Vitali–Hahn–Saks theorem


In mathematics, the Vitali–Hahn–Saks theorem, introduced by ,, and, 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
This gives rise to a metric space by identifying two sets when. Thus a point with representative is the set of all such that.
Proposition: with the metric defined above is a complete metric space.
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 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
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.