# More natural example of measurable but not progressive process

All examples of measurable but not progressive processes I have ever seen seemed to be based on the huge difference between $$\mathcal{F}$$ and $$\mathcal{F}_\infty$$. Here is what I mean.

Consider Example 19.5 from Counterexamples in Probability by Stoyanov: let $$X_t ( \omega ) = 1_{t = \omega}$$ be a process relative to $$(\mathbb{R}_+, \mathcal{F}, \mathcal{F}_t)$$, where $$\mathcal{F} = \mathcal{B} ( \mathbb{R}_+ )$$, whereas $$\mathcal{F}_t$$ does not depend on $$t$$ and is generated by points of $$\mathbb{R}^d$$ (that is, $$A \in \mathcal{F}_t$$ iff $$A$$ or $$A^c$$ is countable). Then it is easy to see that

1. $$X^{-1} ( \{ 1 \} ) = \{ ( t, \omega ) \ \colon \ t = \omega \in \mathbb{R}_+ \}$$ belongs to $$\mathcal{B}(\mathbb{R}_+) \otimes \mathcal{F}$$, which means that $$X$$ is measurable,
2. $$X_t$$ is $$\mathcal{F}_t$$-adapted,
3. $$X^{-1} ( \{ 1 \} ) \cap [0, t] \times \Omega$$ does not belong to $$\mathcal{B}( [0, t] ) \otimes \mathcal{F}_t$$, which means that $$X$$ is not progressive.

Essentially the same example one can find in Capacités et processus stochastiques by Claude Dellacherie (page 47): or a slight variation thereof on p.47 here (Example 1.38). In the latter $$\mathcal{F}$$ is the Lebesgue sigma-algebra on $$[0, 1]$$ and $$\mathcal{F}_t = \mathcal{L}_0$$ is generated by the null-sets.

Both examples seem somewhat artificial, because they only work because $$\mathcal{F}$$ is much bigger and most importantly has little to do with $$\mathcal{F}_t$$.

My question is: are there examples of measurable but not progressive processes with respect to $$\mathcal{F} = \mathcal{F}_\infty$$?

• In your example, why can't you let $\mathcal{F}:=\mathcal{F}_1$? Why do you need for $X^{-1}(\{1\})$ to belong to $\mathcal{B}(\mathbb{R}_+)\otimes\mathcal{F}$? Commented Jan 2, 2023 at 14:47
• @IosifPinelis, in all three examples I cite $\mathcal{F}_t$ is constant in $t$. $X^{-1} ( \{ 1 \} ) \in \mathcal{B} ( \mathbb{R}_+ ) \otimes \mathcal{F}$ means measurability of the process. I am interested in the gap between measurability and progressive measurability. Commented Jan 2, 2023 at 14:52
• I think measurable processes are those with measurable paths. Your definition seems to be of jointly measurable processes. Commented Jan 2, 2023 at 15:22
• @IosifPinelis, I think you are wrong here. See Definition 1.14 in Revuz & Yor, Definition 1.6 in Karatzas & Shreve (1991) or right below D45 in Meyer's Probabilities and Potential. The exact same definition in Sharpe (1988), Rao (1995), Chung & Walsh (2005) and others. As a matter of fact, I've never seen such definitions as you say. Do you mean $\mathbb{P} \{ t \mapsto X_t \ \text{is measurable} \} = 1$? Commented Jan 2, 2023 at 18:01
• See e.g. Definition 3.9 at uregina.ca/~kozdron/Teaching/Regina/862Winter06/Handouts/… . I had certainly seen such a definition elsewhere as well. See also the use of the term "jointly measurable" in the first paragraph at individual.utoronto.ca/jordanbell/notes/… . Anyhow, in the example in my answer below, the process is jointly measurable, that is, measurable in your sense. Commented Jan 2, 2023 at 18:10

An example due to G. Letta in "Un exemple de processus mesurable adapté non progressif" [Séminaire de Probabilités XXII, 449–453, Lecture Notes in Math. 1321, 1988] is illuminating.

Take $$\Omega$$ to be the space of lower semi-continuous functions $$\omega:[0,\infty)\to\Bbb R$$, such that $$\int_0^t|\omega(s)|\,ds<\infty$$ for each $$t>0$$. Let $$H_s(\omega):=\omega(s)$$ be the coordinate process on $$\Omega$$, and $$(\mathcal F_t)_{t\ge0}$$ the filtration generated by $$(H_t)_{t\ge 0}$$. Let $$\mathcal J$$ denote the collection of compact intervals $$J\subset[0,\infty)$$ and for $$J\in\mathcal J$$ define $$H_J(\omega):=\inf\{H_s(\omega): s\in J\}$$. Finally, define $$\mathcal G:=\sigma\{H_J: J\in\mathcal J\}$$. Then for $$c\in\Bbb R$$, because $$s\mapsto H_s(\omega)$$ is lower semi-continuous, $$\{(s,\omega)\in[0,\infty)\times\Omega: H_s(\omega)>c\}=\cup_{J\in\mathcal J_0}\left(J\times\{H_J>c\}\right),$$ where $$\mathcal J_0$$ is the subcollection of $$\mathcal J$$ comprising those intervals with rational endpoints. It follows that $$H$$ is $$\mathcal B[0,\infty)\otimes\mathcal G$$ measurable, and clearly $$\mathcal F_t\subset\mathcal G$$ for each $$t>0$$.

Consider now, for $$t>0$$ fixed, the integral $$X_t:=\int_0^t H_s\,ds$$. If $$H$$ were $$(\mathcal F_t)$$-progressive then $$X_t$$ would be $$\mathcal F_t$$ measurable. In this case there would be a countable set $$S\subset(0,t]$$ with $$X_t$$ measurable over $$\sigma\{H_s:s\in S\}$$. If $$\omega$$ and $$\omega'$$ were distinct elements of $$\Omega$$ with $$\omega(s)=\omega'(s)$$ for all $$s\in S$$, then we would clearly have $$X_t(\omega)=X_t(\omega')$$. The following choice of $$\omega$$ and $$\omega'$$ leads to a contradiction: $$\omega(s)=1$$ for all $$s>0$$; $$\omega'$$ the indicator of an open set $$U\supset S$$ of Lebesgue measure strictly less than $$t$$. Thus, $$H$$ is not $$(\mathcal F_t)$$-progressive.

On the other hand, if we define the $$\sigma$$-field $$\mathcal G_t$$ as we did $$\mathcal G$$ except that the intervals $$J$$ must be contained in $$[0,t]$$, then $$(\mathcal G_t)_{t\ge 0}$$ is a filtration, $$\mathcal F_t\subset\mathcal G_t$$ for each $$t\ge 0$$, and $$\vee_t\mathcal G_t=\mathcal G$$. Arguing as before, using lower semi-continuity, it can be shown that the restriction of $$H$$ to $$[0,t]\times\Omega$$ is $$\mathcal B[0,t]\otimes\mathcal G_t$$ measurable, for each $$t\ge 0$$. That is, $$H$$ is $$(\mathcal G_t)$$-progressive.

As is the other examples of the phenomena, a "gap" between the filtration generated by $$H$$ and a $$\sigma$$-field $$\mathcal G$$ for which $$H$$ is $$\mathcal B[0,\infty)\otimes\mathcal G$$ measurable seems to be crucial.

• I actually like this example a lot! It shows a gap between the natural filtration and the smallest filtration with respect to which the process is progressive in somewhat more natural (generic) setup. Commented Jan 6, 2023 at 15:00
• Is it possible to give a construction of the set $S$? Commented Sep 2, 2023 at 10:16

$$\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}\newcommand\R{\mathbb R}\newcommand\om\omega\newcommand\Om\Omega$$Understanding your question:

are there examples of measurable but not progressive processes with respect to $$\mathcal{F} = \mathcal{F}_\infty$$?

literally, such an example can be obtained by the following slight modification of the example in your post:

Let $$X(t,\om):= 1(t=\om)$$ for $$(t,\om)\in\R_+\times\Om$$, $$\Om:=\R_+$$, and $$\F:=\B(\R_+)$$. For each $$t\in[0,1]$$, let $$\F_t$$ be the $$\sigma$$-algebra generated by the singleton subsets of $$\R_+$$ (or by the singleton subsets of $$[0,t]$$); for each $$t\in(1,\infty)$$, let $$\F_t:=\F$$.

Then $$\F_\infty=\F$$, the function $$X\colon\R_+\times\Om\to\R$$ is $$\B(\R_+)\otimes\F$$-measurable, and $$(X(t,\cdot))$$ is $$(\F_t)$$-adapted but not progressive.

In general, it seems that the filtration $$(\F_t)$$ has to be discontinuous in some sense for such examples to exist.

• Okay, right, I see the trick! Well, I got what I asked for :) What I wanted is to understand whether $\mathcal{F}_t$ indeed has to be this discontinuous. Commented Jan 2, 2023 at 18:09

It seems that I have found an example for my own question, although I have some minor doubts. The following explanation is somewhat handwavy.

Let us take the same process $$X_t ( \omega ) = 1_{t = \omega}$$ on $$\Omega = T = \mathbb{R}_+$$ and let the "continuous" part of the filtration lag in time. Take some continuous function $$\varphi ( t )$$ satisfying

1. $$\varphi(0) = 0$$
2. $$0 < \varphi ( t ) < t$$ (strictly) for all $$t > 0$$
3. $$\varphi ( t )$$ monotonically increases to $$\infty$$

and define the filtration as follows: $$\mathcal{F}_t = \mathcal{G}_{\varphi(t)} \vee \mathcal{H}_t,$$ where $$\mathcal{G}_t = \mathcal{B} [0, t]$$ and $$\mathcal{H}_t$$ is the natural filtration of $$X_t$$, that is, the sigma-algebra generated by singletons of $$[0, t]$$.

The function $$\varepsilon ( t ) = t - \varphi ( t )$$ can then be interpreted as the time lag of $$\mathcal{G}_t$$ at time $$t$$.

One possible interpretation would be that we receive the detailed continuous information about the process with some time lag due to imperfections of our measuring equipment, whereas crude single point information is available right away.

Then, if I am not mistaken, $$X_t$$ is $$\mathcal{F}_t$$-adapted and (jointly) measurable with respect to $$\mathcal{F} = \mathcal{F}_\infty$$.

It isn't progressive, because $$\mathcal{F}_t$$ lacks a tiny piece of continuous information.

I thought of modifying this example a bit further. Take any process $$X_t$$ and consider two different filtrations:

1. its natural filtration $$\mathcal{H}_t$$, with respect to which I assume $$X$$ to be not progressive,
2. its enlargement $$\mathcal{G}_t$$, with which $$X$$ is progressive.

Then, take a function $$\varphi ( t )$$ as described above and define $$\mathcal{F}_t = \mathcal{G}_{\varphi ( t )} \vee \mathcal{H}_t.$$

The definition means that the information which makes $$X$$ progressive arrives with a certain lag. Since it ultimately does, the process is measurable, but at a given $$t$$ we do not have enough information about it.

I didn't think the example through thoroughly, but the idea makes sense to me. Please tell me if something's wrong.

Continuing the handwavy intuitive description of the example, let's say the information about the process comes from two sensors: one is rough but quick (information is available right now), another is detailed but slow (lag in time). The first makes the process adapted. Using the information from the second, we'll ultimately obtain the full and detailed picture.

P.S. My "lagging filtration" example seems to be in a sense a refutation of the following remark in Dellacherie & Meyer, Volume A, page 141:

• Concerning your modification, if $\mathcal{H}_t$ is generated by all finite subfamilies of $(X_s)_{s \in [0, t]}$, then $\mathcal{H}_t$ is the same as $\sigma$-algebra as the one generated by all $X_s$ with $s \in [0, t]$. Your first example seems fine to me. Commented Jan 2, 2023 at 20:21
• @IosifPinelis, shouldn't it only work if $X$ has some continuity (right-continuity or at least separability for instance)? Commented Jan 2, 2023 at 20:56
• @IosifPinelis, I mean, isn't the sigma-algebra generated by singletons in the first example of exactly the same nature? Commented Jan 2, 2023 at 20:59
• No, not of the same nature. Commented Jan 2, 2023 at 21:00
• @IosifPinelis, I'm clearly missing something... A singleton in the first example is a preimage of $X_t$, right? What is the difference then? Commented Jan 2, 2023 at 21:03