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.)