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We derive conditions under which a measure on a -algebra is uniquely determined by the values on a generator. On our way, we will learn about Dynkin systems and their relation to -algebras. Finally, we will prove the uniqueness theorem for measure continuation.
Continuation of functions on sets to measures are usually done in a way that the as many subsets of a basic set enter the -algebra, as possible (while sustaining some nice properties). Often, one requires additional conditions, for instance when constructing a geometric volume, that cuboids in get assigned their geometric volume. In general, it is not clear whether such a measure with the desired properties even exists. If it does, we might have several -algebras where it can be defined on.
The general construction of a measure is as follows: we start with a small set system , such that the function on sets restricted to it fulfills the desired properties. For example, for defining a geometric volume, we choose as the set system of cuboids and as the set function that assigns to each cuboid its geometric volume.
Using the existence of a continuation - theorem we know when a function can be continued to a measure on the -algebra generated by . For the proof of the continuation theorem, one possible continuation has been explicitly stated via outer measures. But can there also be other ways to continue to a measure on ? In other words, we are interested in whether a measure on the -algebra is already uniquely determined by its values on the smaller set system .
To make them easier to check, uniqueness statements are often re-formulated in mathematics: Suppose, and are measures on the -algebra generated by a set system . Further, let and coincide on , which means for all . Then uniqueness just means on the whole algebra .
In the following, we derive conditions for when this is the case.
We will proceed step by step to find conditions on the generator and the two measures and under which uniqueness holds. For this we consider the system of "good sets"
It contains all sets from on which and coincide. Uniqueness would mean that all sets in are good, i.e. .
Actually, this is equivalent to saying that is a -algebra: Since by assumption, is satisfied for all , we have holds. But already generates , i.e. there is no smaller -algebra which contains . So if is a -algebra, then (which was contained in ) must be the entire -algebra (see: monotonicity and idempotence of the -operator).
This type of approach is often used to show that a given property is satisfied for all sets of a set system (like a -algebra). It is called the "principle of good sets" and it works like this:
Suppose one can only make statements about a generator of , perhaps because can only be characterized in terms of the generator. An example is the Borel -algebra, which is extremely large and can only be written down by means of generators and relations. Then it might be smart to proceed indirectly when showing a property of all sets in .
We define the set system of "good sets". Then we show:
is a -algebra.
contains a generator of .
It follows that , i.e. all sets in are "good". Hence, we gained control over the extremely tedious set by only using properties of the simple sets in .
Theorem (The principle of good sets)
Let be a -algebra and let be the system of all sets from for which a given property holds. Assume that
is a -algebra.
contains a generator of .
Then the property holds for all sets from , i.e., we have that .
The principle of good sets is very powerful. It can also be applied to other kinds of set systems (not only for -algebras). For instance, it can also be applied to the ring or -ring generated by a set system.
In our case, the sets of good sets contains all the sets on which the measures and coincide. The equality of the measures on the generator is known, so holds. Now we are still looking for conditions so that becomes a -algebra. So we want to find conditions on the generator and the two measures such that:
Every -algebra contains the basic set , so should be in the set of good sets . That is, it should hold that for both measures and . In general, this need not be the case even if the two measures agree on :
Example (Measure of the basic set is not uniquely determined)
Let be the set system of all one-element subsets of :
Then and are equal on the generator, but not on all of . Thus, they are not yet uniquely determined by specifying the values on . In particular .
We can still generalize this example: Let be a set and let . The -algebra generated by is . Let be the zero measure on it, and for a let the measure be defined by . Then, we have , so and agree on . However, .
So with which conditions on the generator or the two measures can we enforce ?
Idea and definition of an inner approximation[Bearbeiten]
We know that the measures and coincide on the sets in . The idea is to cover the basic set with at most countably many sets from , i.e. their union should be .
Example (Coverings of )
The sets are neither pairwise disjoint nor contained in each other.
The sets are pairwise disjoint.
The sets are contained in each other in ascending order.
Now we want to infer from for all (which holds since these sets are from ), that holds as well. For this the should be either pairwise disjoint or contained in each other in ascending order:
If the are pairwise disjoint, we can use the -additivity of measures:
If they are contained in each other in ascending order (i.e. ), they form a monotonic set sequence with limit . Then we can use the continuity of the measures and :
If neither is the case, then we can not necessarily conclude from for all . We consider the case where the are contained in each other in ascending order and define the notion of exhaustion:
Let be a set and let be subsets for each with . If , then the sequence is called an exhaustion of .
For and there exists an exhaustion with .
Since measures are continuous, íf contains an exhaustion of , then we have .
To conclude, w have found the following first condition on the set system :
To ensure that the basic set is in the "set of good sets" , that is, that holds, we require that contains an exhaustion of .
The condition is automatically satisfied if holds, e.g. if and are probability measures: Then one can assume a priori .
The inner approximating sets have finite measure[Bearbeiten]
By requiring that has an exhaustion by sets from , we ensured that lies in the "set of good sets" .
It remains to investigate under which conditions is closed under formation of differences and countable unions.
For this purpose let us first examine under which operations is already closed with our previous assumptions:
Let and be two sets from the set of good sets . Thus we have and the same for . Now, we take their union. Things look good if : in this case , so is definitely also in the set of good sets . The same happens for
Similarly, the measure value of the union of and is uniquely determined if the two sets are disjoint: In that case, from additivity of the measures and it follows that
So the disjoint union is also uniquely measurable and lies again in .
If we additionally exploit the -additivity of measures, then we even know the measure of countably infinite unions of disjoint sets. Given a sequence of pairwise disjoint sets , the -additivity of measures and implies
Let again and be two sets from , i.e. let , hold. As in the case of the union, the difference of the two sets is again in if and are disjoint. In that case and we have that . (Likewise with the roles on and exchanged).
In the case , the difference of and is equal to . Since , it lies again in . Moreover, because of the additivity of the measure , we have that
In the first equation, we used that is a subset of . The same is true for the measure instead of . Rearranging the above formula together with and yields
This equation is dangerous! If and have infinite measure, we get the ill-defined expression "", which cannot be sensibly defined:
Example ( cannot be defined)
We consider the measure defined by the geometric length in . (We know that such a measure exists by the continuation theorem.) For the two sets and , we have that . So it follows , and in this case, "".
By contrast, for the sets , we have that . So and we get "".
This shows that in general we cannot make sense of the expression , without further assumptions.
Differences of sets with infinite measure[Bearbeiten]
A way out of this problem is to approximate and by ascending sequences of sets of finite measure and take a limit. For this, the sets of the set sequence should also be good sets. We can then calculate the measure of the differences as above, since and are both finite. But we have to be careful: For this to work, the subset relation must also hold for the set sequences for all . So we cannot just choose the sequences and in an arbitrary manner. They need to grow "equally fast" and at the same time approximate and equally fast.
Recall: we assumed that there is an exhaustion of the basic set with sets from the set system , i.e., a monotonically growing set sequence in with limit . (This was to guarantee .)
Then the sets also form an increasing set sequence with limit
The same is true for the sequence . Moreover, because , we also have , so the subset relation is satisfied for every member of the sequence. Because of the sequence of is also monotonically increasing.
Let us now use the same calculation as for the differences of sets with finite measure
and then turn to the limit . For this we need:
and for all . That is, intersections of sets with the are said to lie in .
Each set of the exhaustion has finite measure, i.e., for all . Only then, because of monotonicity, we can be sure that and also holds, ´which is our goal.
Note that the "finiteness" of an exhaustion depends on the considered measure. For example, the exhaustion is finite with respect to , but not with respect to the measure that assigns the value to every set except .
If and satisfy these conditions, we can calculate the original difference ():
Note that we were only able to swap difference and measure because the sets had finite measure. We could do that for the sets with infinite measure. For this we had to construct .
The following examples shows that the sets of the exhaustion indeed need to have finite measure in order to get uniqueness.
We consider the set system of half-open intervals in
It generates a -algebra (the so-called Borel -algebra), which also contains all one-element subsets of . Now consider the two measures defined on and by
for all . ( is thus the trivial measure that takes only the values and , while counts the number of rational numbers contained in ). Because is dense in , the measures and coincide on the set system . Furthermore, the exhaustion of exists with for all .
The two measures and are not equal on the -algebra generated by : For all one-element sets , we have that holds, but or . So the measures or are not yet uniquely determined by their values on .
The "set of good sets" is already closed under (at most countably infinitely many) disjoint unions; unions of sets which are subsets of each other; differences of disjoint sets; and differences of sets which are subsets of each other and have finite measure.
To ensure that the basic set is in , that is, , we require that contains an exhaustion of .
In order for differences of sets of infinite measure, which are subsets of each other, are in , we require, that the exhaustion sets all have finite measure and that for all the intersections lie again in .
Apart from these conditions and , we make no requirements on , , and .
The last condition is somewhat unsatisfactory, because it involves (which may include complicated sets). But we want to find conditions that refer only to the a priori given measures and the generator , respectively. We will still work on this and weaken the condition later.
In particular, the exhaustion requirements are directly satisfied if holds This is always the case if and are probability measures. Then is a monotonically increasing sequence of sets with finite measure, which obviously converges to .
As before, let be the system of good sets on which the two measures and coincide. Assuming that the conditions from the previous intermediate result are satisfied, we now know that the two measures are equal on the following sets:
The basic set : this is guaranteed by the exhaustion .
Unions of finitely or countably infinitely many pairwise disjoint sets in : this holds because of -additivity and continuity of the measures and .
Differences of sets from , where one is contained in the other: this is guaranteed by finiteness of the exhaustion and by the condition that intersections of sets from with sets of the exhaustion are again in .
Thus we can already characterize the set system of "good sets" more precisely. It contains the basic set and is closed under the operations "disjoint union" and "difference of sets contained in each other".
A set system with these properties is called a "Dynkin system".
A set system is called a Dynkin system if it holds that:
for every two sets with we also have .
for each countably many pairwise disjoint sets we also have .
It follows directly from the definition that every -algebra is a Dynkin system. The converse is not true: A Dynkin system is not always a -algebra, because it lacks closure under non-disjoint countable unions.
An equivalent characterization of a Dynkin system is the following.
A set system is a Dynkin system if and only if:
for every countably many , we also have .
for each countably many pairwise disjoint sets we also have .
Let be a Dynkin system in the sense of the definition above. Points 1. and 3. in the theorem are identical to the definition. We only need to show point 2. So let be arbitrary. We have that and , so it follows from property 2. of the definition that also .
Conversely, let the three properties from the theorem be satisfied. We show that then is a Dynkin system in the sense of the definition above. Again, we only need to show point 2. So let be arbitrary with . Let . The union on the right hand side is disjoint because of . By assumptions 2. and 3. the set thus indeed lies in .
So with the preconditions from the intermediate result of the previous section, is already a Dynkin system. All that is still missing for a algebra is closure under arbitrary countable unions.
Moreover, since the two measures and agree on the generator , we have that holds. So the Dynkin system generated by also lies in the set of good sets , which is defined analogously to the generated -algebra:
Definition (Generated Dynkin system)
Let be a set and be a set system. The Dynkin system
is then called the Dynkin system generated by . In other words, is the smallest Dynkin system containing .
As with the definition of the generated -algebra, we need to convince ourselves that is well-defined. This can be done completely analogously to the proof we already gave for generated -algebras.
The intersection in the above definition is not empty. Furthermore, the intersection of any number of Dynkin systems is again a Dynkin system.
Just as for generated -algebras, the following properties hold for the Dynkin system generated by a set system :
Minimality: is the smallest Dynkin system containing . If is a Dynkin system, then .
It follows directly from the definition that any -algebra is also a Dynkin system.
However, not every Dynkin system is a -algebra:
Example (Subsets with even cardinality)
Let be an even number and be a set with exactly elements. The set system
is a Dynkin system:
Since is even, we have that .
If are even with , then is even since are even.
Since is finite, we need only consider unions of finitely many disjoint sets for the third condition. Let be pairwise disjoint sets, each containing an even number of elements. Then, because of disjointness, the union also contains an even number of elements and so lies in .
However, is not a -algebra: Let be three distinct elements (these exist, since by assumption ). Then holds, but their union contains an odd number of elements, so it is not contained in .
We considered conditions under which the set system of "good" sets is a Dynkin system, i.e.
it contains the basic set,
it is closed under taking complements,
it is closed under taking (countable) disjoint unions.
Since we assume that the measures and coincide on the generator , holds. Since is itself a Dynkin system, therefore the Dynkin system generated by is also contained in .
We would like to have , such that and agree on their whole domain of definition. So should not just a Dynkin system, but a -algebra. So what about closedness under non-disjoint finite/countably infinite unions? Let us first look at finite unions. Let be good sets (i.e., and ), non-disjoint, and neither nor .
Initially, we have and , but it is not yet clear whether also . Actually every value may appear on the lift-hand side, as long as we do not violate the monotonicity of , i.e. as long as and .
What other conditions must be satisfied for to hold? It suffices if the intersection is again a good set, i.e. : If and both have finite measure, then we have
If either set has infinite measure, the equality holds anyway. Intersections of good sets should therefore be good sets again.
A set system, which is not left, when taking arbitrary intersections between its sets is called "cut-stable":
Definition (Cut stability of a set system)
A set system is called cut stable, if
By induction, the set system is then closed under intersection of any finite number of sets (not only 2), See below.
Is the cut-stability of the system of good sets in addition to our previous conditions already enough for it to be an -algebra? Suppose is cut-stable. The previous reasoning for two sets can be extended by induction to any finite union of good sets: Let be good sets, i.e., for all . We also assume again that all have finite measure (otherwise the equality holds anyway). Then
(with "i.a." meaning "induction assumption") Making use of the cut-stability, we have that is closed under arbitrary finite unions.
Countably infinite unions can be made "artificially" disjoint by cutting the preceding ones out of each set: Define
Then is a disjoint union of sets. If the are from , then so are the , provided that is cut-stable: According to the preceding, the finite union lies in , and so does the complement of this set (since is a Dynkin system). And by cut-stability also the intersection of it with lies in . Now, is closed under countable disjoint unions (Dynkin system), so also is a good set.
These considerations show: If the Dynkin system is additionally cut stable, it is also closed under arbitrary, at most countable unions, i.e., it is a sigma-algebra. We summarize this in a theorem:
Cut stable Dynkin systems are -algebras.
Let be a cut stable Dynkin system. Then and complement stability is given, as we have a Dynkin system. The remaining -algebra property is the union stability with respect to countable unions. So far, this is only given for pairwise disjoint unions. We first note that because , difference stability follows from complement stability and intersection stability of .
Now let be any sequence of sets in . We define by , for all .
Since for all we have that
With a simple induction argument it follows that and finally .
In particular, it follows from the construction of the sequence that elements of this sequence are pairwise disjoint.
By definition, . Suppose now that holds (induction assumption). Then with stability with respect to disjoint unions it follows that . With difference stability, we further have . Then, by induction, it also holds that for all .
But since the were additionally pairwise disjoint, it follows from the disjoint countable union stability that .
We have found conditions on the measures and the generator by which the set of good sets is a Dynkin system. In particular, thus contains the Dynkin system generated by , since . We want to hold as well. For this it suffices to find additional conditions under which is cut-stable: Since every cut-stable Dynkin system is a -algebra, it then follows that , and we are done.
So under what conditions is the Dynkin system generated by cut-stable? This apparently depends only on the properties of the set system , not on the measures or . In fact, it is sufficient if is cut-stable. This has to do with the fact that the cut operation is compatible with the union and complement operations of a Dynkin system, and thus the cut-stability is inherited from the generator to the generated Dynkin system. We show this in the following theorem.
Any Dynkin system generated by a cut-stable set system is itself cut-stable.
For we define , i.e., the set of sets in our Dynkin system that are cut-stable "with respect to ". By definition, holds for any .
Now we show two things:
is a Dynkin system for every .
for all .
For once we have shown these statements, it follows that , so . Here we used monotonicity in the second step and idempotence of the operator in the third step.
Now if are arbitrary, then because of we know in particular that holds, so . This shows the desired cut-stability.
So let us now show the necessary statements.
Proof step: is a Dynkin system for all .
Let be arbitrary. Obviously , so .
Let be arbitrary. Then holds. Furthermore, because of , of course . From this we conclude
This, as the complement of a disjoint union of two elements of , is also an element of and so , i.e. is complement stable.
Now let be a sequence of pairwise disjoint sets in . Then for all , it holds that . The sequence is also a sequence of pairwise disjoint sets in . From the stability of under countable disjoint unions, it follows that
So . Thus the three properties of a Dynkin system are satisfied and we are done.
Proof step: for all .
Let and be arbitrary. Then for all , due to the cut-stability of , it holds that . So, in particular, . Since was arbitrary, we also have . As is a Dynkin system (as shown before), we get . Further, holds. That is, and so . Since was chosen arbitrary, it finally follows that .
Since every average-stable Dynkin system is a algebra, it follows directly:
If is a cut-stable set system, then holds.
This relation between Dynkin systems and -algebras is very useful and simplifies many proofs about measures. This is because for Dynkin systems one can exploit the -additivity of the measure, since only disjoint unions need to be considered. In the proof of the uniqueness theorem we will see in a moment a first example where this enables to perform a proof.
We can now formulate and prove the uniqueness theorem.
Theorem (Uniqueness of measure continuations)
Let and be two measures on the -algebra over the basic set . Let there be a generator of with the following properties:
and coincide on , i.e. for all ,
There exists in an exhaustion of with sets of finite measure: a monotonically increasing sequence with limit and ,
is cut-stable, i.e. .
Then holds on all of . So in particular a measure is then already uniquely determined by the values on .
Proof (Uniqueness of measure continuations)
We perform the proof with the "principle of good sets" and define the set system as containing "good sets". It contains those sets from on which and coincide. By assumption, holds. We still have to show that is a -algebra. It is enough to show that is a Dynkin system since its generator is cut-stability . Now
where we have exploited in the second equality that the Dynkin system generated by a cut-stable set system is already a -algebra. We now establish the properties of a Dynkin system for in two steps: First assuming that are finite measures (i.e. for all ), then generalizing to the infinite measure case.
Proof step: Proof for finite
We have : Let be the exhaustion of from the assumption. For this exhaustion, we have equality for all . Then from the continuity of the two measures we get .
is complement stable: let . Because of for finite measures, we can exploit the subtractivity and obtain
is also stable under disjoint unions: Let be a sequence of pairwise disjoint sets in . Taking advantage of the -additivity of and we obtain
Proof step: Proof in the general case
Now, we turn to non-finite measures.
We define for the dimensions with and , where are the exhaustion sets from the assumption.
Since by assumption is cut-stable and holds on , the measures and also coincide on . Furthermore, because of and an analogous statement for , the two measures , are finite. So we can apply the statement already proved for the finite case and obtain that holds on all for all . The limit transition gives that also and on all are equal.
This theorem is a good example of the usefulness of Dynkin systems: Because of the disjointness of all unions, we could conveniently exploit the -additivity of the measures and in the proof.
Because of the cut-stability of the sequence of sets exhausting need not be monotonically increasing. It is sufficient to require that there exists a sequence such that and holds: If there exists such a sequence, we can define and obtain a monotonically increasing sequence with limit . Moreover, these sets also satisfy , as we saw in the section on cut-stability: Cut-stability ensures that and coincide even on finite (possibly non-disjoint) unions.
One therefore sometimes finds the following formulation of the uniqueness theorem:
Let and be two measures on the -algebra over the basic set . Let there be a generator of with the following properties:
and coincide on , i.e. for all ,
There is a sequence with and ,
is cut-stable, i.e. .
Then holds on all of .
If and are probability measures, then the second condition is always automatically satisfied: Because of one can assume without restriction and choose the constant sequence . In probability theory, therefore, one often finds the following version of the uniqueness theorem:
Theorem (Uniqueness theorem for probability measures)
Let and be two probability measures on the -algebra over the basic set . Let there be a cut-stable generator of on which and coincide. Then holds on all of .
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