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Assuming countable choice, Borel probability measures on Polish spaces are inner regular
Statement
Assume countable choice. If is Polish and is a Borel probability measure on , then for every Borel and there is a compact with .
Facts & Assumptions
Given: A complete separable metric presentation , a Borel probability , and .
A Polish space has a complete compatible metric and a countable dense set. (Polish spaces are separable completely metrizable spaces)
Measures are continuous from below and countably subadditive. (Continuity from below for measures, Finite and countable subadditivity of measures)
A closed totally bounded subset of a complete metric space is compact. (A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed, A complete, totally bounded metric space is compact, proved from countable choice used exactly once)
A lambda-system containing a pi-system contains the sigma-algebra it generates. (Dynkin's pi-lambda theorem)
Proof
The empty is immediate, so suppose and enumerate a countable dense set as . For each , continuity from below chooses a finite union of -balls centered at the with . Countable choice makes these choices simultaneously.
The set is closed. Countable subadditivity gives . For every scale, one of the finite closed-ball covers from step 1.1 covers ; choosing one point of from each nonempty member of that finite cover and doubling the radius gives a finite net with centres in . Hence is totally bounded and [F3] makes it compact.
Every open contains a compact losing less than : for use ; otherwise intersect with the increasing closed sets . Their union is , so [F2] gives one with the required loss.
Let be the Borel sets which, for every , have compact with open and . Step 3.1 puts every open set in . The tight compact set from step 2.1 shows that complements remain in : from , use and bound the loss by .
For pairwise disjoint , continuity from below makes the measure of the tail of arbitrarily small. A finite union of compact inner approximants handles that tail, while the union of the open outer approximants has loss bounded by the summable errors. Thus is a lambda-system. Open sets are a pi-system generating the Borel sigma-algebra, so [F4] gives . The compact inner approximant for proves the statement.
Depends on
- Polish spaces are separable completely metrizable spaces
- Probability measures and probability spaces
- The Borel sigma-algebra of a topological space
- A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$
- Open ball, closed ball and sphere in a metric space
- Finite $\varepsilon$-net and totally bounded metric space
- Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed
- Continuity from below for measures
- Finite and countable subadditivity of measures
- Measure of a set difference when the smaller set has finite measure
- A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed
- A complete, totally bounded metric space is compact, proved from countable choice used exactly once
- Dynkin's pi-lambda theorem
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right
Used by
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Sources
- Biskup, MATH 275D notes, Lemma 2.7 (standard reference, not scraped)