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Boundary-dominated windows fail the Van Hove condition

Example

The expanding rectangles $W_{n} = [0, n + 1] \times [0, 1] \subset R^{2}, n \geq 0,$ have volume tending to infinity, but their boundary remains a fixed proportion of the volume under a fixed-width enlargement.

Facts & Assumptions

Given: AC $_{ω}$ as required by the Van Hove definition, the definition of a Van Hove window sequence, and the compact set $K = [- 1, 1]^{2}$ .

[F1]

Van Hove sequences of spatial windows, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable, and Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, and Measures on sigma-algebras: under AC $_{ω}$ the Borel rectangles are Lebesgue measurable, have the stated box volumes, and their restricted outer measure is a countably additive measure; Van Hove requires $∣ (W_{n} + K) ∖ W_{n} ∣ / ∣ W_{n} ∣ \to 0$ for every compact $K$ .

Verification

1.1F1givenalgebra

Each rectangle is bounded and Borel, and $∣ W_{n} ∣ = n + 1 > 0$ , so its volume tends to infinity as required in the first Van Hove condition.

2.1F1givenalgebra∎

Minkowski addition gives $W_{n} + K = [- 1, n + 2] \times [- 1, 2]$ . The two closed rectangles are nested, have finite volume by [F1], and Lebesgue measure is a measure by [F1], so countable additivity on $W_{n} ⊔ ((W_{n} + K) ∖ W_{n}) = W_{n} + K$ gives $\frac{∣ (W_{n} + K) ∖ W_{n} ∣}{∣ W_{n} ∣} = \frac{3 (n + 3) - (n + 1)}{n + 1} = 2 + \frac{6}{n + 1} ⟶ 2.$ This limit is not zero, contradicting [F1] for the compact $K$ ; hence the sequence is not Van Hove.

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