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Carathéodory's extension theorem extends a premeasure on a ring of subsets to a measure on a sigma algebra generated by the ring.

One popular proof of the theorem is a contructive one (please correct me if I am wrong):

  • first extend the premeasure on the ring to an outer measure,
  • then construct a measure space from the outer measure, with the measure being restriction of the outer measure on the sigma algebra of the measure space. The measure space is complete wrt its measure, by the way.
  • third prove the sigma algebra of the measure space contains the ring, so it contains the sigma algebra generated by the ring. So the sigma algebra generated by the ring and the measure restricted on it will be the resulting measure space in the theorem.

Questions:

  1. I was wondering if there is a different proof which is also constructive but not using an outer measure? References are perhaps just enough.

  2. When, i.e. under what conditions on the ring and the premeasure on it, will the measure space constructed by restricting "the complete measure space directly out of the outer measure" to "the sigma algebra generated from the ring" happen to be also complete?

    When it is, will it coincide with the complete measure space directly out of the outer measure?

Thanks!

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Tim
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  • Since one can "compose" steps even for general spaces (at least under $\sigma$-finiteness), the answer depends very much on how you do the steps – Michael Greinecker Jan 14 '13 at 09:05
  • @MichaelGreinecker: Thanks! (1) "one can "compose" steps even for general spaces (at least under σ-finiteness)", do you mean the transfinite induction way of repeating the operations under which $\sigma$ algebras are closed? How is $\sigma$ finiteness helpful? (2) as far as you know, are there no known extension theorems for the premeasure from the semiring to the Borel sigma algebra? – Tim Jan 14 '13 at 13:12
  • The most general version of the Caratheodory extension theorem takes a premeasure on a semi-ring $\mathcal{R}$ and extends it to a complete measure on the $\sigma$-algebra that forms the completion of $\sigma(\mathcal{R})$. One needs additional assumption for this extension to be unique, and some form of $\sigma$-finiteness usually does the trick. – Michael Greinecker Jan 14 '13 at 13:16
  • @MichaelGreinecker: Is it necessary to consider the existence of the extension of the premeasure from the semiring to the Borel sigma algebra instead of to the Lebesgue sigma algebra? Any sources for that and known to you? – Tim Jan 14 '13 at 13:21
  • You can always extend to the Lebesgue measurable sets and then restrict the measure to the Borel sets. – Michael Greinecker Jan 14 '13 at 13:28
  • @MichaelGreinecker: Thanks! The Borel sigma algebra is contained in the Lebesgue one, because the Borel one is generated by the topology, and the Lebesgue one contains the topology. Usually, how do we see that the Borel sigma algebra may not equal the Lebesgue one – Tim Jan 14 '13 at 13:53
  • By a cardinality argument. You can show that there are as many Borel sets as real numbers, and as much Lebesgue measurable sets as there are sets of real numbers. I think it is discussed in some questions on MSE. – Michael Greinecker Jan 14 '13 at 14:07

1 Answers1

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There is an approach that works without outer measures and relies on transfinite induction. A relatively dense eplanation of the approach is given in the paper Measure theory based on lattices and transfinite recursion by Oliver Deiser. He also has a more gentle paper, Ordinalzahlen in der Analysis und Maßtheorie, but that paper is written in German.

Michael Greinecker
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  • Thanks! Is that transfinite induction method based on the transfinite induction method for generating a sigma algebra from any collection of subsets? Note that the transfinite induction method for generating a sigma algebra is the rawest way which repeats and alternates between adding complements and adding countable unions transfinitely many times? – Tim Jan 14 '13 at 20:49
  • @Tim Essentially, yes. But, Deiserworks $\sigma$-rings instead of $\sigma$-algebras. The more involved version also includes certain regularity properties. – Michael Greinecker Jan 14 '13 at 20:52
  • (1) Can I claim that in all math text books that we/you have seen, there is no other way without using an outer measure to prove CET (the one using transfinite recursion is a paper not in a book), and no constructive way as elegant as the popular one using an outer measure? (2) When will the measure space constructed by CET complete? – Tim Jan 14 '13 at 20:56
  • @Tim Deiser actually does prove the CET. Completing a measure is actually quite easy, but it is automatic in Caratheodory's oroginal proof. – Michael Greinecker Jan 14 '13 at 20:58
  • In CET, the final measure space may not be complete, but the intermediate measure space directly from the outer measure is coplete. My question was when the final measure space is also complete? – Tim Jan 14 '13 at 21:00
  • @Tim The usual way by means of an outer emasure gives you a complete measure. The proof by transfinite induction does not, but you can easily complete the measure afterwards if you so desire. – Michael Greinecker Jan 14 '13 at 21:01