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 ( 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,


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


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


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


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 ( 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


As ESi(EiS) and ESci(EiSc),


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