# 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 |
---|---|

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 |