Completions of Measures
We’ve shown that we can uniquely extend a -finite measure
on a ring
to a unique
-finite measure on the
-ring
. But, of course, we actually found that we could restrict the outer measure
to the
-ring
of
-measurable sets, which may be larger than
. Luckily, we can get this extra ground without having to go through the outer measure.
What’s the essential difference? What do we know about that we don’t know about
? The measure on
is complete. The smaller
-ring
may not contain all negligible sets.
So, let’s throw them in; if is a measure on a
-ring
, define
to be the class of all sets
, where
, and
is a negligible with respect to the measure
, or “
-negligible”. This collection
is a
-algebra, and the set function
defined on
by
is a complete measure called the “completion” of
.
First, given sets and
with
, we have the two equations
These tell us that any set that can be written as the symmetric difference of a set in and a measurable set can also be written as the union of two other such sets, and vice versa. That is, we can also characterize
as the class of sets of the form
instead of
.
This characterization makes it clear that is closed under countable unions. Indeed, just write each set in a sequence as a union of a set
from
and a
-negligible set
. The countable union of the
is still in
, and the countable union of the negligible sets is still negligible. Thus
is a
-ring.
Now, if we have two ways of writing a set in , say
, then we also have the equation
. And, therefore,
. Therefore,
, and the above definition of
is unambiguous.
Using the characterization of in terms of unions, it’s easy to verify that
is a measure. We only need to check countable additivity, which is perfectly straightforward; the union of a pairwise disjoint sequence of sets
is formed by taking the union of the
and the
. The measure
is countably additive on the
, and the union of the
is still negligible.
Finally, we must show that is complete. But if we have a set
with
, then
and
with
. Thus any subset
is also a subset of
, with
. Writing it as
, we find
, and so
is complete.
There’s just one fly in the ointment: we don’t really know that this complete measure is the same as the one we get by restricting from to
-measurable sets. It turns out that if
is
-finite on
, then the completion of the extension of
to
is the same as this restriction.
Since we’ve been using for the completion, let’s write
for the class of
-measurable sets. Since
restricted to
is complete, it follows that
, and that
and
coincide on
. We just need to show that
.
In light of the fact that is also
-finite on
, we just need to show that if
has finite outer measure, then
. But in this case
has a measurable cover
with
. Since these are finite, we find that
. But
also has a measurable cover
, with
. And so we can write
, showing that
.
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