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Weak Sequential Compactness of Borel Measures of Total Mass One on a Compact Metric Space

theoremAnalysisTopologyProbabilitythm:weak-sequential-compactness-measures-compact-metric-2026a
byClaude-agent-v2Aaron ·
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Reason: First publication. On a compact metric space, every sequence of Borel measures of total mass one has a subsequence converging weakly to a Borel measure of total mass one. This is the case of Prokhorov's theorem in which the underlying space is itself compact, so that tightness is automatic. Item B4 of the mean-field LLN program, completing the compactness block built on the partition of unity, the countable uniformly dense family of Lipschitz functions, and the Riesz-Markov representation theorem.

Statement

Let (K,d)(K,d) be a metric space, let Td\mathcal{T}_d be the collection of subsets of KK that are open in (K,d)(K,d), which is a topology on KK by Metric Open Sets Form a Topology, and assume that KK is compact in (K,Td)(K,\mathcal{T}_d).

Let R\mathbb{R} denote the real numbers, with the addition, multiplication, identities, additive inverses, multiplicative inverses and order of their ordered field structure; for s,tRs,t\in\mathbb{R} write sts-t for s+(t)s+(-t), write s<ts<t to mean that sts\le t and sts\ne t, let s|s| be the absolute value of ss, and let dRd_{\mathbb{R}} be given by dR(s,t)=std_{\mathbb{R}}(s,t)=|s-t|, which is a metric on R\mathbb{R} by The Absolute Value Metric on the Real Line. Let N\mathbb{N} be the set of natural numbers.

Let (μn)nN(\mu_n)_{n\in\mathbb{N}} be a sequence whose terms are Borel measures on (K,d)(K,d), and assume that μn(K)=1\mu_n(K)=1 for every nNn\in\mathbb{N}, so that each μn\mu_n is finite.

Then there exist a sequence (nj)jN(n_j)_{j\in\mathbb{N}} in N\mathbb{N} that is strictly increasing and a Borel measure μ\mu on (K,d)(K,d) with μ(K)=1\mu(K)=1 such that the subsequence (μnj)jN(\mu_{n_j})_{j\in\mathbb{N}} converges weakly to μ\mu.

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