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proof of Carathéodory’s extension theorem

proof of Carathéodory’s extension theorem


The first step is to extend the set functionMathworldPlanetmath μ0 to the power setMathworldPlanetmath P(X). For any subset SX the value of μ*(S) is defined by taking sequences Si in A which cover S,

μ*(S)inf{i=1μ0(Si):SiA,Si=1Si}. (1)

We show that this is an outer measureMathworldPlanetmathPlanetmath (http://planetmath.org/OuterMeasure2). First, it is clearly non-negative. Secondly, if S= then we can take Si= in (1) to obtain μ*(S)iμ0()=0, giving μ*()=0. It is also clear that μ* is increasing, so that if ST then μ*(S)μ*(T). The only remaining property to be proven is subadditivity. That is, if Si is a sequence in P(X) then

μ*(iSi)iμ*(Si). (2)

To prove this inequalityMathworldPlanetmath, choose any ϵ>0 and, by the definition (1) of μ*, for each i there exists a sequence Si,jA such that SijSi,j and,

j=1μ0(Si,j)μ*(Si)+2-iϵ.

As iSii,jSi,j, equation (1) defining μ* gives

μ*(iSi)i,jμ0(Si,j)=ijμ0(Si,j)i(μ*(Si)+2-iϵ)=iμ*(Si)+ϵ.

As ϵ>0 is arbitrary, this proves subadditivity (2). So, μ* is indeed an outer measure.

The next step is to show that μ* agrees with μ0 on A. So, choose any SA. The inequality μ*(S)μ0(S) follows from taking S1=S and Si= in (1), and it remains to prove the reverse inequality. So, let Si be a sequence in A covering S, and set

Si=(SSi)j=1i-1SjA.

Then, Si are disjoint sets satisfying j=1iSj=Sj=1iSj and, therefore, iSi=S. By the countable additivityMathworldPlanetmath of μ0,

iμ0(Si)=i(μ0(Si)+μ0(SiSi))iμ0(Si)=μ0(S).

As this inequality hold for any sequence SiA covering S, equation (1) gives μ*(S)μ0(S) and, by combining with the reverse inequality, shows that μ* does indeed agree with μ0 on A.

We have shown that μ0 extends to an outer measure μ* on the power set of X. The final step is to apply Carathéodory’s lemma on the restrictionPlanetmathPlanetmathPlanetmath of outer measures. A set SX is said to be μ*-measurable if the inequality

μ*(E)μ*(ES)+μ*(ESc) (3)

is satisfied for all subsets E of X. Carathéodory’s lemma then states that the collectionMathworldPlanetmath of μ*-measurable setsMathworldPlanetmath is a σ-algebraMathworldPlanetmath (http://planetmath.org/SigmaAlgebra) and that the restriction of μ* to is a measureMathworldPlanetmath. To completePlanetmathPlanetmathPlanetmathPlanetmath the proof of the theorem it only remains to be shown that every set in A is μ*-measurable, as it will then follow that contains 𝒜=σ(A) and the restriction of μ* to 𝒜 is a measure.

So, choosing any SA and EX, the proof will be complete once it is shown that (3) is satisfied. Given any ϵ>0, equation (1) says that there is a sequence Ei in A such that EiEi and

iμ0(Ei)μ*(E)+ϵ.

As ESi(EiS) and ESci(EiSc),

μ*(ES)+μ*(ESc)iμ0(EiS)+iμ0(EiSc)=iμ0(Ei)μ*(E)+ϵ.

Since ϵ is arbitrary, this shows that (3) is satisfied and S is μ*-measurable.

Title proof of Carathéodory’s extension theorem
Canonical name ProofOfCaratheodorysExtensionTheorem
Date of creation 2013-03-22 18:33:28
Last modified on 2013-03-22 18:33:28
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Proof
Classification msc 28A12
Related topic CaratheodorysLemma
Related topic Measure
Related topic OuterMeasure2








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