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45

Measure Theory

Field: Real Analysis

The study of measures, which generalize the notions of length, area, and volume.

Sequence of Expressions

Definition

Definition

A triple (X,Σ,μ)(X, \Sigma, \mu) where XX is a set, \Sigma is a \sigma-algebra, and \mu is a measure.
Let μ\mu be a measure. If E1E_{1} and E2E_{2} are measurable sets with E1⊆E2E_{1}\subseteq E_{2} then μ(E1)≤μ(E2).\mu (E_{1})\leq \mu (E_{2}). For any countablesequence E1,E2,E3,…E_{1},E_{2},E_{3},\ldots of (not necessarily disjoint) measurable sets EnE_{n} in Σ:\Sigma : μ(⋃i=1∞Ei)≤∑i=1∞μ(Ei).\mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)\leq \sum _{i=1}^{\infty }\mu (E_{i}). If E1,E2,E3,…E_{1},E_{2},E_{3},\ldots are measurable sets that are increasing (meaning that E1⊆E2⊆E3⊆…E_{1}\subseteq E_{2}\subseteq E_{3}\subseteq \ldots ) then the union of the sets EnE_{n} is measurable and μ(⋃i=1∞Ei) = lim⁡i→∞μ(Ei)=sup⁡i≥1μ(Ei).\mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)~=~\lim _{i\to \infty }\mu (E_{i})=\sup _{i\geq 1}\mu (E_{i}). If E1,E2,E3,…E_{1},E_{2},E_{3},\ldots are measurable sets that are decreasing (meaning that E1⊇E2⊇E3⊇…E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq \ldots ) then the intersection of the sets EnE_{n} is measurable; furthermore, if at least one of the EnE_{n} has finite measure then μ(⋂i=1∞Ei)=lim⁡i→∞μ(Ei)=inf⁡i≥1μ(Ei).\mu \left(\bigcap _{i=1}^{\infty }E_{i}\right)=\lim _{i\to \infty }\mu (E_{i})=\inf _{i\geq 1}\mu (E_{i}). This property is false without the assumption that at least one of the EnE_{n} has finite measure. For instance, for each n∈N,n\in \mathbb {N} , let En=[n,∞)⊆R,E_{n}=[n,\infty )\subseteq \mathbb {R} , which all have infinite Lebesgue measure, but the intersection is empty.
A measurable set XX is called a null set if μ(X)=0.\mu (X)=0. A subset of a null set is called a negligible set. A negligible set need not be measurable, but every measurable negligible set is automatically a null set. A measure is called complete if every negligible set is measurable. A measure can be extended to a complete one by considering the σ-algebra of subsets YY which differ by a negligible set from a measurable set X,X, that is, such that the symmetric difference of XX and YY is contained in a null set. One defines μ(Y)\mu (Y) to equal μ(X).\mu (X). If f:X→[0,+∞]f:X\to [0,+\infty ] is (Σ,B([0,+∞]))(\Sigma ,{\cal {B}}([0,+\infty ])) -measurable, then μ{x∈X:f(x)≥t}=μ{x∈X:f(x)>t}\mu \{x\in X:f(x)\geq t\}=\mu \{x\in X:f(x)>t\} for almost all t∈[−∞,∞].t\in [-\infty ,\infty ]. This property is used in connection with Lebesgue integral. Proof Both F(t):=μ{x∈X:f(x)>t}F(t):=\mu \{x\in X:f(x)>t\} and G(t):=μ{x∈X:f(x)≥t}G(t):=\mu \{x\in X:f(x)\geq t\} are monotonically non-increasing functions of t,t, so both of them have at most countably many discontinuities and thus they are continuous almost everywhere, relative to the Lebesgue measure. If t<0t<0 then {x∈X:f(x)≥t}=X={x∈X:f(x)>t},\{x\in X:f(x)\geq t\}=X=\{x\in X:f(x)>t\}, so that F(t)=G(t),F(t)=G(t), as desired. If tt is such that μ{x∈X:f(x)>t}=+∞\mu \{x\in X:f(x)>t\}=+\infty then monotonicity implies μ{x∈X:f(x)≥t}=+∞,\mu \{x\in X:f(x)\geq t\}=+\infty , so that F(t)=G(t),F(t)=G(t), as required. If μ{x∈X:f(x)>t}=+∞\mu \{x\in X:f(x)>t\}=+\infty for all tt then we are done, so assume otherwise. Then there is a unique t0∈{−∞}∪[0,+∞)t_{0}\in \{-\infty \}\cup [0,+\infty ) such that FF is infinite to the left of tt (which can only happen when t0≥0t_{0}\geq 0 ) and finite to the right. Arguing as above, μ{x∈X:f(x)≥t}=+∞\mu \{x\in X:f(x)\geq t\}=+\infty when t<t0.t<t_{0}. Similarly, if t0≥0t_{0}\geq 0 and F(t0)=+∞F\left(t_{0}\right)=+\infty then F(t0)=G(t0).F\left(t_{0}\right)=G\left(t_{0}\right). For t>t0,t>t_{0}, let tnt_{n} be a monotonically non-decreasing sequence converging to t.t. The monotonically non-increasing sequences {x∈X:f(x)>tn}\{x\in X:f(x)>t_{n}\} of members of Σ\Sigma has at least one finitely μ\mu -measurable component, and {x∈X:f(x)≥t}=⋂n{x∈X:f(x)>tn}.\{x\in X:f(x)\geq t\}=\bigcap _{n}\{x\in X:f(x)>t_{n}\}. Continuity from above guarantees that μ{x∈X:f(x)≥t}=lim⁡tn↑tμ{x∈X:f(x)>tn}.\mu \{x\in X:f(x)\geq t\}=\lim _{t_{n}\uparrow t}\mu \{x\in X:f(x)>t_{n}\}. The right-hand side lim⁡tn↑tF(tn)\lim _{t_{n}\uparrow t}F\left(t_{n}\right) then equals F(t)=μ{x∈X:f(x)>t}F(t)=\mu \{x\in X:f(x)>t\} if tt is a point of continuity of F.F. Since FF is continuous almost everywhere, this completes the proof. Measures are required to be countably additive. However, the condition can be strengthened as follows. For any set II and any set of nonnegative rir_{i} where i∈Ii\in I define: ∑i∈Iri=sup⁡{∑i∈Jri:∣J∣<∞,J⊆I}.\sum _{i\in I}r_{i}=\sup \left\lbrace \sum _{i\in J}r_{i}:|J|<\infty ,J\subseteq I\right\rbrace . That is, we define the sum of the rir_{i} to be the supremum of all the sums of finitely many of them. A measure μ\mu on Σ\Sigma is κ\kappa -additive if for any λ<κ\lambda <\kappa and any family of disjoint sets Xα,α<λX_{\alpha },\alpha <\lambda the following hold: ⋃α∈λXα∈Σ\bigcup _{\alpha \in \lambda }X_{\alpha }\in \Sigma μ(⋃α∈λXα)=∑α∈λμ(Xα).\mu \left(\bigcup _{\alpha \in \lambda }X_{\alpha }\right)=\sum _{\alpha \in \lambda }\mu \left(X_{\alpha }\right). The second condition is equivalent to the statement that the ideal of null sets is κ\kappa -complete. A measure space (X,Σ,μ)(X,\Sigma ,\mu ) is called finite if μ(X)\mu (X) is a finite real number (rather than ∞\infty ). Nonzero finite measures are analogous to probability measures in the sense that any finite measure μ\mu is proportional to the probability measure 1μ(X)μ.{\frac {1}{\mu (X)}}\mu . A measure μ\mu is called σ-finite if XX can be decomposed into a countable union of measurable sets of finite measure. Analogously, a set in a measure space is said to have a σ-finite measure if it is a countable union of sets with finite measure. For example, the real numbers with the standard Lebesgue measure are σ-finite but not finite. Consider the closed intervals [k,k+1][k,k+1] for all integers k;k; there are countably many such intervals, each has measure 1, and their union is the entire real line. Alternatively, consider the real numbers with the counting measure, which assigns to each finite set of reals the number of points in the set. This measure space is not σ-finite, because every set with finite measure contains only finitely many points, and it would take uncountably many such sets to cover the entire real line. The σ-finite measure spaces have some very convenient properties; σ-finiteness can be compared in this respect to the Lindelöf property of topological spaces. They can be also thought of as a vague generalization of the idea that a measure space may have 'uncountable measure'. Let XX be a set, let A{\cal {A}} be a sigma-algebra on X,X, and let μ\mu be a measure on A.{\cal {A}}. We say μ\mu is semifinite to mean that for all A∈μpre{+∞},A\in \mu ^{\text{pre}}\{+\infty \}, P(A)∩μpre(R>0)≠∅.{\cal {P}}(A)\cap \mu ^{\text{pre}}(\mathbb {R} _{>0})\neq \emptyset . Semifinite measures generalize sigma-finite measures, in such a way that some big theorems of measure theory that hold for sigma-finite but not arbitrary measures can be extended with little modification to hold for semifinite measures. (To-do: add examples of such theorems; cf. the talk page.) - Every sigma-finite measure is semifinite. - Assume A=P(X),{\cal {A}}={\cal {P}}(X), let f:X→[0,+∞],f:X\to [0,+\infty ], and assume μ(A)=∑a∈Af(a)\mu (A)=\sum _{a\in A}f(a) for all A⊆X.A\subseteq X. - We have that μ\mu is sigma-finite if and only if f(x)<+∞f(x)<+\infty for all x∈Xx\in X and fpre(R>0)f^{\text{pre}}(\mathbb {R} _{>0}) is countable. We have that μ\mu is semifinite if and only if f(x)<+∞f(x)<+\infty for all x∈X.x\in X. - Taking f=X×{1}f=X\times \{1\} above (so that μ\mu is counting measure on P(X){\cal {P}}(X) ), we see that counting measure on P(X){\cal {P}}(X) is - sigma-finite if and only if XX is countable; and - semifinite (without regard to whether XX is countable). (Thus, counting measure, on the power set P(X){\cal {P}}(X) of an arbitrary uncountable set X,X, gives an example of a semifinite measure that is not sigma-finite.) - Let dd be a complete, separable metric on X,X, let B{\cal {B}} be the Borel sigma-algebra induced by d,d, and let s∈R>0.s\in \mathbb {R} _{>0}. Then the Hausdorff measure Hs∣B{\cal {H}}^{s}|{\cal {B}} is semifinite. - Let dd be a complete, separable metric on X,X, let B{\cal {B}} be the Borel sigma-algebra induced by d,d, and let s∈R>0.s\in \mathbb {R} _{>0}. Then the packing measure Hs∣B{\cal {H}}^{s}|{\cal {B}} is semifinite. The zero measure is sigma-finite and thus semifinite. In addition, the zero measure is clearly less than or equal to μ.\mu . It can be shown there is a greatest measure with these two properties: Theorem (semifinite part)—For any measure μ\mu on A,{\cal {A}}, there exists, among semifinite measures on A{\cal {A}} that are less than or equal to μ,\mu , a greatest element μsf.\mu _{\text{sf}}. We say the semifinite part of μ\mu to mean the semifinite measure μsf\mu _{\text{sf}} defined in the above theorem. We give some nice, explicit formulas, which some authors may take as definition, for the semifinite part: - μsf=(sup⁡{μ(B):B∈P(A)∩μpre(R≥0)})A∈A.\mu _{\text{sf}}=(\sup\{\mu (B):B\in {\cal {P}}(A)\cap \mu ^{\text{pre}}(\mathbb {R} _{\geq 0})\})_{A\in {\cal {A}}}. - μsf=(sup⁡{μ(A∩B):B∈μpre(R≥0)})A∈A}.\mu _{\text{sf}}=(\sup\{\mu (A\cap B):B\in \mu ^{\text{pre}}(\mathbb {R} _{\geq 0})\})_{A\in {\cal {A}}}\}. - μsf=μ∣μpre(R>0)∪{A∈A:sup⁡{μ(B):B∈P(A)}=+∞}×{+∞}∪{A∈A:sup⁡{μ(B):B∈P(A)}<+∞}×{0}.\mu _{\text{sf}}=\mu |_{\mu ^{\text{pre}}(\mathbb {R} _{>0})}\cup \{A\in {\cal {A}}:\sup\{\mu (B):B\in {\cal {P}}(A)\}=+\infty \}\times \{+\infty \}\cup \{A\in {\cal {A}}:\sup\{\mu (B):B\in {\cal {P}}(A)\}<+\infty \}\times \{0\}. Since μsf\mu _{\text{sf}} is semifinite, it follows that if μ=μsf\mu =\mu _{\text{sf}} then μ\mu is semifinite. It is also evident that if μ\mu is semifinite then μ=μsf.\mu =\mu _{\text{sf}}. Every 0−∞0-\infty measure that is not the zero measure is not semifinite. (Here, we say 0−∞0-\infty measure to mean a measure whose range lies in {0,+∞}\{0,+\infty \} : (∀A∈A)(μ(A)∈{0,+∞}).(\forall A\in {\cal {A}})(\mu (A)\in \{0,+\infty \}). ) Below we give examples of 0−∞0-\infty measures that are not zero measures. - Let XX be nonempty, let A{\cal {A}} be a σ\sigma -algebra on X,X, let f:X→{0,+∞}f:X\to \{0,+\infty \} be not the zero function, and let μ=(∑x∈Af(x))A∈A.\mu =(\sum _{x\in A}f(x))_{A\in {\cal {A}}}. It can be shown that μ\mu is a measure. - μ={(∅,0)}∪(A∖{∅})×{+∞}.\mu =\{(\emptyset ,0)\}\cup ({\cal {A}}\setminus \{\emptyset \})\times \{+\infty \}. - X={0},X=\{0\}, A={∅,X},{\cal {A}}=\{\emptyset ,X\}, μ={(∅,0),(X,+∞)}.\mu =\{(\emptyset ,0),(X,+\infty )\}. - Let XX be uncountable, let A{\cal {A}} be a σ\sigma -algebra on X,X, let C={A∈A:A is countable}{\cal {C}}=\{A\in {\cal {A}}:A{\text{ is countable}}\} be the countable elements of A,{\cal {A}}, and let μ=C×{0}∪(A∖C)×{+∞}.\mu ={\cal {C}}\times \{0\}\cup ({\cal {A}}\setminus {\cal {C}})\times \{+\infty \}. It can be shown that μ\mu is a measure. Measures that are not semifinite are very wild when restricted to certain sets. Every measure is, in a sense, semifinite once its 0−∞0-\infty part (the wild part) is taken away. — A. Mukherjea and K. Pothoven, Real and Functional Analysis, Part A: Real Analysis (1985) Theorem (Luther decomposition)—For any measure μ\mu on A,{\cal {A}}, there exists a 0−∞0-\infty measure ξ\xi on A{\cal {A}} such that μ=ν+ξ\mu =\nu +\xi for some semifinite measure ν\nu on A.{\cal {A}}. In fact, among such measures ξ,\xi , there exists a least measure μ0−∞.\mu _{0-\infty }. Also, we have μ=μsf+μ0−∞.\mu =\mu _{\text{sf}}+\mu _{0-\infty }. We say the 0−∞\mathbf {0-\infty } part of μ\mu to mean the measure μ0−∞\mu _{0-\infty } defined in the above theorem. Here is an explicit formula for μ0−∞\mu _{0-\infty } : μ0−∞=(sup⁡{μ(B)−μsf(B):B∈P(A)∩μsfpre(R≥0)})A∈A.\mu _{0-\infty }=(\sup\{\mu (B)-\mu _{\text{sf}}(B):B\in {\cal {P}}(A)\cap \mu _{\text{sf}}^{\text{pre}}(\mathbb {R} _{\geq 0})\})_{A\in {\cal {A}}}. - Let F\mathbb {F} be R\mathbb {R} or C,\mathbb {C} , and let T:LF∞(μ)→(LF1(μ))∗:g↦Tg=(∫fgdμ)f∈LF1(μ).T:L_{\mathbb {F} }^{\infty }(\mu )\to \left(L_{\mathbb {F} }^{1}(\mu )\right)^{*}:g\mapsto T_{g}=\left(\int fgd\mu \right)_{f\in L_{\mathbb {F} }^{1}(\mu )}. Then μ\mu is semifinite if and only if TT is injective. (This result has import in the study of the dual space of L1=LF1(μ)L^{1}=L_{\mathbb {F} }^{1}(\mu ) .) - Let F\mathbb {F} be R\mathbb {R} or C,\mathbb {C} , and let T{\cal {T}} be the topology of convergence in measure on LF0(μ).L_{\mathbb {F} }^{0}(\mu ). Then μ\mu is semifinite if and only if T{\cal {T}} is Hausdorff. - (Johnson) Let XX be a set, let A{\cal {A}} be a sigma-algebra on X,X, let μ\mu be a measure on A,{\cal {A}}, let YY be a set, let B{\cal {B}} be a sigma-algebra on Y,Y, and let ν\nu be a measure on B.{\cal {B}}. If μ,ν\mu ,\nu are both not a 0−∞0-\infty measure, then both μ\mu and ν\nu are semifinite if and only if (μ×cldν)(\mu \times _{\text{cld}}\nu ) (A×B)=μ(A)ν(B)(A\times B)=\mu (A)\nu (B) for all A∈AA\in {\cal {A}} and B∈B.B\in {\cal {B}}. (Here, μ×cldν\mu \times _{\text{cld}}\nu is the measure defined in Theorem 39.1 in Berberian '65.) Localizable measures are a special case of semifinite measures and a generalization of sigma-finite measures. Let XX be a set, let A{\cal {A}} be a sigma-algebra on X,X, and let μ\mu be a measure on A.{\cal {A}}. - Let F\mathbb {F} be R\mathbb {R} or C,\mathbb {C} , and let T:LF∞(μ)→(LF1(μ))∗:g↦Tg=(∫fgdμ)f∈LF1(μ).T:L_{\mathbb {F} }^{\infty }(\mu )\to \left(L_{\mathbb {F} }^{1}(\mu )\right)^{*}:g\mapsto T_{g}=\left(\int fgd\mu \right)_{f\in L_{\mathbb {F} }^{1}(\mu )}. Then μ\mu is localizable if and only if TT is bijective (if and only if LF∞(μ)L_{\mathbb {F} }^{\infty }(\mu ) "is" LF1(μ)∗L_{\mathbb {F} }^{1}(\mu )^{*} ). A measure is said to be s-finite if it is a countable sum of finite measures. S-finite measures are more general than sigma-finite ones and have applications in the theory of stochastic processes. - ^Fremlin, D. H. (2010), Measure Theory, vol. 2 (Second ed.), p. 221 - ^ ^{a}^{b}Mukherjea & Pothoven 1985, p. 90. - ^Folland 1999, p. 25. - ^Edgar 1998, Theorem 1.5.2, p. 42. - ^Edgar 1998, Theorem 1.5.3, p. 42. - ^ ^{a}^{b}Nielsen 1997, Exercise 11.30, p. 159. - ^Fremlin 2016, Section 213X, part (c). - ^Royden & Fitzpatrick 2010, Exercise 17.8, p. 342. - ^Hewitt & Stromberg 1965, part (b) of Example 10.4, p. 127. - ^Fremlin 2016, Section 211O, p. 15. - ^Luther 1967, Theorem 1. - ^Mukherjea & Pothoven 1985, part (b) of Proposition 2.3, p. 90. - ^Fremlin 2016, part (a) of Theorem 243G, p. 159. - ^ ^{a}^{b}Fremlin 2016, Section 243K, p. 162. - ^Fremlin 2016, part (a) of the Theorem in Section 245E, p. 182. - ^Fremlin 2016, Section 245M, p. 188. - ^Berberian 1965, Theorem 39.1, p. 129. - ^Fremlin 2016, part (b) of Theorem 243G, p. 159. Cite error: There are tags on this page, but the references will not show without a {{reflist|group=Note}} template (see the help page).
Let X\text{X} be a non-empty set. A σ\sigma-algebra A\text{A} on X\text{X} is a collection of subsets of X\text{X} satisfying: (i) X∈A\text{X} \in \text{A}; (ii) X∖E∈A\text{X} \setminus E \in \text{A} for all E∈AE \in \text{A}; and (iii) \text{\bigcup}_{i=1}^{\infty} E_i \in \text{A} for any countable sequence (Ei)i=1∞(E_i)_{i=1}^{\infty} with Ei∈AE_i \in \text{A}. The pair (X,A)(\text{X}, \text{A}) is called a measurable space.\n\nGiven a measurable space (X,A)(\text{X}, \text{A}), a measure μ\mu is a function μ:A→[0,∞]\mu: \text{A} \to [0, \infty] such that: (i) μ(∅)=0\mu(\emptyset) = 0; (ii) μ\mu is countably additive: for any sequence of pairwise disjoint sets (Ei)i=1∞(E_i)_{i=1}^{\infty} in A\text{A}, we have μ(⋃i=1∞Ei)=∑i=1∞μ(Ei)\mu(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \mu(E_i).\n\nIf X=[a,b]⊂R\text{X} = [a, b] \subset \mathbb{R} and A=B[a,b]\text{A} = \mathcal{B}_{[a, b]} (the Borel σ\sigma-algebra restricted to [a,b][a, b]), the Lebesgue measure λ\lambda is the unique measure (up to scaling) satisfying λ([x,y])=y−x\lambda([x, y]) = y - x for all x,y∈[a,b]x, y \in [a, b]. The resulting structure (X,A,μ)(\text{X}, \text{A}, \mu) is the foundation of Measure Theory.