# Is every predictable process a pointwise limit of left-continuous, adapted processes?

Define the predictable $$\sigma$$-algebra as $$\mathcal P := \sigma(X: \text{ X is a left-continuous and adapted process }),$$ and say that a stochastic process is predictable if it is measurable w.r.t. $$\mathcal P$$. [The stochastic processes are assumed to take values in some Polish space $$E$$ and be indexed by $$[0,\infty)$$]

As measurablility is preserved under pointwise limits, a pointwise limit of left-continuous and adapted processes will be predictable.

Is it true that any predictable process is a pointwise limit (or even a.s. pointwise limit if we assume our probability space is complete) of a sequence of left-continuous and adapted processes.?

More generally, is it always true that if we define $$\mathcal F := \sigma(f: \text{ f with some property })$$, then every $$\mathcal F$$-measurable function is the limit of a sequence of functions with that same property?

• Regarding your question on predictable processes as pointwise limits: Yes, I think so. Take a look at Theorem 1.2.24 in "Martingales and Stochastic Analysis" by J. Yeh
– saz
Jun 11, 2020 at 14:21
• @saz The Theorem I.2.24 in Yeh's book is not convinced, because its proof depends on the Proposition I.1.12 and there is a bug in the proof of Prop. I.1.12.(How to verify "the V statisfies condition $2^\circ$") Jun 17, 2020 at 2:41

Any pointwise limit of left-continuous adapted processes would have sample paths $$t\mapsto X_t(\omega)$$ that were, at worst, in the fourth Baire class. Thus a deterministic process $$X_t(\omega)=f(t)$$, for a Borel function $$f$$ of the fifth Baire class, is not the pointwise limit of the type you request.

The good news is that any argument you might be trying to make based on such a pointwise approximation "fact" can likely be made through the use of the functional monotone class theorem.

• Could you please expand on why they can be at worst Baire class 4? Is it something that holds true for all left-continuous functions? Jan 29, 2022 at 2:26
• Yes. Think of approximating a left-continuous function by left-continuous step functions. The "4" we erring on the safe side. Jan 29, 2022 at 18:00
• A left-continuous step function (ie the indicator function of some semiopen interval) is Baire 1. A left-continuous function should therefore be Baire 2, and limits of left-continuous Baire 3. Why did you say "4"? Jan 29, 2022 at 22:19
• I mean, I understand that a deterministic Baire 5 function works as a counterexample to the question of the OP, but so does a function in, say, Baire 1000. My question is about optimality. Jan 29, 2022 at 22:24
• @Titti Not sure why 4 specifically — this was written 18 months ago. Guess: I took 2 and doubled it to be safe.. Jan 30, 2022 at 17:06

I don't think the generalization is true because the property could be too narrow. For example, consider functions from $$[0,1]$$ to $$[0,1]$$ equipped with the Borel sigma algebra. If we let $$\mathcal{F} := \sigma(f : f(x) = x \text{ for all x \in [0,1]})$$ then $$\mathcal{F}$$ is the Borel sigma algebra on $$[0,1]$$, but the only function that can be obtained by limits of functions with that property is the identity.

• Great counterexample! Thanks! Jun 11, 2020 at 15:23