LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generated

How statement and proof provenance work

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

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

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

A probability measure on the line has no nonzero translation period

Statement

Let $μ$ be a Borel probability measure on $R$ . If a real number $a$ satisfies $μ (A + a) = μ (A)$ for every Borel set $A \subseteq R$ , then $a = 0$ .

Statement references

A probability measure has total mass one and is countably additive

Translation of a subset of the real line

Facts & Assumptions

Given: A Borel probability measure $μ$ on $R$ and an $a \in R$ such that $μ (A + a) = μ (A)$ for every Borel set $A$ .

[F1]

A probability measure is countably additive and has total mass one. Probability measures and probability spaces

[F2]

$A + a = {x + a : x \in A}$ is the translate of $A$ . Translation of a subset of $R^{n}$

[F3]

The real line is a complete ordered field, hence its positive reals satisfy the Archimedean property. Dedekind-complete ordered real field, Every complete ordered field is Archimedean

Proof

Given: The objects and hypotheses in the Statement.

1.1givenF2construct

Suppose $a \neq 0$ and put $ℓ = ∣ a ∣ > 0$ . Invariance under translation by $a$ implies invariance under translation by $ℓ$ : when $a = - ℓ$ , apply the given identity to $A + ℓ$ . The half-open intervals $I_{k} = [k ℓ, (k + 1) ℓ)$ , $k \in Z$ , partition $R$ and are translates of $I_{0}$ , so they all have the same measure $c = μ (I_{0})$ .

2.1step 1.1F1F3contradiction

If $c > 0$ , choose a natural $n > 1 / c$ by [F3]. The first $n$ intervals are disjoint, so finite additivity gives $1 \geq μ (⋃_{k = 0}^{n - 1} I_{k}) = n c > 1$ , a contradiction. Hence $c = 0$ .

3.1step 1.1step 2.1F1discharge-contradiction∎

Countable additivity over the partition $(I_{k})_{k \in Z}$ now gives $1 = μ (R) = \sum_{k \in Z} μ (I_{k}) = 0$ , again a contradiction. Therefore $a = 0$ .

References

Peters et al., JMLR (2014), Appendix A.5

Depends on

Used by

Dependency tree · two levels

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

Sources

Jonas Peters, Joris Mooij, Dominik Janzing, and Bernhard Schölkopf, Causal Discovery with Continuous Additive Noise Models, Appendix A.5 (standard reference, not scraped)