# measurability of analytic sets

Analytic subsets (http://planetmath.org/AnalyticSet2) of a measurable space $(X,\mathcal{F})$ do not, in general, have to be measurable. See, for example, a Lebesgue measurable but non-Borel set (http://planetmath.org/ALebesgueMeasurableButNonBorelSet). However, the following result is true.

###### Theorem.

All analytic subsets of a measurable space are universally measurable.

Therefore for a universally complete measurable space $(X,\mathcal{F})$ all $\mathcal{F}$-analytic sets are themselves in $\mathcal{F}$ and, in particular, this applies to any complete $\sigma$-finite measure space (http://planetmath.org/SigmaFinite) $(X,\mathcal{F},\mu)$. For example, analytic subsets of the real numbers $\mathbb{R}$ are Lebesgue measurable.

The proof of the theorem follows as a consequence of the capacitability theorem. Suppose that $A$ is an $\mathcal{F}$-analytic set. Then, for any finite measure $\mu$ on $(X,\mathcal{F})$, let $\mu^{*}$ be the outer measure generated by $\mu$. This is an $\mathcal{F}$-capacity and, by the capacitability theorem, $A$ is $(\mathcal{F},\mu^{*})$-capacitable, hence is in the completion of $\mathcal{F}$ with respect to $\mu$ (see capacity generated by a measure (http://planetmath.org/CapacityGeneratedByAMeasure)). As this is true for all such finite measures, $A$ is universally measurable.

Title measurability of analytic sets MeasurabilityOfAnalyticSets 2013-03-22 18:47:24 2013-03-22 18:47:24 gel (22282) gel (22282) 6 gel (22282) Theorem msc 28A05