In
mathematics, the VitaliâHahnâSaks theorem, introduced by
Vitali (
1907),
Hahn (
1922), and
Saks (
1933), proves that under some conditions a sequence of
measures converging point-wise does so uniformly and the limit is also a measure.
Statement of the theorem
If is a
measure space with and a sequence of
complex measures. Assuming that each is
absolutely continuous with respect to and that a for all the finite limits exist Then the absolute continuity of the with respect to is uniform in that is, implies that uniformly in Also is countably additive on
Preliminaries
Given a measure space a distance can be constructed on the set of measurable sets with This is done by defining
Proof: Let
Then
This means that the metric space can be identified with a subset of the
Banach space.
Let , with
Then we can choose a sub-sequence such that exists
almost everywhere and . It follows that for some (furthermore if and only if for large enough, then we have that the
limit inferior of the sequence) and hence Therefore, is complete.
Proof of Vitali-Hahn-Saks theorem
Each defines a function on by taking . This function is well defined, this is it is independent on the representative of the class due to the absolute continuity of with respect to . Moreover is continuous.
For every the set
is closed in , and by the hypothesis we have that
By
Baire category theorem at least one must contain a non-empty open set of . This means that there is and a such that
implies
On the other hand, any with can be represented as with and . This can be done, for example by taking and . Thus, if and then
Therefore, by the absolute continuity of with respect to , and since is arbitrary, we get that implies uniformly in In particular, implies
By the additivity of the limit it follows that is
finitely-additive. Then, since it follows that is actually countably additive.
In
mathematics, the VitaliâHahnâSaks theorem, introduced by
Vitali (
1907),
Hahn (
1922), and
Saks (
1933), proves that under some conditions a sequence of
measures converging point-wise does so uniformly and the limit is also a measure.
Statement of the theorem
If is a
measure space with and a sequence of
complex measures. Assuming that each is
absolutely continuous with respect to and that a for all the finite limits exist Then the absolute continuity of the with respect to is uniform in that is, implies that uniformly in Also is countably additive on
Preliminaries
Given a measure space a distance can be constructed on the set of measurable sets with This is done by defining
Proof: Let
Then
This means that the metric space can be identified with a subset of the
Banach space.
Let , with
Then we can choose a sub-sequence such that exists
almost everywhere and . It follows that for some (furthermore if and only if for large enough, then we have that the
limit inferior of the sequence) and hence Therefore, is complete.
Proof of Vitali-Hahn-Saks theorem
Each defines a function on by taking . This function is well defined, this is it is independent on the representative of the class due to the absolute continuity of with respect to . Moreover is continuous.
For every the set
is closed in , and by the hypothesis we have that
By
Baire category theorem at least one must contain a non-empty open set of . This means that there is and a such that
implies
On the other hand, any with can be represented as with and . This can be done, for example by taking and . Thus, if and then
Therefore, by the absolute continuity of with respect to , and since is arbitrary, we get that implies uniformly in In particular, implies
By the additivity of the limit it follows that is
finitely-additive. Then, since it follows that is actually countably additive.