In mathematics, the Vitali–Hahn–Saks theorem, introduced by , , and , proves that a pointwise convergent sequence of finite measures is uniformly absolutely continuous with respect to any common dominating finite measure, and its limit is also a finite measure. This holds in the more general setting of finite signed measures or complex measures.
Preliminaries
Let be a measurable space and let be a complex measure. The variation of is the finite measure given by
where the supremum is taken over all partitions of into a countable number of disjoint measurable subsets. The total variation is then .
A complex measure is called absolutely continuous with respect to some measure (denoted ), if for all measurable sets with . This is equivalent to the following: For every there is a so that implies . Since
absolute continuity is also equivalent to the following: For every there is a so that implies .
If , then, by the Radon–Nikodym theorem there is an integrable function , such that for every measurable . This function is uniquely defined up to an -null set and called a Radon–Nikodym derivative of with respect to . It is usually denoted by .
Statement of the theorem
Let be a measurable space and let be a sequence of finite complex measures such that the pointwise limit exists in for every measurable set . Define the function by
Then, the following holds: Countable Additivity and Boundedness: is countably additive and thus a finite complex measure. Furthermore, and . Absolute Continuity: Assume that is a finite measure on satisfying for all , such as Then, is also absolutely continuous with respect to , that is , and the sequence is uniformly absolutely continuous with respect to , that is, for every there is a so that implies uniformly for all . Densities: Again, if is a finite measure on satisfying for all , set for all and . Then, is uniformly integrable and converges in with respect to the weak topology , that is, for all bounded measurable .
Preparations for the Proof
Let be a measure space and write to denote the set of measurable sets with :
Write to denote the symmetric difference of the sets . Define the equivalence relation on by if . Write to denote the quotient set of with respect to the equivalence relation :
where is the equivalence class of . Then, induces a well-defined quotient map on which is denoted by : .
Proposition: Define by
Then, is well-defined and is a complete metric space.
Using indicator functions as before, we have
This means that the metric space can be identified with a subset of the space which is a Banach space. Hence, is a well-defined metric. Consider a Cauchy sequence . Then,
Hence, we can choose a subsequence such that converges almost everywhere. Define by
Then, is measurable, has values in almost everywhere and
Therefore, almost everywhere for some and
that is, the Cauchy sequence contains a convergent subsequence and is convergent itself. Therefore, is complete. We remark that is explicitly given by the limit inferior of the sequence :
Proposition: Let be a sequence of complex measures such that is bounded. Then, converges pointwise, that is, the sequence converges for every , if and only if, for every bounded measurable function , the sequence
converges.
The implication is obvious, since, for every , the indicator function with
is bounded and measurable and . For the other implication , we first note that, by linearity of limits, the sequence converges for all measurable simple functions . Write for the bound of total variations:
Given a bounded measurable function and an there is a measurable simple function with . Furthermore, converges, thus there exists such that
for all . Then,
for all . Consequently, the sequence converges.
Let be a measure on . Fix some . A set is called -indivisible, if it cannot be partitioned into two disjoint measurable subsets satisfying and .
Proposition: Let be a finite measure on and fix some . Then, there exists a partition of into at most disjoint -indivisible, measurable sets.
Start with the partition . Assume that is already defined. If all sets in are already -indivisible, then set . If not, pick one set that is not -indivisible. Then, can be partitioned into two disjoint measurable subsets satisfying and . In this case, set . This process must stabilize after at most , since each splitting creates a new set of measure at least .
Proof of Vitali-Hahn-Saks theorem
Let be a finite measure on satisfying for all . If none is given, take for instance
as mentioned above. Since is finite, we have . Furthermore, we write for the quotient set above. Each defines a quotient map on by . This quotient map is well-defined: If , then and thus , since . Therefore, . Moreover, is continuous as a function from the metric space to .
Fix . For every the function is again continuous, so that
is a closed subset of . Consequently,
is also a closed subset of . By the hypothesis the sequence converges for every , so that
Therefore, we obtain
By Baire category theorem at least one of the sets must contain a non-empty open set of . This means that there is and and an such that contains the open ball in with center and radius :
In other words:
Since for all , there is a , so that implies for all . We may choose sufficiently small, so that .
Now, consider a measurable set with . Define and . Then, we have
and
Thus, if , then and, if , then
Therefore, is uniformly absolutely continuous with respect to .
By the linearity of the limit is finitely additive. And by uniformly absolute continuity of the limit is actually countably additive.
Next, we prove boundedness of the total variations: Fix . By uniformly absolute continuity of there is such that implies . By the third proposition above, there is a finite partition of into -indivisible sets with respect to the finite measure . For any and any measurable , we cannot have both, and . If , then, by uniformly absolute continuity,
for all . If, on the other hand , then, again by uniform absolute continuity, for all , which implies
for all . In both cases, for every measurable subset and every :
Thus,
Since converges for every and is finite, we have
and therefore
The remaining assertions of the theorem follow directly using the second proposition above and boundedness of all total variations.
References