This set is due Friday, April 27.
The goal of these problems is to prove Carathéodory‘s theorem that “extracts” a measure from any outer measure. In particular, when applied to Lebesgue outer measure, this construction recovers Lebesgue measure.
Recall that an outer measure on a set is a function such that:
- .
- implies .
- For any subsets of , we have .
Given a set and an outer measure on , let denote the collection of subsets of with the property that
for all .
- Prove that is a -algebra on .
This requires some work. You may want to proceed by stages:
- First, check that is precisely the collection of sets such that, for any , we have
.
- Check that , and that is closed under complements.
- Check that is closed under finite unions. Conclude that it is also closed under set theoretic differences: If , then .
- The crux of the matter, of course, is to verify that is closed under countable unions. Accordingly, suppose that for all , and let .
Let , and note that , where , and, recursively, for . (Note also that for all .)
Then, for , , and for all ,
Conclude that . (Why does this limit exist?)
Also, prove that . (Again, why does this limit exist?)
Conclude from these inequalities and item 1 that . This concludes the proof that is a -algebra.
Now let denote the restriction of to .
- Prove that is a measure space.
In view of what we have proved already, note that this “reduces” to prove that, whenever are pairwise disjoint elements of , then
.
With notation as before, check first that for all , and conclude.
- Prove that is in fact a complete measure. Recall that this means that any subset of a set of -measure 0 is measurable and also has measure 0. In fact, check that if , then , and conclude from this.
- Suppose that . Show that the restriction of to is an outer measure on . Denote by resp. the set defined above, for resp. . Show that if , then . Suppose that is measurable (i.e., that ). Is ? If so, is this the only case where equality holds?
- Prove that if is , Lebesgue outer measure on , then is precisely , Lebesgue measure on . (This may be a bit easier for than in general.)