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Every finite Borel measure on a compact Hausdorff space is regular
Statement
Assuming the Axiom of Countable Choice, the assertion “every finite Borel measure on a compact Hausdorff space is regular” is false.
Facts & Assumptions
Given: The Axiom of Countable Choice and with the resulting Dieudonne probability measure .
Refutation
The space is compact Hausdorff and . For the open Borel set , one has , while every compact is bounded and has .
Hence is not even inner regular on the open set , and therefore is not regular.
Depends on
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- The Dieudonne club-set function is a Borel measure
- Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, $\omega_1$ is countably compact and sequentially compact while $\omega_1 + 1$ is compact
- Regular Borel measure on an LCH space
Used by
- The Dieudonne Borel measure on [0, omega₁] is not regular Counterexample
Dependency tree · two levels
33 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
- Donald L. Cohn, Measure Theory, 2nd ed., Chapter 7 (standard reference, not scraped)