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单词 ProofOfCapacityGeneratedByAMeasure
释义

proof of capacity generated by a measure


For a finite measure space (X,,μ), define

μ*:𝒫(X)+,
μ*(S)=inf{μ(A):A,AS}.

We show that μ* is an -capacity and that a subset SX is (,μ*)-capacitable (http://planetmath.org/ChoquetCapacity) if and only if it is in the completionPlanetmathPlanetmath (http://planetmath.org/CompleteMeasure) of with respect to μ.

Note, first of all, that μ*(S)=μ(S) for any S.That μ* is increasing follows directly from the definition. If An is a decreasing sequence of sets then A=nAn is also in and, by continuity from above (http://planetmath.org/PropertiesForMeasure) for measuresMathworldPlanetmath,

μ*(An)=μ(An)μ(A)=μ*(A)

as n.

Now suppose that Sn is an increasing sequence of subsets of X and set S=nSn. Then, μ*(Sn)μ*(S) for each n and, hence, limnμ*(Sn)μ*(S).

To prove the reverse inequalityMathworldPlanetmath, choose any ϵ>0 and sequenceMathworldPlanetmath An with SnAn and μ(An)μ*(Sn)+2-nϵ.Then, AmAnSn whenever mn and, therefore,

μ(AnAm)=μ(An)-μ(AmAn)μ(An)-μ*(Sn)2-nϵ.

Additivity of μ then gives

μ(mnAm)μ(An)+m=1n-1μ(AmAn)μ*(Sn)+ϵ.

So, by continuity from below for measures,

μ*(S)μ(nAn)=limnμ(mnAm)limnμ*(Sn)+ϵ.

Choosing ϵ arbitrarily small shows that μ*(Sn)μ*(S) and, therefore, μ* is indeed an -capacity.

Now suppose that S is in the completion of with respect to μ, so that there exists A,B with ASB and μ(BA)=0. Then,

μ*(A)=μ(A)=μ(B)μ*(S)

and S is indeed (,μ*)-capacitable.Conversely, let S be (,μ*)-capacitable. Then, there exists An,Bn such that AnSBn and

μ(An)μ*(S)-1/n,μ(Bn)μ*(S)+1/n.

Setting A=nAn and B=nBn gives ASB and

μ(A)μ*(S)μ(B).

So μ(BA)=μ(B)-μ(A)=0, as required.

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更新时间:2025/5/4 21:12:33