measurability of analytic sets

AnalyticPlanetmathPlanetmath subsets ( of a measurable spaceMathworldPlanetmathPlanetmath (X,) do not, in general, have to be measurable. See, for example, a Lebesgue measurable but non-Borel set ( However, the following result is true.


All analytic subsets of a measurable space are universally measurable.

Therefore for a universally complete measurable space (X,) all -analytic setsMathworldPlanetmath are themselves in and, in particular, this applies to any completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath σ-finite measure space ( (X,,μ). For example, analytic subsets of the real numbers are Lebesgue measurable.

The proof of the theorem follows as a consequence of the capacitability theorem. Suppose that A is an -analytic set. Then, for any finite measureMathworldPlanetmath μ on (X,), let μ* be the outer measure generated by μ. This is an -capacity and, by the capacitability theorem, A is (,μ*)-capacitable, hence is in the completion of with respect to μ (see capacity generated by a measure ( As this is true for all such finite measures, A is universally measurable.

Title measurability of analytic sets
Canonical name MeasurabilityOfAnalyticSets
Date of creation 2013-03-22 18:47:24
Last modified on 2013-03-22 18:47:24
Owner gel (22282)
Last modified by gel (22282)
Numerical id 6
Author gel (22282)
Entry type Theorem
Classification msc 28A05