Extensions of Measures
Oops, forgot to post this this earlier…
We can put together what we’ve been doing recently to state the following theorem:
If is a
-finite measure on the ring
, then there is a unique measure
on the
-ring
extending
. That is, if
, then
. Further, the extended measure
is also
-finite.
The existence is straightforward. We can induce an outer measure, and then restrict it to get . It’s straightforward to verify from the definitions that
. And we know that since
is
-finite, so is
, and thus
.
What we need to show is that is unique. To this end, let
and
be two measures on
that both extend
. Let
be the class of sets on which
and
agree; this obviously contains
.
Now, if one of these two measures — say — is finite, and if
is a monotone sequence of sets on which
and
agree, then the limit of this sequence is again in
. Indeed, since measures are continuous, we must have
and similarly for . Since
is finite, and
agrees with
on
, we have a sequence of finite measures
. The limits of these sequences must then agree, and so
as well. Thus
is a monotone class. Since it contains
, it must contain
, and thus
.
On the other hand, neither measure may be finite. In this case, let be some fixed set of finite measure. Now
— the collection of intersections of sets in
with
— is again a ring, and
is the smallest
-ring containing it. Restricting
and
to
gives finite measures, and we can use the argument above.
Now every set can be covered by a countable, pairwise disjoint collection of sets
. For each one, we have
, and so we must find
. From here, countable additivity finishes the theorem.
In light of the uniqueness of this extension, we will just call the extended measure again, rather than
.
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