Measure space

2020 Mathematics Subject Classification: Primary: 28Axx [MSN][ZBL]

$\newcommand{\Om}{\Omega} \newcommand{\A}{\mathcal A} \newcommand{\B}{\mathcal B} \newcommand{\M}{\mathcal M} $ A measure space is a triple $(X,\A,\mu)$ where $X$ is a set, $\A$ a σ-algebra of its subsets, and $\mu:\A\to[0,+\infty]$ a measure. Thus, a measure space consists of a measurable space and a measure. The notation $(X,\A,\mu)$ is often shortened to $(X,\mu)$ and one says that $\mu$ is a measure on $X$; sometimes the notation is shortened to $X$.

## Contents

* 1 Basic notions and constructions
* 2 Completion
* 3 Isomorphism
* 4 Finite and σ-finite
* 5 Perfect and standard
* 6 Atoms and continuity
* 7 Sub-σ-algebras, measure subalgebras, and linear sublattices
* 8 On terminology
* 9 References

#### Basic notions and constructions[edit]

Inner measure $\mu_*$ and outer measure $\mu^*$ are defined for all subsets $A\subset X$ by

$ \mu_*(A) = \max\\{\mu(B):B\in\A,B\subset A\\}\,,\quad \mu^*(A) = \min\\{\mu(B):B\in\A,B\supset A\\}\,;$

$A$ is called a null (or negligible) set if $\mu^*(A)=0$; in this case the complement $X\setminus A$ is called a set of full measure (or conegligible), and one says that $x\notin A$ for almost all $x$ (in other words, almost everywhere). Two sets $A,B\subset X$ are almost equal (or equal mod 0) if $(x\in A)\iff(x\in B)$ for almost all $x$ (in other words, $A\setminus B$ and $B\setminus A$ are negligible). Two functions $f,g:X\to Y$ are almost equal (or equal mod 0, or equivalent) if they are equal almost everywhere.

Equivalence classes of sets of $\A$ (w.r.t. the equivalence relation "almost equal") are the measure algebra of $(X,\A,\mu)$. Equivalence classes of square integrable complex-valued (or real-valued) measurable functions on $X$ are the Hilbert space $L_2(X,\A,\mu)$.

Let $(X_1,\A_1,\mu_1)$, $(X_2,\A_2,\mu_2)$ be measure spaces. A map $f:X_1\to X_2$ is called measure preserving if for every $A_2\in\A_2$ the inverse image $A_1=f^{-1}(A_2)$ satisfies $A_1\in\A_1$, $\mu_1(A_1)=\mu_2(A_2)$. Such $f$ leads to an embedding of the measure algebra of $(X_2,\A_2,\mu_2)$ into the measure algebra of $(X_1,\A_1,\mu_1)$, as follows: the equivalence class of $A_2$ turns into the equivalence class of $A_1=f^{-1}(A_2)$.

The product of two (or finitely many) measure spaces is a well-defined measure space.

A probability space is a measure space $(X,\A,\mu)$ satisfying $\mu(X)=1$. The product of infinitely many probability spaces is a well-defined probability space. (See [D, Sect. 8.2], [B, Sect. 3.5], [P, Sect. 4.8].)

#### Completion[edit]

A subset $A\subset X$ is called measurable (or $\mu$-measurable) if it is almost equal to some $B\in\A$. In this case $\mu_*(A)=\mu^*(A)=\mu(B)$. If $\mu_*(A)=\mu^*(A)<\infty$ then $A$ is $\mu$-measurable. All $\mu$-measurable sets are a σ-algebra $\A_\mu$ containing $\A$.

Every $\A_\mu$-measurable function $X\to\R$ is almost equal to some $\A$-measurable function $X\to\R$. The same holds for arbitrary countably generated measurable space in place of $\R$.

Example. Let $X$ be the real line, $\A$ the Borel σ-algebra and $\mu$ Lebesgue measure on $\A$, then $\A_\mu$ is the Lebesgue σ-algebra.

Let $(X,\A,\mu)$ be a measure space. Both $(X,\A,\mu)$ and $\mu$ are called complete if $\A_\mu=\A$ or, equivalently, if $\A$ contains all null sets. The completion of $(X,\A,\mu)$ is the complete measure space $(X,\A_\mu,\tilde\mu)$ where $\tilde\mu(A)=\mu(B)$ whenever $A\in\A_\mu$ is almost equal to $B\in\A$. The measure algebras of $(X,\A,\mu)$ and $(X,\A_\mu,\tilde\mu)$ are equal (up to the evident isomorphism).

Let $(X,\A,\mu)$ be complete, and $\B\subset\A$ a sub-σ-algebra. Then $(X,\A,\mu)$ is the completion of $(X,\B,\mu|_\B)$ if and only if for every $A\in\A$ there exist $B,C\in\B$ such that $B\subset A\subset C$ and $\mu(C\setminus B)=0$.

Surprisingly, the Borel σ-algebra can be "almost restored" from the Lebesgue σ-algebra in the following sense.

Let $(X,\A,\mu)$ be complete, and $\B_1\subset\A$, $\B_2\subset\A$ two countably generated sub-σ-algebras such that $(X,\A,\mu)$ is both the completion of $(X,\B_1,\mu|_{\B_1})$ and the completion of $(X,\B_2,\mu|_{\B_2})$. Then there exists a set $Y\in\B_1\cap\B_2$ of full measure such that $\B_1|_Y=\B_2|_Y$. (Here $\B_i|_Y=\\{B\cap Y:B\in\B_i\\}=\\{B\in\B_i:B\subset Y\\}$.)

#### Isomorphism[edit]

A strict isomorphism (or point isomorphism, or metric isomorphism) between two measure spaces $(X_1,\A_1,\mu_1)$ and $(X_2,\A_2,\mu_2)$ is a bijection $f:X_1\to X_2$ such that, first, the conditions $A_1\in\A_1$ and $A_2\in\A_2$ are equivalent whenever $A_1\subset X_1$, $A_2\subset X_2$, $A_2=f(A_1)$, and second, $\mu_1(A_1)=\mu_2(A_2)$ under these conditions.

A mod 0 isomorphism (or almost isomorphism) between two measure spaces $(X_1,\A_1,\mu_1)$ and $(X_2,\A_2,\mu_2)$ is a strict isomorphism between some full measure sets $Y_1\in\A_1$ and $Y_2\in\A_2$ treated as measure subspaces.

Thus we have two equivalence relations between measure spaces: "strictly isomorphic" and "almost isomorphic". (See [I, Sect. 2.4], [B, Sect. 9.2].)

If two measure spaces are almost isomorphic then clearly their completions are almost isomorphic. The converse, being wrong in general, surprisingly holds in the following important case.

Let measure spaces $(X_1,\A_1,\mu_1)$, $(X_2,\A_2,\mu_2)$ be such that (a) their completions are almost isomorphic, and (b) measurable spaces $(X_1,\A_1)$, $(X_2,\A_2)$ are countably generated. Then $(X_1,\A_1,\mu_1)$, $(X_2,\A_2,\mu_2)$ are almost isomorphic (under the same isomorphism, restricted to a smaller subset of full measure).

For complete measure spaces the two notions of isomorphism nearly coincide, as explained below.

An almost isomorphism between complete measure spaces $(X_1,\A_1,\mu_1)$, $(X_2,\A_2,\mu_2)$, being a bijection $Y_1\to Y_2$ between full measure sets $Y_1\subset X_1$, $Y_2\subset X_2$, extends readily to a strict isomorphism $X_1\to X_2$, since all maps are measurable on negligible sets $X_1\setminus Y_1$, $X_2\setminus Y_2$. The only possible obstacle is, different cardinalities of these negligible sets. The conclusion follows.

Assume that $(X_1,\A_1,\mu_1)$ is a complete measure space, $X_1$ is of cardinality continuum and contains some negligible set of cardinality continuum. Assume that $(X_2,\A_2,\mu_2)$ satisfies the same conditions. If $(X_1,\A_1,\mu_1)$, $(X_2,\A_2,\mu_2)$are almost isomorphic then they are strictly isomorphic.

Cardinality continuum is typical, but the fact holds in general, under the following condition: for every negligible set in every one of the two measure spaces there exists a negligible set of the same cardinality in the other measure space. (This argument is used, somewhat implicitly, in [F, Vol. 3, Sect. 344I].)

#### Finite and σ-finite[edit]

Let $(X,\A,\mu)$ be a measure space. Both $(X,\A,\mu)$ and $\mu$ are called totally finite if $\mu(X)<\infty$, and σ-finite if $X$ can be split into countably many sets of finite measure, that is, $X=A_1\cup A_2\cup\dots$ for some $A_n\in\A$ such that $\forall n \;\; \mu(A_n)<\infty$. (Totally finite measures are also σ-finite.)

#### Perfect and standard[edit]

Let $(X,\A,\mu)$ be a totally finite measure space. Both $(X,\A,\mu)$ and $\mu$ are called perfect if for every $\mu$-measurable (or equivalently, for every $\A$-measurable) function $f:X\to\R$ the image $f(X)$ contains a Borel (or equivalently, σ-compact) subset $B$ whose preimage $f^{-1}(B)$ is of full measure. (See [B, Sect. 7.5].)

For standard probability spaces see the separate article. Standard measure spaces are defined similarly. They are perfect, and admit a complete classification.

Examples. The real line with Lebesgue measure on Borel σ-algebra is an incomplete σ-finite measure space. The real line with Lebesgue measure on Lebesgue σ-algebra is a complete σ-finite measure space. The unit interval $(0,1)$ with Lebesgue measure on Lebesgue σ-algebra is a standard probability space. The product of countably many copies of this space is standard; for uncountably many factors the product is perfect but nonstandard. The one-dimensional Hausdorff measure on the plane is not σ-finite.

#### Atoms and continuity[edit]

Let $\mu(X)<\infty$. An atom of $(X,\A,\mu)$ (and of $\mu$) is a non-negligible measurable set $A\subset X$ such that every measurable subset of $A$ is either negligible or almost equal to $A$. Both $(X,\A,\mu)$ and $\mu$ are called atomless or nonatomic (or diffused) if they have no atoms; on the other hand, they are called purely atomic if there exists a partition of $X$ into atoms. (See [D, Sect. 3.5], [B, Sect. 1.12(iii)], [M, Sect. 6.4.1].)

If $x\in X$ is such that the single-point set $\\{x\\}$ is a non-negligible measurable set then clearly $\\{x\\}$ is an atom. If $(X,\A,\mu)$ is standard then every atom is almost equal to some $\\{x\\}$, but in general it is not.

Let $\\{x\\}$ be measurable for all $x\in X$. Both $(X,\A,\mu)$ and $\mu$ are called continuous if $\mu(\\{x\\})=0$ for all $x\in X$; on the other hand, they are called discrete if $X$ is almost equal to some finite or countable set. (See [C, Sect. 1.2], [K, Sect. 17.A].) A discrete space cannot be atomless (unless $\mu(X)=0$), but a purely atomic nonstandard space can be continuous. (See [B, Sect. 7.14(v)].)

See also "taxonomy of measure spaces" in [F, Vol. 2, Chapter 21].

#### Sub-σ-algebras, measure subalgebras, and linear sublattices[edit]

Every sub-σ-algebra $\B\subset\A$ leads to a measure space $(X,\B,\mu|_{\B})$, a measure preserving map $x\mapsto x$ from $(X,\A,\mu)$ to $(X,\B,\mu|_{\B})$, and the corresponding embedding of measure algebras; thus, $\B$ leads to a measure subalgebra of the measure algebra of $(X,\A,\mu)$.

Different sub-σ-algebras can lead to the same measure subalgebra. For example, the least (two-element) measure subalgebra $\\{\bszero,\bsone\\}$ corresponds both to the least (two-element) sub-σ-algebra $\\{\emptyset,X\\}$ and the larger sub-σ-algebra of all null sets and their complements.

From now on (till the end of this section) we restrict ourselves to σ-algebras that contain all null sets. That is, a measure space $(X,\A,\mu)$ is assumed to be complete, and whenever we call $\B\subset\A$ a sub-σ-algebra we also assume that $\B$ contains all null sets: $\forall A\in\A \; (\, \mu(A)=0 \,\Longrightarrow A\in\B \,)$. This assumption is stronger than completeness of the measure space $(X,\B,\mu|_\B)$.

This way we get a bijective correspondence and moreover, an isomorphism between two partially ordered (by inclusion) sets: the set of all sub-σ-algebras of $\A$, and the set of all measure subalgebras of the measure algebra of $(X,\A,\mu)$. Both partially ordered sets are complete lattices (generally, not distributive).

Every sub-σ-algebra $\B\subset\A$ leads also to a subspace (that is, closed linear subset) $L_2(X,\B,\mu|_\B)$ of the Hilbert space $L_2(X,\A,\mu)$.

Theorem (Šidák [S]). A subspace $H\subset L_2(X,\A,\mu)$ is of the form $L_2(X,\B,\mu|_\B)$ for some sub-σ-algebra $\B\subset\A$ if and only if $\bsone\in H$ and $\forall f\in H \; (\, |f|\in H \,)$. (Here $\bsone(x)=1$ and $|f|(x)=|f(x)|$ for $x\in X$.)

Due to linearity, the condition $\forall f\in H \; (\, |f|\in H \,)$ is equivalent to the condition $\forall f,g\in H \; (\, f\land g\in H \text{ and } f\lor g\in H \,)$ where $(f\land g)(x)=\min(f(x),g(x))$ and $(f\lor g)(x)=\max(f(x),g(x))$ for $x\in X$; that is, $H$ is a linear sublattice of the linear lattice $L_2(X,\A,\mu)$.

Ultimately, we have three naturally isomorphic complete lattices: of all sub-σ-algebras, of all measure subalgebras, and of all closed linear sublattices containing $\bsone$.

#### On terminology[edit]

The word "isomorphic" (for measure spaces) is interpreted as almost isomorphic in [I, Sect. 2.4] (which is usual according to [B, Sect. 9.2]) but as strictly isomorphic in [F, Vol. 2, Sect. 254, Notes and comments]; there, the notion almost isomorphic is only mentioned in passing as "nearly an isomorphism".

The phrase "separable measure space" is quite ambiguous. Some authors call $(X,\A,\mu)$ separable when the Hilbert space $L_2(X,\A,\mu)$ is separable; equivalently, when $\A$ contains a countably generated sub-σ-algebra $\B$ such that every set of $\A$ is almost equal to some set of $\B$. (See [B, Sect. 7.14(iv)], [M, Sect. IV.6.0].) But in [I, Sect. 3.1] it is required instead that $\B$ separates points and $(X,\A,\mu)$ is complete, while in [H] all these conditions are imposed together.

The phrase "measurable space" is avoided in [F] "as in fact many of the most interesting examples of such objects have no useful measures associated with them" [F, Vol. 1, Sect. 111B].

According to [M, Sect. I.3], all measure spaces are σ-finite (by definition).

#### References[edit]

[T] | Terence Tao, "An introduction to measure theory", AMS (2011).  MR2827917 Zbl 05952932

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[C] | Donald L. Cohn, "Measure theory", Birkhäuser (1993).  MR1454121 Zbl 0860.28001
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[S] | Zbyněk Šidák, "On relations between strict-sense and wide-sense conditional expectations", Teor. Veroyatnost. i Primenen. 2 (1957), 283–288.  MR0092249 Zbl 0078.31101
[H] | Jean Haezendonck, "Abstract Lebesgue-Rohlin spaces", Bull. Soc. Math. de Belgique 25 (1973), 243–258.  MR0335733 Zbl 0308.60006