Movatterモバイル変換


[0]ホーム

URL:


Jump to content
WikipediaThe Free Encyclopedia
Search

Support (measure theory)

From Wikipedia, the free encyclopedia
Concept in mathematics

Inmathematics, thesupport (sometimestopological support orspectrum) of ameasureμ{\displaystyle \mu } on ameasurabletopological space(X,Borel(X)){\displaystyle (X,\operatorname {Borel} (X))} is a precise notion of where in the spaceX{\displaystyle X} the measure "lives". It is defined to be the largest (closed)subset ofX{\displaystyle X} for which everyopenneighbourhood of every point of theset has positive measure.

Motivation

[edit]

A (non-negative) measureμ{\displaystyle \mu } on a measurable space(X,Σ){\displaystyle (X,\Sigma )} is really a functionμ:Σ[0,+].{\displaystyle \mu :\Sigma \to [0,+\infty ].} Therefore, in terms of the usualdefinition ofsupport, the support ofμ{\displaystyle \mu } is a subset of theσ-algebraΣ:{\displaystyle \Sigma :}supp(μ):={AΣ|μ(A)0}¯,{\displaystyle \operatorname {supp} (\mu ):={\overline {\{A\in \Sigma \,\vert \,\mu (A)\neq 0\}}},}where the overbar denotesset closure. However, this definition is somewhat unsatisfactory: we use the notion of closure, but we do not even have a topology onΣ.{\displaystyle \Sigma .} What we really want to know is where in the spaceX{\displaystyle X} the measureμ{\displaystyle \mu } is non-zero. Consider two examples:

  1. Lebesgue measureλ{\displaystyle \lambda } on thereal lineR.{\displaystyle \mathbb {R} .} It seems clear thatλ{\displaystyle \lambda } "lives on" the whole of the real line.
  2. ADirac measureδp{\displaystyle \delta _{p}} at some pointpR.{\displaystyle p\in \mathbb {R} .} Again, intuition suggests that the measureδp{\displaystyle \delta _{p}} "lives at" the pointp,{\displaystyle p,} and nowhere else.

In light of these two examples, we can reject the following candidate definitions in favour of the one in the next section:

  1. We could remove the points whereμ{\displaystyle \mu } is zero, and take the support to be the remainderX{xXμ({x})=0}.{\displaystyle X\setminus \{x\in X\mid \mu (\{x\})=0\}.} This might work for the Dirac measureδp,{\displaystyle \delta _{p},} but it would definitely not work forλ:{\displaystyle \lambda :} since the Lebesgue measure of any singleton is zero, this definition would giveλ{\displaystyle \lambda } empty support.
  2. By comparison with the notion ofstrict positivity of measures, we could take the support to be the set of all points with a neighbourhood of positive measure:{xXNx open such that (xNx and μ(Nx)>0)}{\displaystyle \{x\in X\mid \exists N_{x}{\text{ open}}{\text{ such that }}(x\in N_{x}{\text{ and }}\mu (N_{x})>0)\}} (or theclosure of this). It is also too simplistic: by takingNx=X{\displaystyle N_{x}=X} for all pointsxX,{\displaystyle x\in X,} this would make the support of every measure except the zero measure the whole ofX.{\displaystyle X.}

However, the idea of "local strict positivity" is not too far from a workable definition.

Definition

[edit]

Let(X,T){\displaystyle (X,T)} be atopological space; letB(T){\displaystyle B(T)} denote theBorel σ-algebra onX,{\displaystyle X,} i.e. the smallest sigma algebra onX{\displaystyle X} that contains all open setsUT.{\displaystyle U\in T.} Letμ{\displaystyle \mu } be a measure on(X,B(T)){\displaystyle (X,B(T))}. Then thesupport (orspectrum) ofμ{\displaystyle \mu } is defined as the set of all pointsx{\displaystyle x} inX{\displaystyle X} for which everyopenneighbourhoodNx{\displaystyle N_{x}} ofx{\displaystyle x} haspositive measure:supp(μ):={xXNxT:(xNxμ(Nx)>0)}.{\displaystyle \operatorname {supp} (\mu ):=\{x\in X\mid \forall N_{x}\in T\colon (x\in N_{x}\Rightarrow \mu (N_{x})>0)\}.}

Some authors prefer to take the closure of the above set. However, this is not necessary: see "Properties" below.

An equivalent definition of support is as the largestCB(T){\displaystyle C\in B(T)} (with respect to inclusion) such that every open set which has non-empty intersection withC{\displaystyle C} has positive measure, i.e. the largestC{\displaystyle C} such that:(UT)(UCμ(UC)>0).{\displaystyle (\forall U\in T)(U\cap C\neq \varnothing \implies \mu (U\cap C)>0).}

Signed and complex measures

[edit]

This definition can be extended to signed and complex measures. Suppose thatμ:Σ[,+]{\displaystyle \mu :\Sigma \to [-\infty ,+\infty ]} is asigned measure. Use theHahn decomposition theorem to writeμ=μ+μ,{\displaystyle \mu =\mu ^{+}-\mu ^{-},}whereμ±{\displaystyle \mu ^{\pm }} are both non-negative measures. Then thesupport ofμ{\displaystyle \mu } is defined to besupp(μ):=supp(μ+)supp(μ).{\displaystyle \operatorname {supp} (\mu ):=\operatorname {supp} (\mu ^{+})\cup \operatorname {supp} (\mu ^{-}).}

Similarly, ifμ:ΣC{\displaystyle \mu :\Sigma \to \mathbb {C} } is acomplex measure, thesupport ofμ{\displaystyle \mu } is defined to be theunion of the supports of its real and imaginary parts.

Properties

[edit]

supp(μ1+μ2)=supp(μ1)supp(μ2){\displaystyle \operatorname {supp} (\mu _{1}+\mu _{2})=\operatorname {supp} (\mu _{1})\cup \operatorname {supp} (\mu _{2})} holds.

A measureμ{\displaystyle \mu } onX{\displaystyle X} isstrictly positiveif and only if it has supportsupp(μ)=X.{\displaystyle \operatorname {supp} (\mu )=X.} Ifμ{\displaystyle \mu } is strictly positive andxX{\displaystyle x\in X} is arbitrary, then any open neighbourhood ofx,{\displaystyle x,} since it is anopen set, has positive measure; hence,xsupp(μ),{\displaystyle x\in \operatorname {supp} (\mu ),} sosupp(μ)=X.{\displaystyle \operatorname {supp} (\mu )=X.} Conversely, ifsupp(μ)=X,{\displaystyle \operatorname {supp} (\mu )=X,} then every non-empty open set (being an open neighbourhood of some point in its interior, which is also a point of the support) has positive measure; hence,μ{\displaystyle \mu } is strictly positive.The support of a measure isclosed inX,{\displaystyle X,}as its complement is the union of the open sets of measure0.{\displaystyle 0.}

In general the support of a nonzero measure may be empty: see the examples below. However, ifX{\displaystyle X} is aHausdorff topological space andμ{\displaystyle \mu } is aRadon measure, a Borel setA{\displaystyle A} outside the support hasmeasure zero:AXsupp(μ)μ(A)=0.{\displaystyle A\subseteq X\setminus \operatorname {supp} (\mu )\implies \mu (A)=0.} The converse is true ifA{\displaystyle A} is open, but it is not true in general: it fails if there exists a pointxsupp(μ){\displaystyle x\in \operatorname {supp} (\mu )} such thatμ({x})=0{\displaystyle \mu (\{x\})=0} (e.g. Lebesgue measure). Thus, one does not need to "integrate outside the support": for anymeasurable functionf:XR{\displaystyle f:X\to \mathbb {R} } orC,{\displaystyle \mathbb {C} ,}Xf(x)dμ(x)=supp(μ)f(x)dμ(x).{\displaystyle \int _{X}f(x)\,\mathrm {d} \mu (x)=\int _{\operatorname {supp} (\mu )}f(x)\,\mathrm {d} \mu (x).}

The concept ofsupport of a measure and that ofspectrum of aself-adjoint linear operator on aHilbert space are closely related. Indeed, ifμ{\displaystyle \mu } is aregular Borel measure on the lineR,{\displaystyle \mathbb {R} ,} then the multiplication operator(Af)(x)=xf(x){\displaystyle (Af)(x)=xf(x)} is self-adjoint on its natural domainD(A)={fL2(R,dμ)xf(x)L2(R,dμ)}{\displaystyle D(A)=\{f\in L^{2}(\mathbb {R} ,d\mu )\mid xf(x)\in L^{2}(\mathbb {R} ,d\mu )\}} and its spectrum coincides with theessential range of the identity functionxx,{\displaystyle x\mapsto x,} which is precisely the support ofμ.{\displaystyle \mu .}[1]

Examples

[edit]

Lebesgue measure

[edit]

In the case of Lebesgue measureλ{\displaystyle \lambda } on the real lineR,{\displaystyle \mathbb {R} ,} consider an arbitrary pointxR.{\displaystyle x\in \mathbb {R} .} Then any open neighbourhoodNx{\displaystyle N_{x}} ofx{\displaystyle x} must contain some openinterval(xϵ,x+ϵ){\displaystyle (x-\epsilon ,x+\epsilon )} for someϵ>0.{\displaystyle \epsilon >0.} This interval has Lebesgue measure2ϵ>0,{\displaystyle 2\epsilon >0,} soλ(Nx)2ϵ>0.{\displaystyle \lambda (N_{x})\geq 2\epsilon >0.} SincexR{\displaystyle x\in \mathbb {R} } was arbitrary,supp(λ)=R.{\displaystyle \operatorname {supp} (\lambda )=\mathbb {R} .}

Dirac measure

[edit]

In the case ofDirac measureδp,{\displaystyle \delta _{p},} letxR{\displaystyle x\in \mathbb {R} } and consider two cases:

  1. ifx=p,{\displaystyle x=p,} then every open neighbourhoodNx{\displaystyle N_{x}} ofx{\displaystyle x} containsp,{\displaystyle p,} soδp(Nx)=1>0.{\displaystyle \delta _{p}(N_{x})=1>0.}
  2. on the other hand, ifxp,{\displaystyle x\neq p,} then there exists a sufficiently small open ballB{\displaystyle B} aroundx{\displaystyle x} that does not containp,{\displaystyle p,} soδp(B)=0.{\displaystyle \delta _{p}(B)=0.}

We conclude thatsupp(δp){\displaystyle \operatorname {supp} (\delta _{p})} is the closure of thesingleton set{p},{\displaystyle \{p\},} which is{p}{\displaystyle \{p\}} itself.

In fact, a measureμ{\displaystyle \mu } on the real line is a Dirac measureδp{\displaystyle \delta _{p}} for some pointp{\displaystyle p}if and only if the support ofμ{\displaystyle \mu } is the singleton set{p}.{\displaystyle \{p\}.} Consequently, Dirac measure on the real line is the unique measure with zerovariance (provided that the measure has variance at all).

A uniform distribution

[edit]

Consider the measureμ{\displaystyle \mu } on the real lineR{\displaystyle \mathbb {R} } defined byμ(A):=λ(A(0,1)){\displaystyle \mu (A):=\lambda (A\cap (0,1))}i.e. auniform measure on the open interval(0,1).{\displaystyle (0,1).} A similar argument to the Dirac measure example shows thatsupp(μ)=[0,1].{\displaystyle \operatorname {supp} (\mu )=[0,1].} Note that the boundary points 0 and 1 lie in the support: any open set containing 0 (or 1) contains an open interval about 0 (or 1), which must intersect(0,1),{\displaystyle (0,1),} and so must have positiveμ{\displaystyle \mu }-measure.

A nontrivial measure whose support is empty

[edit]

The space of allcountable ordinals with the topology generated by "open intervals" is alocally compactHausdorff space. The measure ("Dieudonné measure") that assigns measure 1 to Borel sets containing an unbounded closed subset and assigns 0 to other Borel sets is a Borel probability measure whose support is empty.[2]

A nontrivial measure whose support has measure zero

[edit]

On a compact Hausdorff space the support of a non-zero measure is always non-empty, but may have measure0.{\displaystyle 0.} An example of this is given by adding the first uncountable ordinalΩ{\displaystyle \Omega } to the previous example: the support of the measure is the single pointΩ,{\displaystyle \Omega ,} which has measure0.{\displaystyle 0.}

References

[edit]
  1. ^Mathematical methods in Quantum Mechanics with applications to Schrödinger Operators
  2. ^Measure theory, Example 7.1.3
Basic concepts
Sets
Types ofmeasures
Particular measures
Maps
Main results
Other results
ForLebesgue measure
Applications & related
Retrieved from "https://en.wikipedia.org/w/index.php?title=Support_(measure_theory)&oldid=1314963563"
Categories:
Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp