The counting measure is defined by $\mu (S)$ = number of elements in $S.$
The Lebesgue measure on $\mathbb {R}$ is a completetranslation-invariant measure on a σ-algebra containing the intervals in $\mathbb {R}$ such that $\mu ([0,1])=1$; and every other measure with these properties extends Lebesgue measure.
The Haar measure for a locally compacttopological group is a generalization of the Lebesgue measure (and also of counting measure and circular angle measure) and has similar uniqueness properties.
The Hausdorff measure is a generalization of the Lebesgue measure to sets with non-integer dimension, in particular, fractal sets.
Every probability space gives rise to a measure which takes the value 1 on the whole space (and therefore takes all its values in the unit interval [0, 1]). Such a measure is called a probability measure or distribution. See the list of probability distributions for instances.
In physics an example of a measure is spatial distribution of mass (see for example, gravity potential), or another non-negative extensive property, conserved (see conservation law for a list of these) or not. Negative values lead to signed measures, see "generalizations" below.
Liouville measure, known also as the natural volume form on a symplectic manifold, is useful in classical statistical and Hamiltonian mechanics.
If $E_{1},E_{2},E_{3},\ldots$ are measurable sets that are increasing (meaning that $E_{1}\subseteq E_{2}\subseteq E_{3}\subseteq \ldots$) then the union of the sets $E_{n}$ is measurable and
If $E_{1},E_{2},E_{3},\ldots$ are measurable sets that are decreasing (meaning that $E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq \ldots$) then the intersection of the sets $E_{n}$ is measurable; furthermore, if at least one of the $E_{n}$ has finite measure then
This property is false without the assumption that at least one of the $E_{n}$ has finite measure. For instance, for each $n\in \mathbb {N} ,$ let $E_{n}=[n,\infty )\subseteq \mathbb {R} ,$ which all have infinite Lebesgue measure, but the intersection is empty.
Other propertiesedit
Completenessedit
A measurable set $X$ is called a null set if $\mu (X)=0.$ 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 $Y$ which differ by a negligible set from a measurable set $X,$ that is, such that the symmetric difference of $X$ and $Y$ is contained in a null set. One defines $\mu (Y)$ to equal $\mu (X).$
μ{x : f(x) ≥ t} = μ{x : f(x) > t} (a.e.)edit
If $f:X\to [0,+\infty ]$ is $(\Sigma ,{\cal {B}}([0,+\infty ]))$-measurable, then
Both $F(t):=\mu \{x\in X:f(x)>t\}$ and $G(t):=\mu \{x\in X:f(x)\geq t\}$ are monotonically non-increasing functions of $t,$ so both of them have at most countably many discontinuities and thus they are continuous almost everywhere, relative to the Lebesgue measure.
If $t<0$ then $\{x\in X:f(x)\geq t\}=X=\{x\in X:f(x)>t\},$ so that $F(t)=G(t),$ as desired.
If $t$ is such that $\mu \{x\in X:f(x)>t\}=+\infty$ then monotonicity implies
$\mu \{x\in X:f(x)\geq t\}=+\infty ,$
so that $F(t)=G(t),$ as required.
If $\mu \{x\in X:f(x)>t\}=+\infty$ for all $t$ then we are done, so assume otherwise. Then there is a unique $t_{0}\in \{-\infty \}\cup [0,+\infty )$ such that $F$ is infinite to the left of $t$ (which can only happen when $t_{0}\geq 0$) and finite to the right.
Arguing as above, $\mu \{x\in X:f(x)\geq t\}=+\infty$ when $t<t_{0}.$ Similarly, if $t_{0}\geq 0$ and $F\left(t_{0}\right)=+\infty$ then $F\left(t_{0}\right)=G\left(t_{0}\right).$
For $t>t_{0},$ let $t_{n}$ be a monotonically non-decreasing sequence converging to $t.$ The monotonically non-increasing sequence $\{x\in X:f(x)>t_{n}\}$ of members of $\Sigma$ has at least one finitely $\mu$-measurable component, and
The right-hand side $\lim _{t_{n}\uparrow t}F\left(t_{n}\right)$ then equals $F(t)=\mu \{x\in X:f(x)>t\}$ if $t$ is a point of continuity of $F.$ Since $F$ is continuous almost everywhere, this completes the proof.
Additivityedit
Measures are required to be countably additive. However, the condition can be strengthened as follows.
For any set $I$ and any set of nonnegative $r_{i},i\in I$ define:
That is, we define the sum of the $r_{i}$ to be the supremum of all the sums of finitely many of them.
A measure $\mu$ on $\Sigma$ is $\kappa$-additive if for any $\lambda <\kappa$ and any family of disjoint sets $X_{\alpha },\alpha <\lambda$ the following hold:
The second condition is equivalent to the statement that the ideal of null sets is $\kappa$-complete.
Sigma-finite measuresedit
A measure space $(X,\Sigma ,\mu )$ is called finite if $\mu (X)$ is a finite real number (rather than $\infty$). Nonzero finite measures are analogous to probability measures in the sense that any finite measure $\mu$ is proportional to the probability measure ${\frac {1}{\mu (X)}}\mu .$ A measure $\mu$ is called σ-finite if $X$ 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$[k,k+1]$ for all integers$k;$ 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.^{[original research?]} They can be also thought of as a vague generalization of the idea that a measure space may have 'uncountable measure'.
Strictly localizable measuresedit
Semifinite measuresedit
Let $X$ be a set, let ${\cal {A}}$ be a sigma-algebra on $X,$ and let $\mu$ be a measure on ${\cal {A}}.$ We say $\mu$ is semifinite to mean that for all $A\in \mu ^{\text{pre}}\{+\infty \},$${\cal {P}}(A)\cap \mu ^{\text{pre}}(\mathbb {R} _{>0})\neq \emptyset .$^{[2]}
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.)
Basic examplesedit
Every sigma-finite measure is semifinite.
Assume ${\cal {A}}={\cal {P}}(X),$ let $f:X\to [0,+\infty ],$ and assume $\mu (A)=\sum _{a\in A}f(a)$ for all $A\subseteq X.$
We have that $\mu$ is sigma-finite if and only if $f(x)<+\infty$ for all $x\in X$ and $f^{\text{pre}}(\mathbb {R} _{>0})$ is countable. We have that $\mu$ is semifinite if and only if $f(x)<+\infty$ for all $x\in X.$^{[3]}
Taking $f=X\times \{1\}$ above (so that $\mu$ is counting measure on ${\cal {P}}(X)$), we see that counting measure on ${\cal {P}}(X)$ is
sigma-finite if and only if $X$ is countable; and
semifinite (without regard to whether $X$ is countable). (Thus, counting measure, on the power set ${\cal {P}}(X)$ of an arbitrary uncountable set $X,$ gives an example of a semifinite measure that is not sigma-finite.)
Let $d$ be a complete, separable metric on $X,$ let ${\cal {B}}$ be the Borel sigma-algebra induced by $d,$ and let $s\in \mathbb {R} _{>0}.$ Then the Hausdorff measure${\cal {H}}^{s}|{\cal {B}}$ is semifinite.^{[4]}
Let $d$ be a complete, separable metric on $X,$ let ${\cal {B}}$ be the Borel sigma-algebra induced by $d,$ and let $s\in \mathbb {R} _{>0}.$ Then the packing measure${\cal {H}}^{s}|{\cal {B}}$ is semifinite.^{[5]}
Involved exampleedit
The zero measure is sigma-finite and thus semifinite. In addition, the zero measure is clearly less than or equal to $\mu .$ It can be shown there is a greatest measure with these two properties:
Theorem (semifinite part)^{[6]} — For any measure $\mu$ on ${\cal {A}},$ there exists, among semifinite measures on ${\cal {A}}$ that are less than or equal to $\mu ,$ a greatest element $\mu _{\text{sf}}.$
We say the semifinite part of $\mu$ to mean the semifinite measure $\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:
Since $\mu _{\text{sf}}$ is semifinite, it follows that if $\mu =\mu _{\text{sf}}$ then $\mu$ is semifinite. It is also evident that if $\mu$ is semifinite then $\mu =\mu _{\text{sf}}.$
Non-examplesedit
Every $0-\infty$ measure that is not the zero measure is not semifinite. (Here, we say $0-\infty$ measure to mean a measure whose range lies in $\{0,+\infty \}$: $(\forall A\in {\cal {A}})(\mu (A)\in \{0,+\infty \}).$) Below we give examples of $0-\infty$ measures that are not zero measures.
Let $X$ be nonempty, let ${\cal {A}}$ be a $\sigma$-algebra on $X,$ let $f:X\to \{0,+\infty \}$ be not the zero function, and let $\mu =(\sum _{x\in A}f(x))_{A\in {\cal {A}}}.$ It can be shown that $\mu$ is a measure.
Let $X$ be uncountable, let ${\cal {A}}$ be a $\sigma$-algebra on $X,$ let ${\cal {C}}=\{A\in {\cal {A}}:A{\text{ is countable}}\}$ be the countable elements of ${\cal {A}},$ and let $\mu ={\cal {C}}\times \{0\}\cup ({\cal {A}}\setminus {\cal {C}})\times \{+\infty \}.$ It can be shown that $\mu$ is a measure.^{[2]}
Involved non-exampleedit
Measures that are not semifinite are very wild when restricted to certain sets.^{[Note 1]} Every measure is, in a sense, semifinite once its $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)^{[11]}^{[12]} — For any measure $\mu$ on ${\cal {A}},$ there exists a $0-\infty$ measure $\xi$ on ${\cal {A}}$ such that $\mu =\nu +\xi$ for some semifinite measure $\nu$ on ${\cal {A}}.$ In fact, among such measures $\xi ,$ there exists a least measure $\mu _{0-\infty }.$ Also, we have $\mu =\mu _{\text{sf}}+\mu _{0-\infty }.$
We say the $\mathbf {0-\infty }$ part of $\mu$ to mean the measure $\mu _{0-\infty }$ defined in the above theorem. Here is an explicit formula for $\mu _{0-\infty }$: $\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}}}.$
Results regarding semifinite measuresedit
Let $\mathbb {F}$ be $\mathbb {R}$ or $\mathbb {C} ,$ and let $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 $\mu$ is semifinite if and only if $T$ is injective.^{[13]}^{[14]} (This result has import in the study of the dual space of $L^{1}=L_{\mathbb {F} }^{1}(\mu )$.)
Let $\mathbb {F}$ be $\mathbb {R}$ or $\mathbb {C} ,$ and let ${\cal {T}}$ be the topology of convergence in measure on $L_{\mathbb {F} }^{0}(\mu ).$ Then $\mu$ is semifinite if and only if ${\cal {T}}$ is Hausdorff.^{[15]}^{[16]}
(Johnson) Let $X$ be a set, let ${\cal {A}}$ be a sigma-algebra on $X,$ let $\mu$ be a measure on ${\cal {A}},$ let $Y$ be a set, let ${\cal {B}}$ be a sigma-algebra on $Y,$ and let $\nu$ be a measure on ${\cal {B}}.$ If $\mu ,\nu$ are both not a $0-\infty$ measure, then both $\mu$ and $\nu$ are semifinite if and only if $(\mu \times _{\text{cld}}\nu )$$(A\times B)=\mu (A)\nu (B)$ for all $A\in {\cal {A}}$ and $B\in {\cal {B}}.$ (Here, $\mu \times _{\text{cld}}\nu$ is the measure defined in Theorem 39.1 in Berberian '65.^{[17]})
Localizable measuresedit
Localizable measures are a special case of semifinite measures and a generalization of sigma-finite measures.
Let $X$ be a set, let ${\cal {A}}$ be a sigma-algebra on $X,$ and let $\mu$ be a measure on ${\cal {A}}.$
Let $\mathbb {F}$ be $\mathbb {R}$ or $\mathbb {C} ,$ and let $T:L_{\mathbb {F} }^{\infty }(\mu )\to \left(L_{\mathbb {F} }^{1}(\mu )\right)^{*}:g\mapsto T_{g}=\left(\int fgd\mu \right)_{f\in L_{\mathbb {F} }^{1}(\mu )}.$ Then $\mu$ is localizable if and only if $T$ is bijective (if and only if $L_{\mathbb {F} }^{\infty }(\mu )$ "is" $L_{\mathbb {F} }^{1}(\mu )^{*}$).^{[18]}^{[14]}
s-finite measuresedit
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.
For certain purposes, it is useful to have a "measure" whose values are not restricted to the non-negative reals or infinity. For instance, a countably additive set function with values in the (signed) real numbers is called a signed measure, while such a function with values in the complex numbers is called a complex measure. Observe, however, that complex measure is necessarily of finite variation, hence complex measures include finite signed measures but not, for example, the Lebesgue measure.
Measures that take values in Banach spaces have been studied extensively.^{[19]} A measure that takes values in the set of self-adjoint projections on a Hilbert space is called a projection-valued measure; these are used in functional analysis for the spectral theorem. When it is necessary to distinguish the usual measures which take non-negative values from generalizations, the term positive measure is used. Positive measures are closed under conical combination but not general linear combination, while signed measures are the linear closure of positive measures.
Another generalization is the finitely additive measure, also known as a content. This is the same as a measure except that instead of requiring countable additivity we require only finite additivity. Historically, this definition was used first. It turns out that in general, finitely additive measures are connected with notions such as Banach limits, the dual of $L^{\infty }$ and the Stone–Čech compactification. All these are linked in one way or another to the axiom of choice. Contents remain useful in certain technical problems in geometric measure theory; this is the theory of Banach measures.
A charge is a generalization in both directions: it is a finitely additive, signed measure.^{[20]} (Cf. ba space for information about bounded charges, where we say a charge is bounded to mean its range its a bounded subset of R.)
^One way to rephrase our definition is that $\mu$ is semifinite if and only if $(\forall A\in \mu ^{\text{pre}}\{+\infty \})(\exists B\subseteq A)(0<\mu (B)<+\infty ).$ Negating this rephrasing, we find that $\mu$ is not semifinite if and only if $(\exists A\in \mu ^{\text{pre}}\{+\infty \})(\forall B\subseteq A)(\mu (B)\in \{0,+\infty \}).$ For every such set $A,$ the subspace measure induced by the subspace sigma-algebra induced by $A,$ i.e. the restriction of $\mu$ to said subspace sigma-algebra, is a $0-\infty$ measure that is not the zero measure.
Bibliographyedit
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Referencesedit
^Fremlin, D. H. (2010), Measure Theory, vol. 2 (Second ed.), p. 221
^Rao, M. M. (2012), Random and Vector Measures, Series on Multivariate Analysis, vol. 9, World Scientific, ISBN 978-981-4350-81-5, MR 2840012.
^Bhaskara Rao, K. P. S. (1983). Theory of charges: a study of finitely additive measures. M. Bhaskara Rao. London: Academic Press. p. 35. ISBN 0-12-095780-9. OCLC 21196971.