The construction of the stochastic integral given in the previous post made use of a result showing that certain linear maps can be extended to vector valued measures. This result, Theorem 1 below, was separated out from the main argument in the construction of the integral, as it only involves pure measure theory and no stochastic calculus. For completeness of these notes, I provide a proof of this now.
Given a measurable space , denotes the bounded -measurable functions . For a topological vector space V, the term V-valued measure refers to linear maps satisfying the following bounded convergence property; if a sequence (n=1,2,…) is uniformly bounded, so that for a constant K, and converges pointwise to a limit , then in V.
This differs slightly from the definition of V-valued measures as set functions satisfying countable additivity. However, any such set function also defines an integral satisfying bounded convergence and, conversely, any linear map satisfying bounded convergence defines a countably additive set function . So, these definitions are essentially the same, but for the purposes of these notes it is more useful to represent V-valued measures in terms of their integrals rather than the values on measurable sets.
In the following, a subalgebra of is a subset closed under linear combinations and pointwise multiplication, and containing the constant functions.
Theorem 1 Let be a measurable space, be a subalgebra of generating , and V be a complete vector space. Then, a linear map extends to a V-valued measure on if and only if it satisfies the following properties for sequences .
- If then .
- If , then .
The conditions stated are necessary for the existence of an extension to a V-valued measure because, in both cases, is a uniformly bounded sequence converging to 0. Note the similarity of this result to Carathéodory’s extension theorem for standard (real-valued) measures. Suppose that are linear combinations of indicator functions 1A for A in some subset which is closed under finite intersections and under complements. Then, setting , the conditions of the theorem are equivalent to the requirement that whenever are either pairwise disjoint or decreasing to the empty set.
The construction of the stochastic integral only made use of the special case where is the predictable sigma algebra, is the space of random variables (up to almost sure equivalence) under convergence in probability, and is the set of uniformly bounded continuous and adapted processes. However, it is not much more difficult to prove Theorem 1 in full generality, which we do here.
First, if satisfy then so that, by the second condition of the theorem, . It follows that is continuous under the topology of uniform convergence on and, by continuous linear extensions, extends uniquely to a continuous linear function from the closure of under uniform convergence. As is an algebra, for every polynomial p and . The Stone-Weierstrass theorem states that continuous functions on a bounded interval can be uniformly approximated by polynomials, so for all continuous functions and . Furthermore, this extension of still satisfies the conditions of the theorem. If , then choose with . If then will also be decreasing to zero, giving . Alternatively, if then , again giving . In either case, tends to zero as required. So, by passing to the closure if necessary, it may be assumed that is closed under uniform convergence.
Throughout the remainder of this post, it is assumed that is closed under uniform convergence. So, for continuous functions and . In particular, we make use of the fact that the absolute value of any is itself in .
The proof I give here is close to the construction of the stochastic integral given by Bichteler in Stochastic Integration and Lp-Theory of Semimartingales and in his book, `Stochastic Integration with Jumps’. The idea is to define a semimetric D* on the bounded measurable functions, referred to by Bichteler as the Daniell mean. Once it is shown that is continuous under this metric, a continuous linear extension to the whole of can be used. Let us suppose that the topology on V is given by a translation invariant metric . This is indeed true in the construction of the stochastic integral, where and generates the topology of convergence in probability. Generalizing to arbitrary complete vector spaces is straightforward, as demonstrated in the subsection below. By the standard metric properties, D must satisfy and for all u,v in V. Furthermore, replacing D(v) by min(D(v),1) if required, we suppose that D is bounded.
For any , define
Then, define by
Although I have used a slightly different definition, this is the quantity referred to as the Daniell mean by Bichteler. In fact, D* and D0 will agree on , as we show in a moment. Countable subadditivity follows from this definition in a straightforward way, as given by the following lemma. In particular, the triangle inequality is satisfied, so D* defines a translation invariant semimetric on .
Lemma 2 Suppose that satisfy . Then,
Proof: Using (1), for any there exist satisfying and
As was arbitrary, (2) follows.
Let us also show that D* and D0 agree on . In particular, this gives , so that is continuous with respect to the semimetric D*.
Lemma 3 For any ,
Proof: Setting and for n=2,3,… in (1) gives , so only the reverse inequality needs to be shown.
Given any with , setting and gives and . So,
Taking the supremum over all gives the triangle inequality,
Now choose a sequence with and set . For any with , set so that .
Then, and decrease monotonically to zero. So, by the first property of Theorem 1,
Then, applying the triangle inequality for D0, we obtain the following
Taking the infimum over all such and applying (1) gives . Then taking the supremum over gives as required.
Now, let be the closure of with respect to the topology generated by the semimetric D*. That is, if and only if there is a sequence such that . Then, D* satisfies monotone convergence on .
Lemma 4 Let and be such that . Then, and .
Proof: First, let us show that as m,n go to infinity. Choose any and with . Suppose that for some constant K. Replacing by if necessary, we may suppose that . Then, setting ,
So, is bounded below by and above by . Applying the triangle inequality,
Summing over n=2,…,m gives
Now, choose any sequence . As is increasing, has a sum bounded by K. So, any with also have a sum bounded by K and, by the second condition of Theorem 1, . By Lemma 3, we can choose , showing that . Since the sequence was arbitrary, this shows that as m,n go to infinity. So,
As was arbitrary, this shows that as required.
Now choosing such that whenever then, for , countable subadditivity (Lemma 2) gives
which goes to 0 as k increases to infinity.
Putting together the previous lemmas, it is now straightforward to show that is the whole of and that D* satisfies bounded convergence.
Lemma 5 The equality holds. Furthermore, suppose that are uniformly bounded by some constant K, so that , and pointwise as n goes to infinity. Then, .
Now, suppose that satisfy and . The sequence decreases monotonically to zero. So, and Lemma 4 gives . Therefore,
Lemma 6 Let V be a complete vector space, with translation invariant metric D. Then, assuming the conditions of Theorem 1, extends uniquely to a V-valued measure on .
Generalizing to arbitrary complete vector spaces
The proof of Theorem 1 can be extended to arbitrary complete vector spaces, not given by a translation invariant metric. This is not needed for the construction of the stochastic integral, but it is not difficult, and I do this now for the sake of giving a full proof of the theorem as stated. We can use the result that any vector topology is given by a collection of translation invariant semimetrics . As above, replacing Di(v) by min(Di(v),1), we may suppose that Di are bounded. Then, by the above argument, there is a collection of semimetrics on such that for all , and satisfying bounded convergence as in Lemma 5.
Then, the closure of with respect to the topology generated by is a linear space closed under monotone convergence (by Lemma 4) and, by the monotone class theorem, is the whole of . By continuous linear extensions, extends uniquely to a continuous linear function . By Lemma 5, satisfies bounded convergence and hence is a V-valued measure.
The statement that all vector topologies are generated by a set of translation invariant semimetrics needs some explanation. It is a standard result of functional analysis that a vector topology is given by such a semimetric if and only if it has a countable base of neighborhoods of the origin. Then, for an aribitrary vector topology , let be a base of balanced neighborhoods of the origin. For each i, by the properties of vector topologies, there is a sequence of balanced neighborhoods of the origin Ui,n such that and Ui,n=Ui. Again, by standard results of functional analysis, generate a vector topology which will then be generated by some translation invariant semimetric Di. The set generates as required.