Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Cauchy sequences in probability have a measurable limit

Statement

Let (Yn)n1 be real random variables on one probability space. Suppose that for every ε,η>0 there is N such that P(YnYm>ε)<η(n,mN). Then there is a finite measurable real random variable Y such that YnY in probability.

Facts & Assumptions

[F1]

Convergence in probability: For real random variables (Xn) and X on one probability space, write XnX in probability when, for every ε>0, P(XnX>ε)0. This is precisely def-convergence-in-measure for the probability measure.

[F2]

First Borel-Cantelli lemma for events: Let (An)nN be events in a probability space. If n=0P(An)<+, then P(An i.o.)=0. No independence hypothesis is needed.

[F3]

A series converges iff for every ε>0 there is N with am+1++an<ε for all n>mN: Let (ak) be a sequence of reals, with partial sums sn=k<nak (def-series). Then ak converges if and only if for every real ε>0 there is NN such that k=m+1nak<ε for all n>mN. The block k=m+1nak is the finite sum am+1++an of def-finite-sum, and it equals sn+1sm+1. This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

[F4]

Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable: Let (X,A) be a measurable space and let fn:XR be measurable for every nN. Then the functions supnfn,infnfn,lim supnfn,lim infnfn are measurable. The set {x:limnfn(x) exists in R} is measurable. In particular, if fnf pointwise, then f is measurable.

[F5]

Almost-sure convergence implies convergence in probability: If XnX almost surely, then XnX in probability.

[F6]

Finite and countable subadditivity of measures: Let μ be a measure and let (Ek)kN be measurable. Then μ(kNEk)k=0μ(Ek). For every mN one also has μ(k<mEk)k<mμ(Ek), including m=0, where both sides are 0.

Proof

Given: The objects and hypotheses of the statement.

1.1

Set n0=1. For each k1, choose recursively the least integer nk>nk1 for which all pairs of indices at least nk have probability less than 2k of separation exceeding 2k. Such an integer exists by the hypothesis. Thus, for every k1, P(Ynk+1Ynk>2k)<2k. The first Borel–Cantelli lemma gives a measurable probability-one event where these inequalities fail only finitely often.

F2given
2.1

On that event the series of absolute successive differences is finite: its finite initial part is finite because all values are real, and its remaining part is bounded by a geometric series. Therefore the subsequence is Cauchy and has a finite real limit. Its finite convergence event is measurable: intersect the measurable extended-limit event with {supkYnk<}, a countable union of countable intersections. Define Y to be this limit there and zero outside. The corresponding restricted sequence converges everywhere, so measurable limits give a real random variable.

F3F4step 1.1algebra
3.1

This subsequence converges almost surely and hence in probability to Y. Fix ε,η>0, choose N so late-pair errors at ε/2 are below η/2, and then choose k with nkN and P(YnkY>ε/2)<η/2. For all nN, the triangle and union bounds give P(YnY>ε)<η. This proves convergence of the full sequence, including constant sequences.

F5F6F1step 2.1givenalgebra

Depends on

Used by

Dependency tree · two levels

28 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