measurability of analytic sets
Analytic subsets (http://planetmath.org/AnalyticSet2) of a measurable space 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.
All analytic subsets of a measurable space are universally measurable.
Therefore for a universally complete measurable space all -analytic sets are themselves in and, in particular, this applies to any complete -finite measure space (http://planetmath.org/SigmaFinite) . 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 is an -analytic set. Then, for any finite measure on , let be the outer measure generated by . This is an -capacity and, by the capacitability theorem, is -capacitable, hence is in the completion of with respect to (see capacity generated by a measure (http://planetmath.org/CapacityGeneratedByAMeasure)). As this is true for all such finite measures, is universally measurable.