Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-10
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Converging together lemma

Statement

Let Xn and Yn be Borel-measurable random elements with values in a metric space S, on the same probability space for each n, and let X be an S-valued Borel-measurable random element. Suppose d(Xn,Yn) is measurable and P(d(Xn,Yn)>ε)→0 for every ε>0. If Xn⇒X, then Yn⇒X.

Facts & Assumptions

[F1]

Portmanteau theorem: For Borel probabilities μn,μ on a metric space S, the following are equivalent: (i) μn⇒μ; (ii) integrals converge for all bounded uniformly continuous real tests; (iii) lim sup⁡nμn(F)≤μ(F) for every closed F; (iv) lim inf⁡nμn(G)≥μ(G) for every open G; (v) μn(A)→μ(A) for every Borel A with μ(∂A)=0.

[F2]

Continuity from above when one set has finite measure: Let (En)n∈N be a decreasing sequence of measurable sets for a measure μ. If μ(En0)<+∞ for some n0, then

μ(⋂n∈NEn)=inf⁡n∈Nμ(En).

[F3]

∣d(x,A)−d(y,A)∣≤d(x,y), so the distance to a fixed nonempty set is 1-Lipschitz: Let (X,d) be a metric space (def-metric-space), let A⊆X be nonempty and let x,y∈X. Then

∣d(x,A)−d(y,A)∣≤d(x,y),

with d(⋅,A) the distance to a nonempty set (def-metric-bounded-diameter). Thus the real-valued function u↦d(u,A) changes by at most d(u,v) between u and v: it is 1-Lipschitz.

Proof

Given: The objects, hypotheses and definitions in the statement. Its conclusions are to be established below.

1.1givenalgebraF3

By F3, distance to nonempty F is continuous; its sublevel sets are closed. For nonempty closed F put F[ε]={x:d(x,F)≤ε}, which is closed. If Yn lies in F and d(Xn,Yn)≤ε, then Xn lies in this enlargement. Thus P(Yn∈F)≤P(Xn∈F[ε])+P(d(Xn,Yn)>ε).

2.1F1F2∎

F1 and the probability hypothesis give lim sup⁡nP(Yn∈F)≤PX(F[ε]). For m∈N, the sets Em:=F[1/(m+1)] decrease to F, so F2 makes their probabilities decrease to P_X(F). Empty F has probability zero without an enlargement. The resulting closed-set bound is again F1, now proving Yn⇒X.

Depends on

Used by

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Sources