Difference between revisions of "Measurable space"
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In particular, a set of the form $ \{(x,y)\in\R^2:x-y\in B\} $ belongs to $ \A_{1,1}$ if and only if $B\in\B_1$ (rather than $B\in\A_1$). | In particular, a set of the form $ \{(x,y)\in\R^2:x-y\in B\} $ belongs to $ \A_{1,1}$ if and only if $B\in\B_1$ (rather than $B\in\A_1$). | ||
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+ | ''Example and warning.'' Given a one-to-one map $f:\R\to\R$, we consider the indicator of its graph: $g(x,y)=1$ when $y=f(x)$, otherwise $g(x,y)=0$. The function $y\mapsto g(x,y)$ is Borel measurable for every $x$ (being just the indicator of a single point), and the function $x\mapsto g(x,y)$ is Borel measurable for every $y$. In other words, $g(x,y)$ is Borel measurable in $x$ and $y$ ''separately.'' Nevertheless $g$ is not Borel measurable, unless $f$ is. In other words, $g(x,y)$ need not be ''jointly'' measurable in $x$ and $y$. | ||
''Example and warning.'' The set $\R^X$ of all functions $X\to\R$ may be thought of as the product of copies of $\R$; the corresponding σ-algebra on $\R^X$ is generated by the evaluation maps $f\mapsto f(x)$ for all $x\in X$. However, for uncountable $X$ this approach is less useful than it may seem, because $f(x)$ fails to be jointly measurable in $f$ and $x$ even if all subsets of $X$ are measurable. That is, the map $(f,x)\mapsto f(x)$ from $\R^X\times X$ to $\R$ is not measurable. | ''Example and warning.'' The set $\R^X$ of all functions $X\to\R$ may be thought of as the product of copies of $\R$; the corresponding σ-algebra on $\R^X$ is generated by the evaluation maps $f\mapsto f(x)$ for all $x\in X$. However, for uncountable $X$ this approach is less useful than it may seem, because $f(x)$ fails to be jointly measurable in $f$ and $x$ even if all subsets of $X$ are measurable. That is, the map $(f,x)\mapsto f(x)$ from $\R^X\times X$ to $\R$ is not measurable. |
Revision as of 20:51, 8 January 2012
Also: Borel space
2020 Mathematics Subject Classification: Primary: 28A05 Secondary: 03E1554H05 [MSN][ZBL]
$ \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\Om}{\Omega} \newcommand{\A}{\mathcal A} \newcommand{\B}{\mathcal B} \newcommand{\P}{\mathbf P} \newcommand{\D}{\mathrm d} \newcommand{\M}{\mathcal M} $ A measurable space is a set with a distinguished σ-algebra of subsets (called measurable). More formally, it is a pair $(X,\A)$ consisting of a set $X$ and a σ-algebra $\A$ of subsets of $X$.
Examples: $\R^n$ with the Borel σ-algebra; $\R^n$ with the Lebesgue σ-algebra.
Basic notions
Let $(X,\A)$ and $(Y,\B)$ be measurable spaces.
- A map $f:X\to Y$ is called measurable if $f^{-1}(B) \in \A$ for every $B\in\B$.
- These two measurable spaces are called isomorphic if there exists a bijection $f:X\to Y$ such that $f$ and $f^{-1}$ are measurable (such $f$ is called an isomorphism).
Let $X$ be a set, $(Y,\B)$ a measurable space, and $(f_i)_{i\in I}$ a family of maps $f_i:X\to Y$. The σ-algebra generated by these maps is defined as the smallest σ-algebra $\A$ on $X$ such that all $f_i$ are measurable from $(X,\A)$ to $(Y,\B)$. More generally, one may take measurable spaces $(Y_i,\B_i)$ and maps $f_i:X\to Y_i$. On the other hand, if $Y$ is $\R$ (or $\C$, $\R^n$ etc.) then $\B$ is by default the Borel σ-algebra.
Given a family of measurable spaces $(X_i,\A_i)$ for $i\in I$, their product is defined as the measurable space $(X,\A)$ where $X=\prod_i X_i$ is the direct product of sets, and $\A$ is generated by the projection maps $p_i:X\to X_i$ [3, Sect. 10.B].
Example and warning. Denoting the Borel σ-algebra on $\R^n$ by $ \B_n $ and the Lebesgue σ-algebra on $\R^n$ by $ \A_n $ we have $ (\R^m,\B_m) \times (\R^n,\B_n) = (\R^{m+n},\B_{m+n}) $ but $ (\R^m,\A_m) \times (\R^n,\A_n) \ne (\R^{m+n},\A_{m+n}) $ [1, Exercise 1.7.19]. Denoting $ (\R^m,\A_m) \times (\R^n,\A_n) = (\R^{m+n},\A_{m,n}) $ we have for every $ A \in \A_{m,n} $ \[ \{y\in\R^n:(x,y)\in A\} \in \B_n \quad \text{for almost all } x\in\R^m. \] In particular, a set of the form $ \{(x,y)\in\R^2:x-y\in B\} $ belongs to $ \A_{1,1}$ if and only if $B\in\B_1$ (rather than $B\in\A_1$).
Example and warning. Given a one-to-one map $f:\R\to\R$, we consider the indicator of its graph: $g(x,y)=1$ when $y=f(x)$, otherwise $g(x,y)=0$. The function $y\mapsto g(x,y)$ is Borel measurable for every $x$ (being just the indicator of a single point), and the function $x\mapsto g(x,y)$ is Borel measurable for every $y$. In other words, $g(x,y)$ is Borel measurable in $x$ and $y$ separately. Nevertheless $g$ is not Borel measurable, unless $f$ is. In other words, $g(x,y)$ need not be jointly measurable in $x$ and $y$.
Example and warning. The set $\R^X$ of all functions $X\to\R$ may be thought of as the product of copies of $\R$; the corresponding σ-algebra on $\R^X$ is generated by the evaluation maps $f\mapsto f(x)$ for all $x\in X$. However, for uncountable $X$ this approach is less useful than it may seem, because $f(x)$ fails to be jointly measurable in $f$ and $x$ even if all subsets of $X$ are measurable. That is, the map $(f,x)\mapsto f(x)$ from $\R^X\times X$ to $\R$ is not measurable.
Given a measurable space $(X,\A)$, an equivalence relation $\stackrel{\A}{\sim}$ on $X$, defined by \[ x\stackrel{\A}{\sim}y \quad \text{means} \quad \forall A\in\A \; (\,x\in A \Longleftrightarrow y\in A\,), \] leads to a partition of $X$ into equivalence classes. Every measurable set is saturated (that is, $x\sim y$ implies $x\in A \Longleftrightarrow y\in A$). If the set of equivalence classes is finite or countable then all saturated sets are measurable. But in general saturated sets are more than a σ-algebra; an arbitrary (not just countable) union of saturated sets is a saturated set.
Some classes of measurable spaces
A measurable space $(X,\A)$ (as well as its σ-algebra $\A$) is called countably generated if $\A$ is generated by some countable subset of $\A$.
The product of a finite or countable family of countably generated measurable spaces is countably generated.
If $(X,\A)$ is countably generated then the cardinality of $\A$ is at most continuum [1, Exercise 1.4.16].
Example: $\R^n$ with the Borel σ-algebra is countably generated; $\R^n$ with the Lebesgue σ-algebra is not. Every countably generated sub-σ-algebra $\A_0$ of the Lebesgue σ-algebra is almost Borel in the following sense: there exists a Borel set $B_0$ of full measure such that $A\cap B_0$ is a Borel set for every $A\in\A_0$. The Borel σ-algebra is of cardinality continuum; the Lebesgue σ-algebra is of higher cardinality (since it contains all subset of a null set of cardinality continuum).
A measurable space $(X,\A)$ is called
- separated (in other words, separating points) if the corresponding equivalence relation is the equality, that is, $\{A\in\A:x\in A\}=\{A\in\A:y\in A\}$ implies $x=y$ for $x,y\in X$;
- countably separated if there exists a sequence of sets $A_n\in\A$ such that $\{n:x\in A_n\}=\{n:y\in A_n\}$ implies $x=y$ for $x,y\in X$.
If $(X,\A)$ is separated and $X$ is finite or countable then all subsets of $X$ are measurable.
Example: $\R^n$ with the Borel σ-algebra is countably separated; the same holds for the Lebesgue σ-algebra.
Let $(X,\A)$, $(Y,\B)$ be measurable spaces, $f:X\to Y$ a measurable map, and $(Y,\B)$ countably separated. Then the graph $\{(x,f(x)):x\in X\}$ of $f$ is a measurable subset of $X\times Y$. (See [3, Sect. 12.A].)
A much deeper theory is available for standard measurable spaces (see the separate article).
Relations to measures and integrals
An integral in one variable is measurable in the other variable(s) in the following sense.
Let $(X,\A)$ and $(Y,\B)$ be measurable spaces, $\mu$ a finite measure on $(Y,\B)$, and $f:X\times Y\to\R$ a bounded measurable function. Then the function $g:X\to\R$ defined by \[ g(x) = \int f(x,y) \, \mu(\D y) \] is measurable.
Moreover, the integral is jointly measurable in $x$ and $\mu$ in the following sense.
The formula \[ G(x,\mu) = \int f(x,y) \, \mu(\D y) \] defines a measurable function $G:X\times\M(Y)\to\R$, where $\M(Y)$ is the set of all finite measures on $Y$, endowed with the σ-algebra generated by the maps $\mu\mapsto\mu(B)$ for all $B\in\B$. (See [3, Sect. 17.E].)
Relations to topological spaces
Every topology generates a σ-algebra, called Borel σ-algebra. That is, the Borel σ-algebra on a topological space is, by definition, generated by the open sets.
Example. The following three σ-algebras on a separable Hilbert space $H$ are equal:
- the σ-algebra generated by the linear functionals $ x \mapsto \langle x,y \rangle $ for $y\in H$;
- the Borel σ-algebra corresponding to the norm topology on $H$;
- the Borel σ-algebra corresponding to the weak topology on $H$.
That is instructive: topological spaces are not a prerequisite to measurable spaces.
A Borel measurable map is generally not continuous, and a Borel isomorphism is generally not a homeomorphism. However, every Borel measurable homomorphism between Polish groups is continuous. Accordingly, the topology of a Polish group is uniquely determined by its Borel σ-algebra (see [3, Sect. 9.C], [4, Sect. 1.2]).
Example. Consider again a separable Hilbert space $H$.
- Every Borel measurable linear functional $H\to\C$ is continuous.
- Every Borel measurable linear operator $H\to H$ is continuous.
- A set $U\subset H$ contains a neighborhood of the origin (in the norm topology) if and only if there exists a sequence of Borel sets $A_1,A_2,\dots\subset H$ such that $A_1\cup A_2\cup\dots=H$ and $A_n-A_n\subset U$ for all $n$ (that is, $x-y\in U$ for all $x,y\in A_n$).
The Borel σ-algebra is not the only bridge between topological and measurable spaces. All sets having the Baire property (sometimes called Baire sets, which may be confusing) are a σ-algebra (generated by open sets together with meager sets) greater than Borel [3, Sect.8.F]. On the other hand, all compact $G_\delta$ subsets of a compact Hausdorff topological space generate a σ-algebra (smaller than Borel) of sets called Baire sets in [5, Sect.7.1], [7, Sect.51]. For more general (in particular, uncountable discrete) topological spaces the definitions of [5] and [7] disagree. Note also the σ-algebra of universally measurable sets.
On terminology
"Borel space" and "measurable space" are often used as synonyms. But according to [1, Sect. 12.A] a Borel space is a countably generated measurable space that separates points (or equivalently, a measurable space isomorphic to a separable metric space with the Borel σ-algebra), in which case "Borel" instead of "measurable" applies also to sets and maps.
"Separated" and "countably separated" are used in [6].
Weaker assumptions on $\A$ were usual in the past. For example, according to [7], $\A$ need not contain the whole $X$, it is a σ-ring, not necessarily a σ-algebra. According to [8], a measurable space is not a pair $(X,\A)$ but a measure space $(X,\A,\mu)$ such that $X\in\A$ (and again, $\A$ is generally a σ-ring).
References
[1] | Terence Tao, "An introduction to measure theory", AMS (2011) | MR2827917 | Zbl 05952932 |
[2] | David Pollard, "A user's guide to measure theoretic probability", Cambridge (2002) | MR1873379 | Zbl 0992.60001 |
[3] | Alexander S. Kechris, "Classical descriptive set theory", Springer-Verlag (1995) | MR1321597 | Zbl 0819.04002 |
[4] | Howard Becker and Alexander S. Kechris, "The descriptive set theory of Polish group actions", Cambridge (1996) | MR1425877 | Zbl 0949.54052 |
[5] | Richard M. Dudley, "Real analysis and probability", Wadsworth&Brooks/Cole (1989) | MR0982264 | Zbl 0686.60001 |
[6] | George W. Mackey, "Borel structure in groups and their duals", Trans. Amer. Math. Soc. 85 (1957), 134–165 | MR0089999 | Zbl 0082.11201 |
[7] | Paul R. Halmos, "Measure theory", v. Nostrand (1950) | MR0033869 | Zbl 0040.16802 |
[8] | Walter Rudin, "Principles of mathematical analysis", McGraw-Hill (1953) | MR0055409 | Zbl 0052.05301 |
Measurable space. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Measurable_space&oldid=20100