Universally measurable set

Remarks

Every $A \in \Sigma$ is universally measurable, since $\Sigma \subseteq \Sigma_\mu$ for every $\mu$. The universally measurable sets form a σ-algebra, the intersection of all the $\Sigma_\mu$. Let $X = [0, 1]$ with its Borel σ-algebra, and let $\lambda$ be Lebesgue measure on it. Then $\Sigma_\lambda$ is the σ-algebra of Lebesgue measurable subsets of $[0, 1]$, and a Vitali set $V \subseteq [0, 1]$ is not in it. So $V$ is not universally measurable. The definition is usually applied when $X$ is a Polish space and $\Sigma$ is its Borel σ-algebra; then every analytic set is universally measurable. Replacing probability measures by σ-finite measures gives the same family of sets.