2020 Mathematics Subject Classification: Primary: 28A05 Secondary: 03E1554H05 [MSN][ZBL]
$\newcommand{\A}{\mathcal A} \newcommand{\B}{\mathcal B}$ The term "universally measurable" may be applied to
Definition 1. Let $(X,\A)$ be a measurable space. A subset $A\subset X$ is called universally measurable if it is $\mu$-measurable for every finite measure $\mu$ on $(X,\A)$. In other words: $\mu_*(A)=\mu^*(A)$ where $\mu_*,\mu^*$ are the inner and outer measures for $\mu$, that is,
(See [C, Sect. 8.4], [S, p. 170].)
Universally measurable sets evidently are a σ-algebra that contains the σ-algebra $\A$ of measurable sets.
Warning. Every measurable set is universally measurable, but an universally measurable set is generally not measurable! This terminological anomaly appears because the word "measurable" is used differently in two contexts, of measurable spaces and of measure spaces.
Definition 2. A separable metric space is called universally measurable if it is a universally measurable subset (as defined above) of its completion. Here the completion, endowed with the Borel σ-algebra, is treated as a measurable space. (See [S, p. 170], [D, Sect. 11.5].)
Definition 3. A measurable space is called universally measurable if it is isomorphic to some universally measurable metric space (as defined above) with the Borel σ-algebra. (See [S, p. 171].)
Thus, the phrase "universally measurable space" is ambiguous; it can be interpreted as "universally measurable metric space" or "universally measurable measurable space"! The latter can be replaced with "universally measurable Borel space", but the ambiguity persists. Fortunately, the ambiguity is rather harmless by the following result.
Theorem 1 (Shortt [S, Theorem 1]). The following two conditions on a separable metric space are equivalent:
Evidently, (a) implies (b); surprisingly, also (b) implies (a), which involves a Borel isomorphism (rather than isometry or homeomorphism) between two metric spaces.
Theorem 2 (Shortt [S, Lemma 4]). A countably generated separated measurable space $(X,\A)$ is universally measurable if and only if for every finite measure $\mu$ on $(X,\A)$ there exists a subset $A\in\A$ of full measure (that is, $\mu(X\setminus A)=0$) such that $A$ (treated as a subspace) is itself a standard Borel space.
Every standard Borel space evidently is universally measurable. And moreover:
Theorem 3. Every analytic Borel space is universally measurable.
In [M, Sect. 6] universally measurable spaces are called metrically standard Borel spaces.
In [K, Sect. 21.D] universally measurable subsets of a standard (rather than arbitrary) measurable space are defined.
In [N, Sect. 1.1] an absolute measurable space is defined as a separable metrizable topological space such that every its homeomorphic image in every such space (with the Borel σ-algebra) is a universally measurable subset. The corresponding measurable space (with the Borel σ-algebra) is also called an absolute measurable space in [N, Sect. B.2].
[S] | Rae M. Shortt, "Universally measurable spaces: an invariance theorem and diverse characterizations", Fundamenta Mathematicae 121 (1984), 169–176. MR0765332 Zbl 0573.28018 |
[C] | Donald L. Cohn, "Measure theory", Birkhäuser (1993). MR1454121 Zbl 0860.28001 |
[D] | Richard M. Dudley, "Real analysis and probability", Wadsworth&Brooks/Cole (1989). MR0982264 Zbl 0686.60001 |
[M] | George W. Mackey, "Borel structure in groups and their duals", Trans. Amer. Math. Soc. 85 (1957), 134–165. MR0089999 Zbl 0082.11201 |
[K] | Alexander S. Kechris, "Classical descriptive set theory", Springer-Verlag (1995). MR1321597 Zbl 0819.04002 |
[N] | Togo Nishiura, "Absolute measurable spaces", Cambridge (2008). MR2426721 Zbl 1151.54001 |