CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-10

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Bounded continuous cannot be replaced by all bounded measurable functions

Statement refuted

Weak convergence need not give convergence of integrals for all bounded Borel tests. For $n \geq 1$ , take $μ_{n} = δ_{1 / n}$ , and set $μ = δ_{0}$ and $h = 1_{{0}}$ on the real line.

Facts & Assumptions

[F1]

Weak convergence of borel probability measures: For Borel probability measures $μ_{n}, μ$ on a metric space S, write $μ_{n} \Rightarrow μ$ if $\int f d μ_{n} \to \int f d μ$ for every bounded continuous real function f on S. Continuity is def-metric-continuity. Such f is Borel measurable (inverse images of open sets are open) and $\int ∣ f ∣ d μ \leq ∥ f ∥_{\infty} μ (S) < \infty$ , so the integrals are finite in def-integrable-real-and-complex-functions-and-their-integrals. No completeness or coupling is required.

Counterexample

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

1.1

For every $n \geq 1$ and each bounded continuous f, the point-mass integrals are f(1/n) and f(0), so continuity proves convergence and hence weak convergence by F1.

2.1

The singleton {0} is closed, so h is Borel measurable and bounded between zero and one. But $\int h d μ_{n} = h (1 / n) = 0$ for every $n \geq 1$ , whereas $\int h d μ = h (0) = 1$ . This explicit bounded Borel test fails the conclusion.

givenalgebra∎

Depends on

Weak convergence of borel probability measures

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

van Gaans, Definition 3.1; Dirac test example (standard reference, not scraped)