# Construct a measure space from a premeasure on a ring? (completely updated)

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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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 9:05
@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
@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