When dealing with stochastic processes, it is often important to work with versions of these processes whose paths are as regular as possible. In that direction, the Kolmogorov’s continuity theorem (see Lecture 6) provided a sufficient condition allowing to work with continuous versions of stochastic processes. For martingales, the possibility of working with regular versions, is related to the regularity properties of the filtration with respect to which the martingale property is satisfied.

**Definition:*** Let be a filtration on a probability space . If the following assumptions are fulfilled:*

- If satisfies , then every subset of is in ;
- The filtration is right continuous, that is for every

the filtered probability space

*is said to satisfy the usual conditions.
*

The above set of assumptions are called the usual conditions because, as we will see it in the next Lectures, these are the conditions under which it is convenient to work in order to properly define the stochastic integral.

Let be a filtration on a probability space and let be a (sub, super) martingale with respect to the filtration whose paths are right continuous and left limited. The filtered probability space

may canonically be enlarged into a filtered probability space

that satisfies the usual conditions. Indeed, can be taken to be the -completion of and

where is the set of events whose probability is zero. Moreover is a (sub, super) martingale with respect to the filtration (this is not straightforward and let to the reader as an exercise ). The filtered probability space

is called the usual completion of

**Exercise:** *Let be a filtered probability space that satisfies the usual conditions and let be stochastic process adapted to the filtration whose paths are left limited and right continuous. Let be a compact subset of . Show that the random time*

*is a stopping time of the filtration .
*

**Theorem (Doob’s regularization theorem):** * Let be a filtered probability space that satisfies the usual conditions and let be a supermartingale with respect to the filtration . Let us assume that the function is right continuous. There exists a modification of such that:*

- is adapted to the filtration ;
- The paths of are locally bounded, right continuous and left limited;
- is a supermartingale with respect to the filtration .

**Proof:**

As for the proof of Doob’s convergence theorem (see Lecture 9), the idea is to study the oscillations of . In what follows, we will use the notations introduced in the proof of this theorem that we remind below.

For , and , we denote

and

For , let be the greatest integer for which we can find elements of ,

such that

Let now be the set of such that exists and is finite.

It is easily seen that:

Therefore we have . We may prove, as we proved the Doob's convergence theorem that .

For , we define in the following way:

- If ,
- If ,

It is clear that the paths of are locally bounded, right continuous and left limited. Let us now show that this process is the expected modification of .

We first observe that for , the random variable is measurable with respect to . Furthermore, has a zero probability and is therefore in , according to the usual conditions. This shows that the process is adapted to the filtration .

We now show that is a modification of .

Let . We have almost surely

Let us prove that this convergence also holds in . To prove this, it is enough to check that for every decreasing family such that and that converges toward , the family is uniformly integrable.

Let . Since is assumed to be right continuous, we can find such that and such that for every ,,

For and , we have:

Now, since , we can find such that for every that satisfies , we have But for ,

From Jensen inequality, it is then seen that the process is a submartingale, and therefore

We deduce that for ,

It is thus possible to find such that for every ,

For , we have then

This implies that for every decreasing family such that and that converges toward , the family is uniformly integrable. The convergence

thus also holds in . Now, since is a supermartingale, for we have

This implies,

and

Hence, since is adapted to the filtration

Due to the fact that the function is right continuous, we have

But from the convergence, we also have

This gives

The random variable is therefore non-negative and has a zero expectation. This implies that almost surely . The stochastic process is therefore a modification of . Finally, since a modification of a supermartingale is still a supermartingale, this concludes the proof of the theorem.

The following exercise shows that martingales naturally appear when studying equivalent measures on a filtered probability space. We assume here the reader familiar with the Radon-Nikodym theorem.

**Exercise.*** Let be a filtered probability space that satisfies the usual conditions. We denote*

and for , is the restriction of to . Let be a probability measure on such that for every ,

- Show that there exists a right continuous and left limited martingale such that for every ,
- Show that the following properties are equivalent:

1) ;

2) The martingale is uniformly integrable;

3) converges in ;

4) almost surely converges to an integrable and measurable random variable such that