PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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.

Depends on

Used by

A non-sigma-finite premeasure has distinct Borel extensions Counterexample
A pointwise limit of integrable functions need not be integrable Counterexample
Counting measure on [0,1] shows the dominating measure needs sigma-finiteness Counterexample
Counting-measure tails decrease to the empty set while every term has infinite measure Counterexample
Jensen's inequality can fail on an infinite measure space without normalization Counterexample
Lebesgue plus counting measure has no Lebesgue decomposition relative to Lebesgue measure Counterexample
Assuming countable choice, counting measure is sigma-finite exactly on countable sets Example
Counting measure is an outer measure for which every subset is measurable Example
Counting measure represents finite-support summation on a discrete LCH space Example
Diagonal operator on ell p is compact iff diagonal tends to zero Example
Finite counting measure on n points recovers ℝⁿ p-norms Example
Finite-rank truncations of a square-integrable kernel Example
Integrating against counting measure recovers a series Example
Spectrum of the unilateral shift Example
Every translation-invariant measure on the Borel sets of ℝ is a nonnegative multiple of Lebesgue measure False statement
FALSE: continuity from above needs no finiteness hypothesis False statement
FALSE: every Borel measure on ℝ is finite on compact sets False statement
FALSE: Jensen's inequality holds on an infinite measure space without normalization False statement
FALSE: Lᵖ(μ) is separable for every measure μ and every 1 ≤ p < ∞ False statement
FALSE: pointwise limits of integrable functions are integrable False statement
FALSE: the extension of a premeasure is always unique False statement
FALSE: the Radon-Nikodym theorem holds without sigma-finiteness False statement
Counting measure on a discrete group is Haar, Haar measures there are its multiples, and integrals against them are sums Lemma
The p-functional need not be a norm for 0 < p < 1 Proposition
ℓᵖ is the Lᵖ space of counting measure Remark
Assuming countable choice, every measure is the sum of its semifinite part and a zero-infinity-valued measure Theorem
ℓᵖ includes into ℓʳ for p < r Theorem

Cited to discharge well-definedness by Counting measure on an arbitrary set.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Axler, Measure, Integration & Real Analysis, Example 2.55 (standard reference, not scraped)