

Facts & Assumptions

Given: A measure $μ$ on $(X, A)$ and measurable sets $A \subseteq B$ .

[L1]

A measure is nonnegative and countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

Proof

technique · direct

1.1given

The sets $A$ and $B ∖ A$ are measurable, disjoint, and have union $B$ .

2.1step 1.1L1algebra∎

Countable additivity applied to these two sets and empty sets thereafter gives $μ (B) = μ (A) + μ (B ∖ A) \geq μ (A)$ ; no subtraction is used, so the argument also covers $μ (B) = + \infty$ and the degenerate cases $A = \emptyset$ and $A = B$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Measure of a set difference when the smaller set has finite measure

Statement

Let $μ$ be a measure and let $A \subseteq B$ be measurable with $μ (A) < + \infty$ . Then

$μ (B) = μ (A) + μ (B ∖ A) .$

If $μ (B) < + \infty$ , all terms are real and hence $μ (B ∖ A) = μ (B) - μ (A)$ . If $μ (B) = + \infty$ , then $μ (B ∖ A) = + \infty$ .

Facts & Assumptions

Given: Measurable sets $A \subseteq B$ for a measure $μ$ , with $μ (A) < + \infty$ .

[L1]

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

If $C \subseteq D$ are measurable, then $μ (C) \leq μ (D)$ (Measures are monotone).

Proof

technique · direct

1.1givenL1

The disjoint measurable sets $A$ and $B ∖ A$ have union $B$ , so $μ (B) = μ (A) + μ (B ∖ A)$ .

2.1givenstep 1.1L2algebra

If $μ (B) < + \infty$ , monotonicity makes both summands in step 1.1 finite, and cancellation in $R$ gives $μ (B ∖ A) = μ (B) - μ (A)$ .

3.1givenstep 1.1step 2.1algebra∎

If $μ (B) = + \infty$ , then $μ (B ∖ A)$ cannot be finite, because its sum with the finite number $μ (A)$ would be finite; hence it is $+ \infty$ . Together with steps 1.1 and 2.1 this proves every asserted case, including $A = \emptyset$ and $A = B$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Finite and countable subadditivity of measures

Statement

Let $μ$ be a measure and let $(E_{k})_{k \in N}$ be measurable. Then

$μ (⋃_{k \in N} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k}) .$

For every $m \in N$ one also has

$μ (⋃_{k < m} E_{k}) \leq \sum_{k < m} μ (E_{k}),$

including $m = 0$ , where both sides are $0$ .

Facts & Assumptions

Given: A measure $μ$ and a sequence $(E_{k})$ of measurable sets.

[L1]

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

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

[L3]

A nonnegative extended series is the supremum of its finite partial sums, beginning with the empty sum $0$ (Series in the nonnegative extended real line).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1given

Define $F_{k} : = E_{k} ∖ ⋃_{j < k} E_{j}$ . Then every $F_{k}$ is measurable, the $F_{k}$ are pairwise disjoint, and $F_{k} \subseteq E_{k}$ .

1.2givenL4

The unions of the two sequences agree: if $x \in ⋃_{k} E_{k}$ , then the nonempty set ${k : x \in E_{k}}$ has a least member $r$ , and the definition gives $x \in F_{r}$ ; the reverse inclusion follows from $F_{k} \subseteq E_{k}$ .

2.1step 1.1step 1.2L1L2L3

Countable additivity, monotonicity, and the definition of a nonnegative series give $μ (⋃_{k} E_{k}) = \sum_{k} μ (F_{k}) \leq \sum_{k} μ (E_{k})$ .

3.1step 2.1L1L3∎

For $m \in N$ , apply step 2.1 to the sequence $E_{0}, \dots, E_{m - 1}, \emptyset, \emptyset, \dots$ ; its union and sum are the displayed finite union and finite sum, and when $m = 0$ they are both empty and equal to $0$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Continuity from below for measures

Statement

Let $(E_{n})_{n \in N}$ be an increasing sequence of measurable sets for a measure $μ$ , so $E_{n} \subseteq E_{n + 1}$ . Then

$μ (⋃_{n \in N} E_{n}) = \sup_{n \in N} μ (E_{n}) .$

No finiteness hypothesis is required.

Facts & Assumptions

Given: A measure $μ$ and measurable sets $E_{0} \subseteq E_{1} \subseteq \dots$ ; write $E = ⋃_{n} E_{n}$ .

[L1]

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

If $A \subseteq B$ are measurable and $μ (A) < + \infty$ , then $μ (B) = μ (A) + μ (B ∖ A)$ , with the corresponding finite and infinite cases stated explicitly (Measure of a set difference when the smaller set has finite measure).

[L3]

A nonnegative extended series is the supremum of its finite partial sums (Series in the nonnegative extended real line).

[L4]

Every subset of $\overline{R}$ has a supremum there (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

Proof

technique · direct

1.1given

Define $D_{0} : = E_{0}$ and $D_{k + 1} : = E_{k + 1} ∖ E_{k}$ . The sets $D_{k}$ are measurable and pairwise disjoint, $E_{n} = ⋃_{k < n + 1} D_{k}$ , and $E = ⋃_{k} D_{k}$ .

2.1step 1.1L1

Countable additivity gives $μ (E) = \sum_{k} μ (D_{k})$ .

2.2step 1.1L1L4

If $μ (E_{n}) = + \infty$ for some $n$ , then $μ (E) = + \infty$ by step 1.1 and [L1], while $\sup_{k} μ (E_{k}) = + \infty$ because that value occurs.

2.3step 1.1L2algebra

If every $μ (E_{n})$ is finite, then [L2] gives $μ (D_{0}) = μ (E_{0})$ and $μ (D_{k + 1}) = μ (E_{k + 1}) - μ (E_{k})$ ; hence the first $n + 1$ terms telescope to $μ (E_{n})$ .

3.1step 2.1step 2.2step 2.3L3∎

In the finite-valued case, [L3], steps 2.1 and 2.3 give $μ (E) = \sup_{n} μ (E_{n})$ ; step 2.2 gives the same equality in the remaining case, including $E_{0} = \emptyset$ and a sequence that stabilizes.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Continuity from above when one set has finite measure

Statement

Let $(E_{n})_{n \in N}$ be a decreasing sequence of measurable sets for a measure $μ$ . If $μ (E_{n_{0}}) < + \infty$ for some $n_{0}$ , then

$μ (⋂_{n \in N} E_{n}) = \inf_{n \in N} μ (E_{n}) .$

Facts & Assumptions

Given: Measurable sets $E_{0} \supseteq E_{1} \supseteq \dots$ , an index $n_{0}$ with $μ (E_{n_{0}}) < + \infty$ , and $E = ⋂_{n} E_{n}$ .

[L1]

For increasing measurable $A_{n}$ , $μ (⋃_{n} A_{n}) = \sup_{n} μ (A_{n})$ (Continuity from below for measures).

[L2]

If $A \subseteq B$ and $μ (A) < + \infty$ , then $μ (B) = μ (A) + μ (B ∖ A)$ , with real subtraction valid when $μ (B) < + \infty$ (Measure of a set difference when the smaller set has finite measure).

[L3]

Every subset of $\overline{R}$ has an infimum there (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

Proof

technique · direct

1.1given

For $k \in N$ put $H_{k} : = E_{n_{0}} ∖ E_{n_{0} + k}$ . Then $(H_{k})$ increases and $⋃_{k} H_{k} = E_{n_{0}} ∖ E$ .

1.2givenL2

Every $E_{n_{0} + k}$ and $E$ has finite measure by inclusion in $E_{n_{0}}$ , and [L2] gives $μ (H_{k}) = μ (E_{n_{0}}) - μ (E_{n_{0} + k})$ and $μ (E_{n_{0}} ∖ E) = μ (E_{n_{0}}) - μ (E)$ .

2.1step 1.1step 1.2L1L3algebra

Apply continuity from below to $(H_{k})$ and substitute step 1.2: taking the supremum of the left differences is the same as subtracting the infimum of the decreasing finite values, so cancellation of the finite number $μ (E_{n_{0}})$ yields $μ (E) = \inf_{k} μ (E_{n_{0} + k})$ .

3.1step 2.1L3∎

A decreasing sequence has the same infimum as any of its tails, so step 2.1 gives $μ (E) = \inf_{n} μ (E_{n})$ ; this includes $E = \emptyset$ , $μ (E) = 0$ , and a sequence that is constant from $n_{0}$ onward.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Counting measure on an arbitrary set

Definition

Let $X$ be a set. The counting set function on $P (X)$ is

$#_{X} (E) : = {\begin{cases} ∣ E ∣, & E is finite, \\ + \infty, & E is infinite, \end{cases}$

where finite means equinumerous with a natural number (Finite, countably infinite, countable, uncountable) and $+ \infty$ belongs to the extended real line (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). The two branches are exhaustive and disjoint. The fact that this set function is a measure, and hence deserves the name counting measure, is proved in Counting measure is a measure ↗.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Counting measure is a measure

Statement

For every set $X$ , the counting set function $#_{X}$ of Counting measure on an arbitrary set is a measure on $(X, P (X))$ .

Facts & Assumptions

Given: A set $X$ and a pairwise disjoint sequence $(E_{k})$ of subsets of $X$ .

[L1]

Counting measure assigns a finite set its finite cardinality and an infinite set the value $+ \infty$ (Counting measure on an arbitrary set).

[L2]

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

[L3]

A nonnegative extended series is the supremum of its finite partial sums (Series in the nonnegative extended real line).

[L4]

A set is finite when it is equinumerous with a natural number, and otherwise it may be countably infinite or uncountable (Finite, countably infinite, countable, uncountable).

Proof

technique · direct

1.1givenL1

One has $#_{X} (\emptyset) = 0$ . For every $n$ , disjointness gives $#_{X} (⋃_{k < n} E_{k}) = \sum_{k < n} #_{X} (E_{k})$ whenever all those sets are finite.

1.2givenL1L3

If some $E_{r}$ is infinite, then $⋃_{k} E_{k}$ is infinite and both $#_{X} (⋃_{k} E_{k})$ and $\sum_{k} #_{X} (E_{k})$ are $+ \infty$ .

2.1givenstep 1.1L1L3L4

Suppose every $E_{k}$ is finite. If $E : = ⋃_{k} E_{k}$ is finite, only finitely many pairwise disjoint $E_{k}$ can be nonempty, and step 1.1 gives $#_{X} (E) = \sum_{k} #_{X} (E_{k})$ .

2.2givenstep 1.1L1L3L4

Suppose every $E_{k}$ is finite but $E$ is infinite. For every $m \in N$ , the set $E$ contains more than $m$ distinct points; the finitely many indices of the $E_{k}$ containing those points have a strict upper bound $n$ (take one more than their maximum), so step 1.1 gives $\sum_{k < n} #_{X} (E_{k}) > m$ . Hence the partial sums are unbounded and their supremum is $+ \infty = #_{X} (E)$ .

3.1step 1.1step 1.2step 2.1step 2.2L2∎

Steps 1.2, 2.1 and 2.2 cover all possibilities for the union, so countable additivity holds; with $#_{X} (\emptyset) = 0$ from step 1.1, [L2] proves that $#_{X}$ is a measure.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

The two-set measure identity $μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B)$

Statement

For measurable sets $A$ and $B$ in a measure space,

$μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B) .$

The equality is valid in $[0, + \infty]$ , including when one or both sides equal $+ \infty$ .

Facts & Assumptions

Given: A measure $μ$ and measurable sets $A, B$ .

[L1]

A measure is additive on every finite pairwise disjoint measurable family (Measures on sigma-algebras).

Proof

technique · direct

1.1given

Put $C = A ∖ B$ , $D = A \cap B$ , and $F = B ∖ A$ . These sets are measurable and pairwise disjoint, with $A = C \cup D$ , $B = D \cup F$ , and $A \cup B = C \cup D \cup F$ .

2.1step 1.1L1

Finite additivity gives $μ (A) = μ (C) + μ (D)$ , $μ (B) = μ (D) + μ (F)$ , and $μ (A \cup B) = μ (C) + μ (D) + μ (F)$ .

3.1step 2.1algebra∎

Adding $μ (D) = μ (A \cap B)$ to the last equality and regrouping nonnegative extended sums gives the displayed identity; no subtraction occurs, so infinite values and the cases $A = \emptyset$ , $B = \emptyset$ , or $A = B$ are included.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Inclusion-exclusion for a nonempty finite family of finite-measure sets

Statement

Let $m \geq 1$ be natural and let $A_{0}, \dots, A_{m - 1}$ be measurable sets of finite measure. Then

$μ (⋃_{i < m} A_{i}) = \sum_{\emptyset \neq J \subseteq {0, \dots, m - 1}} (- 1)^{∣ J ∣ + 1} μ (⋂_{j \in J} A_{j}) .$

The finite sum on the right uses the following recursive order. For a one-index family it lists ${0}$ . To pass from the order for the nonempty subsets of ${0, \dots, m - 1}$ to the order for those of ${0, \dots, m}$ , retain the existing list, then append ${m}$ , and then append the sets $J \cup {m}$ for nonempty $J \subseteq {0, \dots, m - 1}$ in that existing order. Thus the displayed formula for a family of size $m$ uses only subsets of ${0, \dots, m - 1}$ . This convention fixes the sum without invoking an unproved permutation rule.

Facts & Assumptions

Given: A nonempty finite list $A_{0}, \dots, A_{m - 1}$ of measurable sets, each of finite measure.

[L1]

For measurable $A, B$ , $μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B)$ (The two-set measure identity $μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B)$ ).

[L2]

The measure of a finite union is at most the sum of the member measures (Finite and countable subadditivity of measures).

[L3]

Finite sums start with the empty sum and satisfy additivity, splitting, and telescoping laws (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

Natural powers satisfy $a^{0} = 1$ and $a^{n + 1} = a^{n} a$ (Integer powers $a^{m}$ ).

[L5]

A property true at $0$ and inherited by successors holds for every natural number (The principle of mathematical induction).

Proof

technique · induction

1.1givenL5

Let $P (r)$ be the displayed inclusion-exclusion formula for a list of $r + 1$ finite-measure measurable sets. It suffices to prove $P (r)$ for every $r \in N$ .

1.2baseL3L4

For $r = 0$ , both sides of $P (0)$ are $μ (A_{0})$ , since the only nonempty subset of ${0}$ is ${0}$ and $(- 1)^{2} = 1$ .

1.3ih

Fix $r$ and assume $P (r)$ for every list of $r + 1$ such sets.

2.1givenstep 1.3L1L2algebra

Put $U = ⋃_{i < r + 1} A_{i}$ . By [L2], $μ (U) < + \infty$ , and [L1] applied to $U$ and $A_{r + 1}$ gives $μ (U \cup A_{r + 1}) = μ (U) + μ (A_{r + 1}) - μ (U \cap A_{r + 1})$ in $R$ .

2.2givenstep 1.3L2L3

The induction hypothesis expands $μ (U)$ over the nonempty subsets of ${0, \dots, r}$ and expands $μ (U \cap A_{r + 1}) = μ (⋃_{i < r + 1} (A_{i} \cap A_{r + 1}))$ over the same subsets; all intersections remain finite-measure.

3.1step 2.1step 2.2L3L4

In step 2.1, the terms from $μ (U)$ are indexed by nonempty subsets not containing $r + 1$ , the term $μ (A_{r + 1})$ is indexed by ${r + 1}$ , and the negated terms from step 2.2 are indexed in the stated recursive order by the sets $J \cup {r + 1}$ and acquire the sign $(- 1)^{∣ J ∣ + 2} = (- 1)^{∣ J \cup {r + 1} ∣ + 1}$ . Finite-sum splitting therefore gives $P (r + 1)$ .

4.1step 1.1step 1.2step 2.1step 3.1L5discharge-induction∎

By induction, $P (r)$ holds for every $r$ , hence the stated formula holds for every nonempty finite family; the one-set boundary is step 1.2, and finiteness was used exactly in step 2.1 to permit subtraction.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Sets whose symmetric difference is null have the same measure

Statement

If $A$ and $B$ are measurable and $μ (A △ B) = 0$ , where $A △ B = (A ∖ B) \cup (B ∖ A)$ , then $μ (A) = μ (B)$ .

Facts & Assumptions

Given: Measurable sets $A, B$ with $μ (A △ B) = 0$ .

[L1]

Measures are monotone: $C \subseteq D$ implies $μ (C) \leq μ (D)$ (Measures are monotone).

[L2]

For measurable $C, D$ , $μ (C \cup D) + μ (C \cap D) = μ (C) + μ (D)$ (The two-set measure identity $μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B)$ ).

Proof

technique · direct

1.1givenL1

Monotonicity gives $μ (A ∖ B) = μ (B ∖ A) = 0$ , since both sets lie in $A △ B$ .

2.1step 1.1L2

Applying the two-set identity to $A \cap B$ and $A ∖ B$ , which are disjoint and have union $A$ , gives $μ (A) = μ (A \cap B) + 0$ ; the same argument gives $μ (B) = μ (A \cap B) + 0$ .

3.1step 2.1∎

Hence $μ (A) = μ (B)$ , including when the common value $μ (A \cap B)$ is $+ \infty$ .

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Countable additivity and continuity of finitely additive set functions

Statement

Let $ϕ : A \to [0, + \infty]$ be a finitely additive nonnegative set function on a sigma-algebra. The following are equivalent:

$ϕ$ is countably additive, and hence is a measure;
whenever $E_{n} ↑ E$ , one has $ϕ (E) = \sup_{n} ϕ (E_{n})$ .

If in addition $ϕ (X) < + \infty$ , these conditions are also equivalent to:

whenever $E_{n} ↓ \emptyset$ , one has $\inf_{n} ϕ (E_{n}) = 0$ .

Facts & Assumptions

Given: A finitely additive nonnegative set function $ϕ$ on a sigma-algebra $A$ over $X$ .

[L1]

Finite additivity means $ϕ (\emptyset) = 0$ and $ϕ (A \cup B) = ϕ (A) + ϕ (B)$ for disjoint measurable $A, B$ (Finitely additive nonnegative set functions).

[L2]

Countable additivity together with the empty-set condition is exactly the definition of a measure (Measures on sigma-algebras).

[L3]

A nonnegative extended series is the supremum of its finite partial sums (Series in the nonnegative extended real line).

Proof

technique · direct

1.1given

For the implication from countable additivity to continuity from below, let $E_{n} ↑ E$ and define $D_{0} = E_{0}$ , $D_{n + 1} = E_{n + 1} ∖ E_{n}$ ; the $D_{n}$ are disjoint, their union is $E$ , and their first $n + 1$ terms have union $E_{n}$ .

1.2givenL1

For the implication from continuity from below to countable additivity, let $(A_{n})$ be disjoint and put $B_{n} = ⋃_{k < n} A_{k}$ ; then $B_{n} ↑ ⋃_{k} A_{k}$ and finite additivity gives $ϕ (B_{n}) = \sum_{k < n} ϕ (A_{k})$ .

1.3givenL1

For the finite-total-mass implications, assume $ϕ (X) < + \infty$ ; then every value of $ϕ$ is finite by finite additivity and nonnegativity.

2.1step 1.1L2L3

Under countable additivity, [L2] and the decomposition in step 1.1 give $ϕ (E) = \sum_{n} ϕ (D_{n}) = \sup_{n} ϕ (E_{n})$ , proving condition 1 implies condition 2.

2.2step 1.2L3

Under condition 2, step 1.2 and [L3] give $ϕ (⋃_{n} A_{n}) = \sup_{n} ϕ (B_{n}) = \sum_{n} ϕ (A_{n})$ , proving condition 2 implies condition 1.

2.3step 1.3L1algebra

For condition 2 implies condition 3 under finite total mass, if $E_{n} ↓ \emptyset$ then $X ∖ E_{n} ↑ X$ ; finite additivity and condition 2 give $ϕ (X) - \inf_{n} ϕ (E_{n}) = \sup_{n} ϕ (X ∖ E_{n}) = ϕ (X)$ , hence the infimum is $0$ .

2.4step 1.3L1algebra

For condition 3 implies condition 2 under finite total mass, if $E_{n} ↑ E$ then $E ∖ E_{n} ↓ \emptyset$ ; finite additivity and condition 3 give $ϕ (E) - \sup_{n} ϕ (E_{n}) = \inf_{n} ϕ (E ∖ E_{n}) = 0$ .

3.1step 2.1step 2.2step 2.3step 2.4∎

Steps 2.1 and 2.2 prove the first equivalence, while steps 2.3 and 2.4 separately prove both directions involving condition 3 under the stated finite-total-mass hypothesis.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The first Borel-Cantelli lemma for measures

Statement

Let $(E_{k})_{k \in N}$ be measurable sets in a measure space. If the nonnegative extended sum satisfies

$\sum_{k = 0}^{\infty} μ (E_{k}) < + \infty,$

then

$μ ({lim sup}_{k \to \infty} E_{k}) = 0.$

No independence hypothesis and no finiteness hypothesis on the whole space are required.

Facts & Assumptions

Given: Measurable sets $(E_{k})$ with $S : = \sum_{k} μ (E_{k}) < + \infty$ .

[L1]

Measures are monotone (Measures are monotone) and countably subadditive (Finite and countable subadditivity of measures).

[L2]

The set limsup is $⋂_{N} ⋃_{k \geq N} E_{k}$ (Limit superior and limit inferior of a sequence of sets).

[L3]

A nonnegative extended sum is the supremum of its partial sums (Series in the nonnegative extended real line), while a convergent real series is the limit of its real partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L4]

A convergent real series and each of its tails converge, with total sum equal to the initial partial sum plus the tail sum (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L5]

Limits of real sequences respect addition and subtraction (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct

1.1givenL3

Since $S$ is finite, every $μ (E_{k})$ and every partial sum $s_{n}$ is real. The $s_{n}$ increase and have supremum $S$ ; given $ε > 0$ , the defining property of the supremum supplies $N$ with $S - ε < s_{N} \leq S$ , and then $S - ε < s_{n} \leq S$ for every $n \geq N$ . Thus the real series $\sum_{k} μ (E_{k})$ converges to $S$ , and its tail sums $T_{N} : = \sum_{k \geq N} μ (E_{k})$ are real.

1.2givenL1L2

For every $N$ , [L2] gives ${lim sup}_{k} E_{k} \subseteq ⋃_{k \geq N} E_{k}$ , so monotonicity and subadditivity give $μ ({lim sup}_{k} E_{k}) \leq T_{N}$ .

2.1step 1.1L4L5

If $s_{N} = \sum_{k < N} μ (E_{k})$ , then $T_{N} = S - s_{N}$ by tail invariance, and $s_{N} \to S$ ; hence $T_{N} \to 0$ by the algebra of limits.

3.1step 1.2step 2.1∎

The nonnegative number $μ ({lim sup}_{k} E_{k})$ is at most every $T_{N}$ by step 1.2, while step 2.1 makes those tails arbitrarily small; therefore it is $0$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

The measure of a set liminf is at most the liminf of the measures

Statement

For every sequence $(E_{k})$ of measurable sets,

$μ ({lim inf}_{k \to \infty} E_{k}) \leq {lim inf}_{k \to \infty} μ (E_{k}),$

where the numerical liminf is taken in $[0, + \infty]$ as in Limit superior and limit inferior of a nonnegative extended-real sequence.

Facts & Assumptions

Given: A measure $μ$ and a sequence $(E_{k})$ of measurable sets.

[L1]

For increasing measurable sets, the measure of the union is the supremum of their measures (Continuity from below for measures).

[L2]

Measures are monotone under inclusion (Measures are monotone).

[L3]

The set liminf is $⋃_{N} ⋂_{k \geq N} E_{k}$ (Limit superior and limit inferior of a sequence of sets).

[L4]

For a nonnegative extended sequence $(a_{k})$ , ${lim inf}_{k} a_{k} = \sup_{N} \inf_{k \geq N} a_{k}$ (Limit superior and limit inferior of a nonnegative extended-real sequence), and all these bounds exist (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

Proof

technique · direct

1.1givenL3

Put $F_{N} : = ⋂_{k \geq N} E_{k}$ . Then $F_{N} \subseteq F_{N + 1}$ and $⋃_{N} F_{N} = {lim inf}_{k} E_{k}$ .

1.2givenL2L4

For every $N$ and every $k \geq N$ , $F_{N} \subseteq E_{k}$ , so $μ (F_{N}) \leq μ (E_{k})$ and hence $μ (F_{N}) \leq \inf_{k \geq N} μ (E_{k})$ .

2.1step 1.1step 1.2L1L4∎

Continuity from below and step 1.2 give $μ ({lim inf}_{k} E_{k}) = \sup_{N} μ (F_{N}) \leq \sup_{N} \inf_{k \geq N} μ (E_{k}) = {lim inf}_{k} μ (E_{k})$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The limsup of the measures is at most the measure of the set limsup under a finite-union bound

Statement

Let $(E_{k})$ be measurable and suppose $μ (⋃_{k} E_{k}) < + \infty$ . Then

${lim sup}_{k \to \infty} μ (E_{k}) \leq μ ({lim sup}_{k \to \infty} E_{k}),$

where the numerical limsup is taken in $[0, + \infty]$ .

Facts & Assumptions

Given: A measure $μ$ , measurable sets $(E_{k})$ , and $μ (⋃_{k} E_{k}) < + \infty$ .

[L1]

For decreasing measurable sets, if one has finite measure, the measure of their intersection is the infimum of their measures (Continuity from above when one set has finite measure).

[L2]

Measures are monotone under inclusion (Measures are monotone).

[L3]

The set limsup is $⋂_{N} ⋃_{k \geq N} E_{k}$ (Limit superior and limit inferior of a sequence of sets).

[L4]

For a nonnegative extended sequence $(a_{k})$ , ${lim sup}_{k} a_{k} = \inf_{N} \sup_{k \geq N} a_{k}$ (Limit superior and limit inferior of a nonnegative extended-real sequence), and all these bounds exist (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

Proof

technique · direct

1.1givenL3

Put $G_{N} : = ⋃_{k \geq N} E_{k}$ . Then $G_{N + 1} \subseteq G_{N}$ , $⋂_{N} G_{N} = {lim sup}_{k} E_{k}$ , and $G_{0} = ⋃_{k} E_{k}$ has finite measure.

1.2givenL2L4

For every $N$ and $k \geq N$ , $E_{k} \subseteq G_{N}$ , so $μ (E_{k}) \leq μ (G_{N})$ and hence $\sup_{k \geq N} μ (E_{k}) \leq μ (G_{N})$ .

2.1step 1.1step 1.2L1L4∎

Continuity from above and step 1.2 give ${lim sup}_{k} μ (E_{k}) = \inf_{N} \sup_{k \geq N} μ (E_{k}) \leq \inf_{N} μ (G_{N}) = μ ({lim sup}_{k} E_{k})$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Measures converge for a convergent sequence of sets contained in one finite-measure set

Statement

Let $(E_{n})$ and $E$ be measurable sets. Suppose there is measurable $D$ with $μ (D) < + \infty$ and $E_{n} \subseteq D$ for every $n$ , and suppose membership converges pointwise to membership in $E$ : for every $x$ there is $N$ such that, for all $n \geq N$ , one has $x \in E_{n}$ if and only if $x \in E$ . Then

${lim inf}_{n} μ (E_{n}) = {lim sup}_{n} μ (E_{n}) = μ (E) .$

In particular the real sequence $μ (E_{n})$ converges to $μ (E)$ .

Facts & Assumptions

Given: Measurable $E_{n}, E, D$ with $E_{n} \subseteq D$ , $μ (D) < + \infty$ , and pointwise convergence of memberships to $E$ .

[L1]

For a sequence of sets, eventual membership characterizes the set liminf and repeated membership characterizes the set limsup (Set liminf means eventual membership, set limsup means repeated membership, and liminf is contained in limsup).

[L2]

One has $μ (lim inf E_{n}) \leq lim inf μ (E_{n})$ (The measure of a set liminf is at most the liminf of the measures).

[L3]

If the union has finite measure, then $lim sup μ (E_{n}) \leq μ (lim sup E_{n})$ (The limsup of the measures is at most the measure of the set limsup under a finite-union bound).

[L4]

The numerical liminf and limsup are the supremum of tail infima and the infimum of tail suprema, respectively (Limit superior and limit inferior of a nonnegative extended-real sequence).

Proof

technique · direct

1.1givenL1

The membership hypothesis and [L1] give ${lim inf}_{n} E_{n} = E = {lim sup}_{n} E_{n}$ .

1.2given

Since every $E_{n} \subseteq D$ , their union is contained in $D$ and has finite measure. Moreover $E \subseteq D$ : every $x \in E$ belongs to $E_{n}$ for all sufficiently large $n$ , and hence belongs to $D$ .

2.1step 1.1step 1.2L2L3L4

Apply [L2] and [L3] using steps 1.1 and 1.2 to obtain $μ (E) \leq {lim inf}_{n} μ (E_{n}) \leq {lim sup}_{n} μ (E_{n}) \leq μ (E)$ .

3.1step 1.2step 2.1∎

All quantities in step 2.1 are therefore equal; they are finite because $E \subseteq D$ , so the usual squeeze criterion gives real convergence to $μ (E)$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Null sets are closed under countable unions and, in a complete space, under arbitrary subsets

Statement

In every measure space, a countable union of measurable null sets is measurable and null, and every measurable subset of a null set is null. If the measure space is complete, every subset of a measurable null set is measurable and null. Thus the null sets of a complete measure space form a sigma-ideal.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ .

[L1]

Countable subadditivity bounds the measure of a countable union by the sum of the member measures (Finite and countable subadditivity of measures).

[L2]

Measures are monotone under inclusion (Measures are monotone).

[L3]

A complete measure space contains every subset of every measurable null set in its sigma-algebra (Complete measure spaces).

Proof

technique · direct

1.1givenL1

If $N_{k} \in A$ and $μ (N_{k}) = 0$ for every $k$ , then $⋃_{k} N_{k}$ is measurable and [L1] gives $0 \leq μ (⋃_{k} N_{k}) \leq \sum_{k} 0 = 0$ .

1.2givenL2

If $S \in A$ and $S \subseteq N$ for a measurable null set $N$ , then $0 \leq μ (S) \leq μ (N) = 0$ .

2.1step 1.2L3

If the space is complete and $S \subseteq N$ for a measurable null set $N$ , then [L3] first makes $S$ measurable and step 1.2 makes it null.

3.1step 1.1step 2.1∎

Step 1.1 gives closure under countable unions, and step 2.1 gives closure under arbitrary subsets in a complete space, including the empty union and the empty subset.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The completion domain and proposed completed set function of a measure space

Definition

Let $(X, A, μ)$ be a measure space. Its completion domain is

$\overline{A} : = {E \subseteq X : E = A \cup N for some A, Z \in A and N \subseteq Z with μ (Z) = 0} .$

For a displayed representation $E = A \cup N$ , define the proposed completed set function by

$\overline{μ} (E) : = μ (A) .$

The value must not depend on the representation. That obligation is discharged by The completed measure is independent of the representing measurable set ↗, and Assuming countable choice, the completion domain is a sigma-algebra ↗ proves, under the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), that $\overline{A}$ is a sigma-algebra. Thus the definite phrases completion sigma-algebra and completed measure are used only with those two results in force. Under the same choice hypothesis, Assuming countable choice, every measure space has a unique complete extension to its completion ↗ proves that $\overline{μ}$ is the unique complete extension on this domain.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Assuming countable choice, the completion domain is a sigma-algebra

Statement

Assume the Axiom of Countable Choice. For every measure space $(X, A, μ)$ , the completion domain $\overline{A}$ of The completion domain and proposed completed set function of a measure space is a sigma-algebra on $X$ containing $A$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and the Axiom of Countable Choice.

[L1]

The completion domain consists of the sets $A \cup N$ with $A, Z \in A$ , $N \subseteq Z$ , and $μ (Z) = 0$ (The completion domain and proposed completed set function of a measure space).

[L2]

A sigma-algebra contains the empty set, is closed under relative complements, and is closed under countable unions (Sigma-algebras).

[L3]

A countable union of measurable null sets is measurable and null (Null sets are closed under countable unions and, in a complete space, under arbitrary subsets).

[L4]

Countable choice selects one member from every nonempty natural-number-indexed family (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenL1L2

Every $A \in A$ belongs to $\overline{A}$ by taking $N = Z = \emptyset$ ; in particular $\emptyset, X \in \overline{A}$ .

1.2givenL1L2

If $E = A \cup N$ with $N \subseteq Z$ , $μ (Z) = 0$ , replace $N$ by $N ∖ A$ without changing $E$ . Then $A$ and $N$ are disjoint, and $E^{c} = (A^{c} ∖ Z) \cup ((A^{c} \cap Z) ∖ N)$ , whose first part is measurable and whose second lies in the measurable null set $Z$ ; hence $E^{c} \in \overline{A}$ .

1.3givenL1L4choose

Let $(E_{k})$ be a sequence in $\overline{A}$ . For each $k$ , the family of triples $(A, N, Z)$ witnessing [L1] is nonempty, so [L4] selects $E_{k} = A_{k} \cup N_{k}$ with $N_{k} \subseteq Z_{k}$ and $μ (Z_{k}) = 0$ .

2.1step 1.3L2L3

Put $A = ⋃_{k} A_{k}$ and $Z = ⋃_{k} Z_{k}$ . Then $A, Z \in A$ , $μ (Z) = 0$ by [L3], and $(⋃_{k} E_{k}) ∖ A \subseteq Z$ , so $⋃_{k} E_{k} \in \overline{A}$ .

3.1step 1.1step 1.2step 2.1L2∎

Step 1.1 gives the empty set and containment of $A$ , step 1.2 gives complements, and step 2.1 gives countable unions; therefore $\overline{A}$ is a sigma-algebra.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The completed measure is independent of the representing measurable set

Statement

If $E = A \cup N = B \cup M$ are two representations in the completion domain, with $N \subseteq Z$ , $M \subseteq W$ , and $Z, W$ measurable null sets, then $μ (A) = μ (B)$ . Consequently $\overline{μ} (E) : = μ (A)$ is well defined.

Facts & Assumptions

Given: Representations $E = A \cup N = B \cup M$ as in the Statement.

[L1]

A completed set is represented by a measurable core together with a subset of a measurable null set (The completion domain and proposed completed set function of a measure space).

[L2]

Measurable sets whose symmetric difference is null have equal measure, including at $+ \infty$ (Sets whose symmetric difference is null have the same measure).

[L3]

In every measure space, a countable union of measurable null sets is measurable and null, and every measurable subset of a null set is null (Null sets are closed under countable unions and, in a complete space, under arbitrary subsets).

Proof

technique · direct

1.1givenL1

If $x \in A ∖ B$ , then $x \in E = B \cup M$ forces $x \in M \subseteq W$ ; similarly $B ∖ A \subseteq Z$ . Hence $A △ B \subseteq Z \cup W$ .

1.2givenL3

The measurable set $Z \cup W$ is null, and the measurable subset $A △ B$ is therefore null.

2.1step 1.1step 1.2L2∎

By [L2], $μ (A) = μ (B)$ ; this also covers empty null envelopes and the case in which the common value is $+ \infty$ , so the proposed value $\overline{μ} (E)$ is independent of every representation.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Assuming countable choice, every measure space has a unique complete extension to its completion

Statement

Assume the Axiom of Countable Choice. Let $(X, A, μ)$ be a measure space, and let $(X, \overline{A}, \overline{μ})$ be its completion construction. Then $\overline{μ}$ is a complete measure on $\overline{A}$ extending $μ$ . It is the unique complete measure on $\overline{A}$ that extends $μ$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and the Axiom of Countable Choice.

[L1]

The completion domain is a sigma-algebra containing $A$ (Assuming countable choice, the completion domain is a sigma-algebra).

[L2]

The value $\overline{μ} (E) = μ (A)$ is independent of the measurable core in a completed representation (The completed measure is independent of the representing measurable set).

[L3]

Countable unions of measurable null sets are null, and completeness makes all their subsets measurable and null (Null sets are closed under countable unions and, in a complete space, under arbitrary subsets).

[L4]

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

[L5]

Countable choice selects witnesses from each nonempty natural-number-indexed family (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L6]

Every member of the completion domain has a representation $E = A \cup N$ with $A$ measurable and $N$ contained in a measurable null set, and the proposed completed value is $μ (A)$ (The completion domain and proposed completed set function of a measure space).

Proof

technique · direct

1.1givenL1L2L4

By [L1] and [L2], $\overline{μ}$ is a function on the sigma-algebra $\overline{A}$ , and $\overline{μ} (A) = μ (A)$ for $A \in A$ by the representation $A = A \cup \emptyset$ ; in particular $\overline{μ} (\emptyset) = 0$ .

1.2givenL3L5L6choose

Let $(E_{k})$ be disjoint in $\overline{A}$ . By [L6] each family of completed representations is nonempty, so [L5] chooses $E_{k} = A_{k} \cup N_{k}$ with measurable $A_{k} \subseteq E_{k}$ and $N_{k}$ contained in a measurable null set $Z_{k}$ . Then the $A_{k}$ are disjoint, $⋃_{k} E_{k} = (⋃_{k} A_{k}) \cup N$ with $N \subseteq ⋃_{k} Z_{k}$ , and $⋃_{k} Z_{k}$ is null by [L3].

1.3givenL2L3L6

If $E \in \overline{A}$ and $\overline{μ} (E) = 0$ , choose a representation $E = A \cup N$ from [L6] with $N \subseteq Z$ null. Then $A \cup Z$ is measurable and null, and every $S \subseteq E$ is represented by the empty measurable core plus the sub-null set $S \subseteq A \cup Z$ .

1.4givenL2L3L4L6

Let $λ$ be any complete measure on $\overline{A}$ extending $μ$ . For a representation $E = A \cup N$ from [L6], with $N \subseteq Z$ and $μ (Z) = 0$ , one has $λ (Z) = 0$ and completeness gives $λ (N) = 0$ . Replacing $N$ by $N ∖ A$ makes the union disjoint without changing $E$ , so finite additivity gives $λ (E) = λ (A) = μ (A) = \overline{μ} (E)$ .

2.1step 1.1step 1.2L2L4

Countable additivity of $μ$ applied to the measurable cores in step 1.2 gives $\overline{μ} (⋃_{k} E_{k}) = μ (⋃_{k} A_{k}) = \sum_{k} μ (A_{k}) = \sum_{k} \overline{μ} (E_{k})$ ; with step 1.1, this proves that $\overline{μ}$ is a measure.

2.2step 1.3L2L3

Step 1.3 shows that every subset of every $\overline{μ}$ -null set belongs to $\overline{A}$ and has value $0$ , so the completed measure space is complete.

3.1step 2.1step 2.2step 1.4∎

Steps 2.1, 2.2 and 1.4 prove respectively that $\overline{μ}$ is an extending measure, is complete, and is the unique complete extension on $\overline{A}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Restriction of a measure to a measurable set

Definition

Let $μ$ be a measure on $(X, A)$ and fix a measurable set $E \in A$ . The restriction of $μ$ to $E$ on the original sigma-algebra is the set function

$μ_{E} : A \to [0, + \infty], μ_{E} (A) : = μ (A \cap E) .$

This is the same-ambient convention: the domain remains $A$ , rather than becoming the trace sigma-algebra on $E$ . The fact that $μ_{E}$ is a measure is The restriction of a measure to a measurable set is a measure ↗.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The restriction of a measure to a measurable set is a measure

Statement

If $μ$ is a measure on $(X, A)$ and $E \in A$ , then $μ_{E} (A) = μ (A \cap E)$ is a measure on $(X, A)$ .

Facts & Assumptions

Given: A measure $μ$ on $(X, A)$ and a measurable set $E$ .

[L1]

The restricted set function is $μ_{E} (A) = μ (A \cap E)$ on the original sigma-algebra (Restriction of a measure to a measurable set).

[L2]

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

Proof

technique · direct

1.1givenL1L2

One has $μ_{E} (\emptyset) = μ (\emptyset \cap E) = 0$ .

1.2given

If $(A_{k})$ is pairwise disjoint in $A$ , then $(A_{k} \cap E)$ is pairwise disjoint and $⋃_{k} (A_{k} \cap E) = (⋃_{k} A_{k}) \cap E$ .

2.1step 1.1step 1.2L1L2∎

Countable additivity of $μ$ applied to step 1.2 gives $μ_{E} (⋃_{k} A_{k}) = \sum_{k} μ (A_{k} \cap E) = \sum_{k} μ_{E} (A_{k})$ ; together with step 1.1 this proves that $μ_{E}$ is a measure, including $E = \emptyset$ and $E = X$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The Dirac set function at a point

Definition

Let $A$ be a sigma-algebra on a nonempty set $X$ (Sigma-algebras) and fix $x_{0} \in X$ . The Dirac set function at $x_{0}$ is

$δ_{x_{0}} (E) : = {\begin{cases} 1, & x_{0} \in E, \\ 0, & x_{0} \notin E, \end{cases} E \in A .$

The two branches are exhaustive and disjoint. The fact that this set function is a probability measure is proved in A Dirac set function is a probability measure ↗.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

A Dirac set function is a probability measure

Statement

For $x_{0} \in X$ , the Dirac set function $δ_{x_{0}}$ is a probability measure on every sigma-algebra on $X$ .

Facts & Assumptions

Given: A sigma-algebra $A$ on a nonempty set $X$ and a point $x_{0} \in X$ .

[L1]

The Dirac set function has value $1$ exactly on the measurable sets containing $x_{0}$ , and value $0$ otherwise (The Dirac set function at a point).

[L2]

A probability measure is a measure whose value on the whole space is $1$ (Probability measures and probability spaces).

[L3]

A nonnegative extended series is the supremum of its finite partial sums, with the empty sum equal to $0$ (Series in the nonnegative extended real line).

Proof

technique · direct

1.1givenL1

One has $δ_{x_{0}} (\emptyset) = 0$ and $δ_{x_{0}} (X) = 1$ .

1.2givenL1L3

If $(E_{k})$ is pairwise disjoint, then $x_{0}$ belongs to at most one $E_{k}$ . If it belongs to none, both $δ_{x_{0}} (⋃_{k} E_{k})$ and $\sum_{k} δ_{x_{0}} (E_{k})$ are $0$ ; if it belongs to the unique $E_{r}$ , both are $1$ .

2.1step 1.1step 1.2L2∎

Step 1.2 proves countable additivity and step 1.1 gives the empty-set and total-mass conditions, so $δ_{x_{0}}$ is a probability measure.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Nonnegative scalar multiples and countable weighted sums of measures

Definition

Let $μ$ be a measure and let $c \in [0, + \infty]$ . Define the scalar set function $c μ$ by three disjoint branches:

$(c μ) (E) : = {\begin{cases} 0, & c = 0, \\ c μ (E), & 0 < c < + \infty, \\ 0, & c = + \infty and μ (E) = 0, \\ + \infty, & c = + \infty and μ (E) > 0. \end{cases}$

Thus neither the zero branch nor the infinite branch forms the undefined extended-real product $0 \cdot (+ \infty)$ .

For measures $(μ_{k})_{k \in N}$ on the same measurable space and weights $c_{k} \in [0, + \infty]$ , their countable weighted sum is the pointwise set function

$(\sum_{k = 0}^{\infty} c_{k} μ_{k}) (E) : = \sum_{k = 0}^{\infty} (c_{k} μ_{k}) (E),$

using Series in the nonnegative extended real line. Finite weighted sums are defined by the corresponding finite partial sum, with the empty weighted sum the zero set function. The fact that all these set functions are measures is Nonnegative scalar multiples and countable weighted sums of measures are measures ↗.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Nonnegative scalar multiples and countable weighted sums of measures are measures

Statement

Let $(μ_{k})$ be measures on one measurable space and let $c_{k} \in [0, + \infty]$ . Each scalar multiple $c_{k} μ_{k}$ , defined by the zero, finite-positive, and positive-infinity branches of Nonnegative scalar multiples and countable weighted sums of measures, is a measure. Every finite or countable weighted sum $\sum_{k} c_{k} μ_{k}$ is also a measure.

Facts & Assumptions

Given: Measures $(μ_{k})$ on $(X, A)$ and coefficients $c_{k} \in [0, + \infty]$ .

[L1]

Scalar multiplication has separate $c = 0$ , $0 < c < + \infty$ , and $c = + \infty$ branches, and weighted sums are pointwise nonnegative extended sums (Nonnegative scalar multiples and countable weighted sums of measures).

[L2]

For a nonnegative extended double sequence, the two iterated sums are equal (Tonelli's theorem for double series of nonnegative extended real numbers).

[L3]

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

Proof

technique · direct

1.1givenL1L3

For $c = 0$ , the set function $c μ$ is the zero measure.

1.2givenL1L3algebra

For $0 < c < + \infty$ , $(c μ) (\emptyset) = 0$ ; for disjoint $(E_{j})$ , multiplying the finite partial-sum identities by $c$ and taking their supremum gives $c μ (⋃_{j} E_{j}) = \sum_{j} c μ (E_{j})$ , whether the common value is finite or $+ \infty$ .

1.3givenL1L3

For $c = + \infty$ , a disjoint union has $μ$ -measure zero exactly when every member has $μ$ -measure zero: this follows directly from countable additivity and nonnegativity. Hence the infinite branch takes value $0$ on the union exactly when every term value is $0$ , and otherwise both it and the series of term values are $+ \infty$ ; this branch is a measure without forming $0 \cdot (+ \infty)$ .

2.1step 1.1step 1.2step 1.3

Steps 1.1, 1.2 and 1.3 prove that every scalar multiple $c_{k} μ_{k}$ is a measure for all possible coefficients.

3.1givenL1step 2.1

Put $ν (E) = \sum_{k} (c_{k} μ_{k}) (E)$ . Then $ν (\emptyset) = 0$ .

3.2step 2.1L1L2L3

If $(E_{j})$ is disjoint, then step 2.1 and Tonelli give $ν (⋃_{j} E_{j}) = \sum_{k} \sum_{j} (c_{k} μ_{k}) (E_{j}) = \sum_{j} \sum_{k} (c_{k} μ_{k}) (E_{j}) = \sum_{j} ν (E_{j})$ .

4.1step 3.1step 3.2L3∎

Steps 3.1 and 3.2 prove that the countable weighted sum is a measure; the same proof for a finite index range, or zero coefficients thereafter, gives every finite weighted sum, including the empty zero measure.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

A measure on a finite sigma-algebra is a finite weighted sum over its atoms

Statement

Let $A$ be a finite sigma-algebra on $X$ . Its atoms are the nonempty measurable sets containing no proper nonempty measurable subset. The atoms form a finite partition of $X$ , every measurable set is the union of the atoms it contains, and every measure $μ$ on $(X, A)$ has the representation

$μ = \sum_{C an atom} μ (C) δ_{x_{C}},$

where $x_{C} \in C$ is chosen once for each atom and scalar multiplication uses the explicit infinite-coefficient branch. If $X = \emptyset$ , there are no atoms and the representation is the empty weighted sum.

Facts & Assumptions

Given: A finite sigma-algebra $A$ on $X$ and a measure $μ$ on it.

[L1]

A measure is finitely additive on disjoint measurable families (Measures on sigma-algebras).

[L2]

A Dirac measure at $x$ has value $1$ exactly on sets containing $x$ (The Dirac set function at a point, A Dirac set function is a probability measure).

[L3]

A coefficient $+ \infty$ times a measure is $0$ on its null sets and $+ \infty$ elsewhere; finite and empty weighted sums are defined pointwise (Nonnegative scalar multiples and countable weighted sums of measures, Series in the nonnegative extended real line, Finite sums and finite products, by recursion).

[L4]

A finite set is equinumerous with some natural number (Finite, countably infinite, countable, uncountable), and a natural-number-indexed finite family of nonempty sets has a choice function in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1given

For $x \in X$ define $C_{x} : = ⋂ {A \in A : x \in A}$ . The intersected family is nonempty because it contains $X$ , and it is finite, so $C_{x} \in A$ and $x \in C_{x}$ .

1.2givenL4choose

Enumerate the finite family of atoms by a natural number. Applying finite choice to that enumeration selects one representative $x_{C} \in C$ for every atom; when $X = \emptyset$ , the family and the choice function are empty.

2.1step 1.1

If $y \in C_{x}$ , then $x$ and $y$ belong to exactly the same members of $A$ : otherwise the complement of a measurable set separating them would contradict $y \in C_{x}$ . Hence $C_{y} = C_{x}$ , while $y \notin C_{x}$ gives $C_{y} \cap C_{x} = \emptyset$ ; the distinct $C_{x}$ form a finite measurable partition of $X$ .

3.1step 1.1step 2.1

Every $C_{x}$ is an atom, and if $A \in A$ and $x \in A$ , then $C_{x} \subseteq A$ by definition; conversely, if $C$ is an atom and $x \in C$ , then the nonempty measurable set $C_{x} \subseteq C$ forces $C_{x} = C$ . Consequently every atom is one of the partition blocks and every measurable $A$ is the disjoint union of the atoms it contains.

4.1step 3.1L1L3

For measurable $A$ , finite additivity and step 3.1 give $μ (A) = \sum_{C \subseteq A} μ (C)$ .

4.2step 3.1step 1.2L2L3

For each atom $C$ , the term $μ (C) δ_{x_{C}} (A)$ equals $μ (C)$ when $C \subseteq A$ and $0$ otherwise; this remains true when $μ (C) = + \infty$ because [L3] uses the null/non-null infinite branch.

5.1step 4.1step 4.2L3∎

Summing step 4.2 over the atoms and comparing with step 4.1 proves the representation. For $X = \emptyset$ both sides are the zero measure and the sum is empty.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Every measure on a countable discrete space is its weighted sum of Dirac measures

Statement

Let $X$ be at most countable and equip it with $P (X)$ . Every measure $μ$ on this discrete measurable space is determined by the weights $w_{x} : = μ ({x})$ and satisfies

$μ (E) = \sum_{x \in E} w_{x}, μ = \sum_{x \in X} w_{x} δ_{x} .$

For finite $X$ this is a finite sum over a bijective finite listing; for countably infinite $X$ it is a series over a bijection $N \to X$ . No point is repeated. Each coefficient, including $+ \infty$ , is uniquely forced by $μ$ .

Facts & Assumptions

Given: An at most countable set $X$ and a measure $μ$ on $(X, P (X))$ .

[L1]

An at most countable set is finite or is in bijection with $N$ (Finite, countably infinite, countable, uncountable).

[L2]

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

[L3]

The Dirac set function $δ_{y}$ has value $1$ on sets containing $y$ and $0$ otherwise (The Dirac set function at a point); Dirac set functions are probability measures (A Dirac set function is a probability measure); and nonnegative finite and countable weighted sums of measures are measures (Nonnegative scalar multiples and countable weighted sums of measures, Nonnegative scalar multiples and countable weighted sums of measures are measures).

Proof

technique · direct

1.1givenL2L3

If $X = \emptyset$ , then $μ$ is the zero measure and the asserted expression is the empty weighted sum.

1.2givenL1L2

If $X$ is finite and nonempty, choose a bijection $e : n \to X$ for some $n \geq 1$ . Every $E \subseteq X$ is the finite disjoint union of the singletons ${e (i)}$ with $e (i) \in E$ , so $μ (E) = \sum_{i < n, e (i) \in E} μ ({e (i)})$ .

1.3givenL1L2

If $X$ is countably infinite, choose a bijection $e : N \to X$ . Every $E \subseteq X$ is the disjoint union of the sequence whose $k$ -th term is ${e (k)}$ when $e (k) \in E$ and $\emptyset$ otherwise, so $μ (E) = \sum_{k : e (k) \in E} μ ({e (k)})$ .

1.4givenL3

Evaluating any asserted representation at the singleton ${x}$ leaves only the Dirac term at $x$ , so its coefficient must be $μ ({x})$ .

2.1step 1.2L3

In the finite case, the weighted Dirac sum with coefficients $w_{e (i)} = μ ({e (i)})$ has the value computed in step 1.2 on every $E$ ; the explicit positive-infinity branch gives $0$ off $e (i)$ and $+ \infty$ on sets containing it.

2.2step 1.3L3

In the countably infinite case, the countable weighted Dirac sum has the value computed in step 1.3 on every $E$ , with the same interpretation of infinite coefficients.

3.1step 1.1step 2.1step 2.2step 1.4L1∎

Steps 1.1, 2.1 and 2.2 prove the representation in the empty, finite nonempty, and countably infinite cases, and step 1.4 proves uniqueness of every coefficient.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Assuming countable choice, an infinite-measure set in a semifinite measure space has arbitrarily large finite-measure subsets

Statement

Assume the Axiom of Countable Choice. Let $μ$ be semifinite and let $E$ be measurable with $μ (E) = + \infty$ . For every real $R > 0$ there is measurable $F \subseteq E$ such that

$R < μ (F) < + \infty .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a semifinite measure $μ$ , a measurable $E$ with $μ (E) = + \infty$ , and a real $R > 0$ .

[L1]

Semifiniteness means that every measurable set of positive measure contains a measurable subset of positive finite measure (Finite, sigma-finite, and semifinite measures).

[L2]

Countable choice selects one member from every nonempty natural-number-indexed family (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L3]

Measures are continuous from below on increasing measurable sequences (Continuity from below for measures) and countably subadditive (Finite and countable subadditivity of measures).

[L4]

If $A \subseteq B$ , $μ (A) < + \infty$ , and $μ (B) = + \infty$ , then $μ (B ∖ A) = + \infty$ (Measure of a set difference when the smaller set has finite measure).

[L5]

[L6]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Proof

technique · contradiction

1.1givenL1L5

Let $M : = \sup {μ (F) : F \subseteq E measurable and μ (F) < + \infty}$ . The family is nonempty because it contains $\emptyset$ , and semifiniteness makes $M > 0$ .

2.1givenstep 1.1L5

If $M = + \infty$ , the definition of supremum directly supplies a finite-measure $F \subseteq E$ with $μ (F) > R$ , so only the case $M < + \infty$ can fail the conclusion.

2.2step 1.1L2assume-contrachoose

Suppose for contradiction that $M < + \infty$ . For each $n$ , the family of measurable finite-measure $F \subseteq E$ with $μ (F) > M - 1 / (n + 1)$ is nonempty; [L2] selects one such $F_{n}$ for every $n$ .

3.1step 1.1step 2.2L3L5L6

Put $G_{n} = ⋃_{k < n + 1} F_{k}$ and $G = ⋃_{n} G_{n}$ . Finite subadditivity makes every $G_{n}$ finite-measure, while $μ (G_{n}) \leq M$ by the definition of $M$ and $μ (G_{n}) > M - 1 / (n + 1)$ because $F_{n} \subseteq G_{n}$ ; [L6] shows these lower bounds approach $M$ , and continuity from below gives $μ (G) = M < + \infty$ .

4.1step 3.1L1L4choose

By [L4], $μ (E ∖ G) = + \infty$ . Semifiniteness supplies measurable $H \subseteq E ∖ G$ with $0 < μ (H) < + \infty$ .

5.1step 2.1step 3.1step 4.1discharge-contradiction∎

The disjoint set $G \cup H \subseteq E$ has finite measure $M + μ (H) > M$ , contradicting the definition of $M$ . Hence $M = + \infty$ , and step 2.1 gives the required $F$ .

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The semifinite part of a measure

Definition

Let $μ$ be a measure on $(X, A)$ . Its semifinite part is the set function

$μ_{s f} (E) : = \sup {μ (F) : F \in A, F \subseteq E, μ (F) < + \infty}, E \in A .$

The set inside the supremum is nonempty because it contains the value $μ (\emptyset) = 0$ , and its supremum exists in $[0, + \infty]$ by Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ . Assuming the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), Assuming countable choice, the semifinite part is a semifinite measure and equals the original measure exactly when it is semifinite ↗ proves that this set function is a semifinite measure and identifies when it equals $μ$ .

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Assuming countable choice, the semifinite part is a semifinite measure and equals the original measure exactly when it is semifinite

Statement

Assume the Axiom of Countable Choice. For every measure $μ$ , its semifinite part $μ_{s f}$ is a semifinite measure and $μ_{s f} \leq μ$ . Moreover,

$μ_{s f} = μ ⟺ μ is semifinite .$

Facts & Assumptions

Given: A measure $μ$ on $(X, A)$ and the Axiom of Countable Choice.

[L1]

The semifinite part is the supremum of the finite values $μ (F)$ over measurable $F \subseteq E$ (The semifinite part of a measure).

[L2]

Under countable choice, an infinite-measure set for a semifinite measure contains finite-measure subsets of arbitrarily large measure (Assuming countable choice, an infinite-measure set in a semifinite measure space has arbitrarily large finite-measure subsets).

[L3]

A measure is countably additive on disjoint measurable sequences (Measures on sigma-algebras), and measures are monotone (Measures are monotone).

[L4]

A natural-number-indexed finite family of nonempty sets has a choice function in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1givenL1L3

One has $μ_{s f} (\emptyset) = 0$ and $μ_{s f} (E) \leq μ (E)$ for every measurable $E$ , since every finite-measure $F \subseteq E$ satisfies $μ (F) \leq μ (E)$ .

1.2givenL1L3

Let $(E_{k})$ be disjoint and $E = ⋃_{k} E_{k}$ . If $F \subseteq E$ is measurable with $μ (F) < + \infty$ , then the $F \cap E_{k}$ are disjoint finite-measure subsets of $E_{k}$ , so $μ (F) = \sum_{k} μ (F \cap E_{k}) \leq \sum_{k} μ_{s f} (E_{k})$ . Taking the supremum over $F$ gives $μ_{s f} (E) \leq \sum_{k} μ_{s f} (E_{k})$ .

1.3givenL1L3L4choose

Conversely, for each finite initial range, the supremum property and finite choice permit finite-measure $F_{k} \subseteq E_{k}$ arbitrarily close to $μ_{s f} (E_{k})$ ; their finite disjoint union has finite measure and lies in $E$ . If one of the finitely many suprema is $+ \infty$ , use an arbitrarily large finite value instead. Hence every finite partial sum $\sum_{k < n} μ_{s f} (E_{k})$ is at most $μ_{s f} (E)$ , and so is their supremum.

1.4givenL1L3

The set function $μ_{s f}$ is semifinite: if $μ_{s f} (E) > 0$ , its defining supremum supplies a measurable $F \subseteq E$ with $0 < μ (F) < + \infty$ , and then $μ_{s f} (F) = μ (F)$ .

1.5givenL1L2

For the reverse direction, suppose $μ$ is semifinite. If $μ (E) < + \infty$ , the choice $F = E$ gives $μ_{s f} (E) = μ (E)$ ; if $μ (E) = + \infty$ , [L2] makes the defining finite values unbounded, so again $μ_{s f} (E) = μ (E)$ .

2.1step 1.1step 1.2step 1.3step 1.4L3

Steps 1.1, 1.2 and 1.3 give the empty-set condition and both countable-additivity inequalities, so $μ_{s f}$ is a measure; step 1.4 makes it semifinite.

3.1step 2.1

For the forward direction of the displayed equivalence, if $μ_{s f} = μ$ , then $μ$ is semifinite because step 2.1 proves that $μ_{s f}$ is semifinite.

4.1step 1.1step 3.1step 1.5∎

Steps 3.1 and 1.5 prove both directions of the equivalence, while step 1.1 records the pointwise inequality $μ_{s f} \leq μ$ .

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-21Open item page →

Assuming countable choice, every measure is the sum of its semifinite part and a zero-infinity-valued measure

Statement

Assume the Axiom of Countable Choice. For a measure $μ$ , define

$ν (E) : = {\begin{cases} 0, & E is sigma-finite for μ, \\ + \infty, & E is not sigma-finite for μ . \end{cases}$

Then $ν$ is a measure taking only the values $0$ and $+ \infty$ , and

$μ = μ_{s f} + ν .$

The zero-infinity summand in such a decomposition need not be unique.

Facts & Assumptions

Given: A measure $μ$ on $(X, A)$ and the Axiom of Countable Choice.

[L1]

Under countable choice, the semifinite part is a semifinite measure and agrees with a measure exactly when that measure is semifinite (Assuming countable choice, the semifinite part is a semifinite measure and equals the original measure exactly when it is semifinite).

[L2]

Sigma-finiteness means admitting a countable measurable cover by finite-measure sets (Finite, sigma-finite, and semifinite measures).

[L3]

Countable choice selects countably many covers (The Axiom of Countable Choice ( ${A C}_{ω}$ )), and a product of two at most countable sets is at most countable (A product of two at most countable sets is at most countable).

[L4]

Under countable choice, a countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L5]

Counting measure is a measure (Counting measure is a measure), and $R$ is uncountable ( $R$ is uncountable (Cantor's nested intervals, 1874)).

[L6]

A countable union has measure at most the nonnegative sum of the measures of its members (Finite and countable subadditivity of measures).

[L7]

The value $μ_{s f} (E)$ is the supremum of the values of finite-measure measurable subsets of $E$ (The semifinite part of a measure).

[L8]

Counting measure assigns a finite set its cardinality and an infinite set $+ \infty$ (Counting measure on an arbitrary set).

Proof

technique · direct

1.1givenL2L3choose

A measurable subset of a sigma-finite set is sigma-finite. Under [L3], a countable union of sigma-finite measurable sets is sigma-finite: select a finite-measure cover for each member and flatten the resulting $N \times N$ family to one countable cover.

1.2givenL1L2L6L7

If $E$ is sigma-finite for $μ$ , restriction of the measure axioms makes $μ ∣_{E}$ a measure, and it is semifinite: a positive-measure subset must meet one member of a finite-measure cover in positive measure, since otherwise [L6] would make it null. The semifinite part of this restriction, evaluated at $E$ , is exactly the supremum in [L7]. Thus [L1] applied to $μ ∣_{E}$ gives $μ_{s f} (E) = μ (E)$ .

1.3givenL1L4L5L8

For nonuniqueness, take counting measure $#$ on the uncountable set $R$ . It is semifinite: every set of positive counting measure is nonempty, so one of its points gives a singleton subset of finite positive measure $1$ by [L8]. Hence [L1] makes its semifinite part itself. Besides the zero measure, define $η (E) = 0$ for countable $E$ and $+ \infty$ for uncountable $E$ . One has $η (\emptyset) = 0$ ; for a disjoint sequence, [L4] makes the union countable exactly when every member is countable, so countable additivity has both sides $0$ in that case and both sides $+ \infty$ otherwise. Thus $η$ is a measure, and both $0$ and $η$ satisfy $# = #_{s f} + 0 = #_{s f} + η$ .

2.1step 1.1L2

Therefore, for a disjoint sequence $(E_{k})$ , the union is sigma-finite exactly when every $E_{k}$ is sigma-finite. Hence $ν (⋃_{k} E_{k}) = 0$ exactly when every $ν (E_{k}) = 0$ , and otherwise both sides of $ν (⋃_{k} E_{k}) = \sum_{k} ν (E_{k})$ are $+ \infty$ ; also $ν (\emptyset) = 0$ .

2.2step 1.2L2

If $E$ is sigma-finite, step 1.2 gives $(μ_{s f} + ν) (E) = μ (E) + 0$ ; if $E$ is not sigma-finite, then $μ (E) = + \infty$ and $(μ_{s f} + ν) (E) = μ_{s f} (E) + (+ \infty) = + \infty$ .

3.1step 2.1

Step 2.1 proves that $ν$ is a zero-infinity-valued measure.

4.1step 2.2step 1.3∎

Step 2.2 proves $μ = μ_{s f} + ν$ , and step 1.3 supplies two distinct zero-infinity summands for the same semifinite part, proving the final assertion.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Finite measures agreeing on a generating pi-system and on the whole space are equal

Statement

Let $P$ be a pi-system on $X$ and let $A = σ_{X} (P)$ . If finite measures $μ$ and $ν$ on $(X, A)$ agree on $P$ and satisfy $μ (X) = ν (X)$ , then $μ = ν$ on $A$ .

The total-mass equality is separate because this library's pi-system convention does not require $X \in P$ .

Facts & Assumptions

Given: A pi-system $P$ generating $A$ , and finite measures $μ, ν$ agreeing on $P$ and on $X$ .

[L1]

A pi-system is a nonempty family closed under binary intersections and need not contain $X$ (Pi-systems).

[L2]

If a lambda-system contains a pi-system $P$ , then it contains $σ_{X} (P)$ (Dynkin's pi-lambda theorem).

[L3]

Measures are finitely and countably additive (Measures on sigma-algebras) and continuous from below (Continuity from below for measures).

[L4]

For finite measures, the value on a relative difference is obtained by subtracting the smaller-set value (Measure of a set difference when the smaller set has finite measure).

[L5]

The generated sigma-algebra is the intersection of all sigma-algebras containing the generating family (The sigma-algebra generated by a family of sets).

Proof

technique · direct

1.1givenL1

Let $D : = {A \in A : μ (A) = ν (A)}$ . Then $X \in D$ by the total-mass hypothesis and $P \subseteq D$ by the agreement hypothesis.

1.2givenL4algebra

If $A \subseteq B$ lie in $D$ , finiteness and [L4] give $μ (B ∖ A) = μ (B) - μ (A) = ν (B) - ν (A) = ν (B ∖ A)$ , so $B ∖ A \in D$ .

1.3givenL3

If $A_{n} ↑ A$ and each $A_{n} \in D$ , continuity from below gives $μ (A) = \sup_{n} μ (A_{n}) = \sup_{n} ν (A_{n}) = ν (A)$ , so $A \in D$ .

2.1step 1.1step 1.2step 1.3

Steps 1.1, 1.2 and 1.3 show that $D$ is a lambda-system containing $P$ .

3.1step 2.1L2L5∎

Dynkin's theorem gives $A = σ_{X} (P) \subseteq D$ , so $μ (A) = ν (A)$ for every $A \in A$ .

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system

Statement

Let $P$ be a pi-system on $X$ generating $A$ , and let $μ, ν$ be measures on $(X, A)$ that agree on $P$ . Suppose there is an increasing sequence $(P_{n})$ in $P$ with

$X = ⋃_{n} P_{n}, μ (P_{n}) = ν (P_{n}) < + \infty (n \in N) .$

Then $μ = ν$ on $A$ .

Facts & Assumptions

Given: Measures $μ, ν$ , a generating pi-system $P$ , and an increasing finite-measure exhaustion $(P_{n})$ as in the Statement.

[L1]

Finite measures agreeing on a generating pi-system and on the whole space are equal (Finite measures agreeing on a generating pi-system and on the whole space are equal).

[L2]

Measures are continuous from below (Continuity from below for measures).

[L3]

A pi-system is closed under binary intersections (Pi-systems).

[L4]

For a measurable $P_{n}$ , the set function $A \mapsto μ (A \cap P_{n})$ is a measure on the original sigma-algebra (The restriction of a measure to a measurable set is a measure).

Proof

technique · direct

1.1givenL4

Fix $n$ and define $μ_{n} (A) : = μ (A \cap P_{n})$ and $ν_{n} (A) : = ν (A \cap P_{n})$ on $A$ . By [L4] these are measures, and their total masses are the common finite value $μ (P_{n}) = ν (P_{n})$ .

1.2givenL3

If $Q \in P$ , then $Q \cap P_{n} \in P$ by [L3], so $μ_{n} (Q) = μ (Q \cap P_{n}) = ν (Q \cap P_{n}) = ν_{n} (Q)$ .

2.1step 1.1step 1.2L1

The finite uniqueness lemma applied to steps 1.1 and 1.2 gives $μ (A \cap P_{n}) = ν (A \cap P_{n})$ for every $A \in A$ and every $n$ .

3.1step 2.1L2∎

For fixed $A$ , the sets $A \cap P_{n}$ increase to $A$ ; continuity from below and step 2.1 give $μ (A) = \sup_{n} μ (A \cap P_{n}) = \sup_{n} ν (A \cap P_{n}) = ν (A)$ . Thus the measures agree everywhere.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

FALSE: every subset of a measure-null set is measurable

Statement

False claim. In every measure space, every subset of a measurable null set is measurable. Equivalently, every measure space is complete in the sense of Complete measure spaces.

Facts & Assumptions

Given: The two-point set $X = {0, 1}$ and the family $A = {\emptyset, X}$ .

[L1]

A measure has value $0$ at the empty set and is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

Completeness requires every subset of every measurable null set to belong to the sigma-algebra (Complete measure spaces).

Refutation

technique · direct

1.1givenL1

The family $A$ is a sigma-algebra on $X$ , and the set function $μ (\emptyset) = μ (X) = 0$ is a measure: both sides of countable additivity are $0$ for every disjoint measurable sequence.

1.2given

The set $X$ is measurable and $μ (X) = 0$ , but ${0} \subseteq X$ and ${0} \notin A$ .

2.1step 1.1step 1.2L2∎

By step 1.2 the measure space of step 1.1 is not complete, so the claimed universal measurability of subsets of null sets is false.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21Open item page →

FALSE: continuity from above needs no finiteness hypothesis

Statement

False claim. For every decreasing sequence $(E_{k})$ of measurable sets, one has $μ (⋂_{k} E_{k}) = \inf_{k} μ (E_{k})$ , without requiring any $E_{k}$ to have finite measure. The valid theorem Continuity from above when one set has finite measure includes precisely that missing hypothesis.

Facts & Assumptions

Given: Counting measure $#$ on $N$ and the tails $E_{k} : = {n \in N : k \leq n}$ .

[L1]

Counting measure assigns $+ \infty$ to every infinite set (Counting measure on an arbitrary set) and is a measure (Counting measure is a measure).

[L2]

Continuity from above is proved when one member of the decreasing sequence has finite measure (Continuity from above when one set has finite measure).

[L3]

The natural order is defined by $m \leq n$ exactly when $m + k = n$ for some natural $k$ (Order on the natural numbers), natural addition is cancellative (Addition is cancellative), and $k < k + 1$ (Discreteness: $σ (n)$ is the immediate successor).

Refutation

technique · direct

1.1givenL3

The tails decrease, $E_{k + 1} \subseteq E_{k}$ , and $E_{0} = N$ .

1.2givenL1L3

Every $E_{k}$ is infinite, because $n \mapsto k + n$ injects $N$ into it; hence $# (E_{k}) = + \infty$ for every $k$ .

1.3givenL3

The intersection is empty: if $n$ belonged to every tail, it would belong to $E_{n + 1}$ , which would say $n + 1 \leq n$ , contrary to discreteness of the natural order.

2.1step 1.2step 1.3L1L2∎

Thus $# (⋂_{k} E_{k}) = # (\emptyset) = 0$ but $\inf_{k} # (E_{k}) = + \infty$ . This refutes the claim and shows why [L2] cannot be applied: no tail has finite counting measure.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21Open item page →

FALSE: agreement on an arbitrary generating family determines a measure

Statement

False claim. If two probability measures agree on any family that generates the sigma-algebra, then they agree everywhere. The valid theorem Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system requires the generating family to be a pi-system.

Facts & Assumptions

Given: The set $X = {N W, N E, S W, S E}$ and the family $G$ of its north row, south row, west column, and east column.

[L1]

A sigma-algebra generated by $G$ is the smallest sigma-algebra containing $G$ (The sigma-algebra generated by a family of sets).

[L2]

A probability measure is a measure of total mass $1$ (Probability measures and probability spaces).

[L3]

A pi-system is nonempty and closed under binary intersections (Pi-systems), and agreement on a generating pi-system with the stated finite exhaustion determines a measure (Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system).

Refutation

technique · direct

1.1givenL2

Define $μ$ by masses $1 / 2$ at $N W, S E$ and $0$ at the other points, and define $ν$ by masses $1 / 2$ at $N E, S W$ and $0$ at the other points. Finite summation over subsets makes both probability measures on $P (X)$ .

1.2givenL1L3

Intersections of a row and a column give every singleton, so $σ_{X} (G) = P (X)$ ; but those singleton intersections are not themselves in $G$ , so $G$ is not a pi-system.

2.1givenstep 1.1algebra

Each row and each column contains one point of mass $1 / 2$ for each measure, so $μ$ and $ν$ agree on every member of $G$ .

3.1step 1.1step 2.1step 1.2L3∎

The measures differ, for example $μ ({N W}) = 1 / 2$ while $ν ({N W}) = 0$ . Together with steps 2.1 and 1.2 this refutes the claim and identifies the missing intersection-closure hypothesis.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21Open item page →

FALSE: every finitely additive nonnegative set function on a sigma-algebra is a measure

Statement

False claim. Every finitely additive nonnegative set function on a sigma-algebra is countably additive and therefore is a measure.

Facts & Assumptions

Given: The power-set sigma-algebra $P (N)$ .

[L1]

Finite additivity requires additivity on disjoint pairs and value $0$ at the empty set (Finitely additive nonnegative set functions).

[L2]

A measure must be countably additive on disjoint measurable sequences (Measures on sigma-algebras).

[L3]

Finite sets are those equinumerous with a natural number (Finite, countably infinite, countable, uncountable).

Refutation

technique · direct

1.1givenL3

Define $m (A) = 0$ when $A \subseteq N$ is finite and $m (A) = + \infty$ when $A$ is infinite; in particular $m (\emptyset) = 0$ .

2.1step 1.1L1

If disjoint $A, B$ have finite union, then both are finite and $m (A \cup B) = 0 = 0 + 0$ ; if their union is infinite, at least one is infinite and both sides of $m (A \cup B) = m (A) + m (B)$ are $+ \infty$ . Thus $m$ is finitely additive.

2.2step 1.1L3

The singletons ${k}$ are disjoint, each has value $0$ , and their union is $N$ , which has value $+ \infty$ .

3.1step 2.1step 2.2L2∎

Step 2.2 violates countable additivity, so the finitely additive set function of step 2.1 is not a measure and the claim is false.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

FALSE: measures are additive on arbitrary countable unions

Statement

False claim. For every measure and every measurable sequence $(E_{k})$ , whether or not it is pairwise disjoint,

$μ (⋃_{k} E_{k}) = \sum_{k} μ (E_{k}) .$

Facts & Assumptions

Given: The one-point measurable space $X = {x}$ with sigma-algebra $P (X)$ and its Dirac probability $δ_{x}$ .

[L1]

Countable additivity in the definition of a measure applies only to pairwise disjoint sequences (Measures on sigma-algebras).

[L2]

The Dirac set function assigns $1$ to measurable sets containing $x$ and $0$ otherwise (The Dirac set function at a point), and it is a probability measure (A Dirac set function is a probability measure).

[L3]

A nonnegative extended series starts at index $0$ and is the supremum of its partial sums (Series in the nonnegative extended real line).

Refutation

technique · direct

1.1given

Define $E_{0} = E_{1} = {x}$ and $E_{k} = \emptyset$ for $k \geq 2$ . Then $⋃_{k} E_{k} = {x}$ .

2.1step 1.1L2L3algebra

The union has Dirac measure $1$ , while the term values are $1, 1, 0, 0, \dots$ , whose nonnegative extended sum is $2$ .

3.1step 2.1L1∎

Since $1 \neq 2$ , the claimed equality fails; the repeated set at indices $0$ and $1$ pinpoints the absent disjointness hypothesis in [L1].

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

FALSE: a measure on an infinite set that vanishes on every singleton is the zero measure

Statement

False claim. If $X$ is infinite and a measure $μ$ on $X$ satisfies $μ ({x}) = 0$ for every $x \in X$ , then $μ$ is the zero measure.

Facts & Assumptions

Given: The Axiom of Countable Choice and the uncountable set $X = R$ .

[L1]

A sigma-algebra is closed under complements and countable unions (Sigma-algebras), and a measure is countably additive on disjoint measurable sequences (Measures on sigma-algebras).

[L2]

Countable means finite or in bijection with $N$ (Finite, countably infinite, countable, uncountable), and under countable choice a countable union of at most countable sets is at most countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L3]

The real line is uncountable ( $R$ is uncountable (Cantor's nested intervals, 1874)).

Refutation

technique · direct

1.1givenL1L2L3

Let $A$ consist of the countable and cocountable subsets of $X$ . Complements exchange the two classes, and [L2] shows that a countable union of countable members is countable; if one member is cocountable, the union is cocountable. Hence $A$ is a sigma-algebra.

2.1step 1.1L1L2

Define $μ (A) = 0$ for countable $A$ and $μ (A) = 1$ for cocountable $A$ . In a disjoint measurable sequence at most one member is cocountable; if none is, the union is countable by [L2], and if one is, the union is cocountable. Thus the value on the union equals the sum of the values, so $μ$ is a probability measure.

3.1step 2.1L2L3

Every singleton is countable and has measure $0$ , while $X$ is cocountable and has measure $1$ .

4.1step 2.1step 3.1∎

The measure in step 2.1 vanishes on every singleton but is not the zero measure by step 3.1, so it refutes the claim.

Measures and Their Basic Properties

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Series in the nonnegative extended real line

Definition

Limit superior and limit inferior of a nonnegative extended-real sequence

Definition

Tonelli's theorem for double series of nonnegative extended real numbers

Statement

Facts & Assumptions

Proof

Measures on sigma-algebras

Definition

Measure spaces

Definition

Finitely additive nonnegative set functions

Definition

Finite, sigma-finite, and semifinite measures

Definition

Probability measures and probability spaces

Definition

Measure-null sets and almost-everywhere statements relative to a measure

Definition

Complete measure spaces

Definition

Measures are monotone

Statement

Facts & Assumptions

Proof

Measure of a set difference when the smaller set has finite measure

Statement

Facts & Assumptions

Proof

Finite and countable subadditivity of measures

Statement

Facts & Assumptions

Proof

Continuity from below for measures

Statement

Facts & Assumptions

Proof

Continuity from above when one set has finite measure

Statement

Facts & Assumptions

Proof

Counting measure on an arbitrary set

Definition

Counting measure is a measure

Statement

Facts & Assumptions

Proof

The two-set measure identity μ(A∪B)+μ(A∩B)=μ(A)+μ(B)μ(A∪B)+μ(A∩B)=μ(A)+μ(B)

Statement

Facts & Assumptions

Proof

Inclusion-exclusion for a nonempty finite family of finite-measure sets

Statement

Facts & Assumptions

Proof

Sets whose symmetric difference is null have the same measure

Statement

Facts & Assumptions

Proof

Countable additivity and continuity of finitely additive set functions

Statement

Facts & Assumptions

Proof

The first Borel-Cantelli lemma for measures

Statement

Facts & Assumptions

Proof

The measure of a set liminf is at most the liminf of the measures

Statement

Facts & Assumptions

Proof

The limsup of the measures is at most the measure of the set limsup under a finite-union bound

Statement

Facts & Assumptions

The two-set measure identity $μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B)$