measurability of analytic sets

Analytic subsets (http://planetmath.org/AnalyticSet2) of a measurable space $(X,ℱ)$ 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 $(X,ℱ)$ all $ℱ$-analytic sets are themselves in $ℱ$ and, in particular, this applies to any complete $\sigma$-finite measure space (http://planetmath.org/SigmaFinite) $(X,ℱ,\mu )$. For example, analytic subsets of the real numbers $ℝ$ are Lebesgue measurable.

The proof of the theorem follows as a consequence of the capacitability theorem.Suppose that $A$ is an $ℱ$-analytic set. Then, for any finite measure $\mu$ on $(X,ℱ)$, let ${\mu }^{*}$ be the outer measure generated by $\mu$. This is an $ℱ$-capacity and, by the capacitability theorem, $A$ is $(ℱ,{\mu }^{*})$-capacitable, hence is in the completion of $ℱ$ 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.