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?

  • 1
    $\begingroup$ 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 $\endgroup$
    – saz
    Jun 11, 2020 at 14:21
  • $\begingroup$ @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 $2^\circ$") $\endgroup$
    – JGWang
    Jun 17, 2020 at 2:41

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – No-one
    Jan 29, 2022 at 2:26
  • $\begingroup$ Yes. Think of approximating a left-continuous function by left-continuous step functions. The "4" we erring on the safe side. $\endgroup$ Jan 29, 2022 at 18:00
  • $\begingroup$ 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"? $\endgroup$
    – No-one
    Jan 29, 2022 at 22:19
  • $\begingroup$ 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. $\endgroup$
    – No-one
    Jan 29, 2022 at 22:24
  • $\begingroup$ @Titti Not sure why 4 specifically — this was written 18 months ago. Guess: I took 2 and doubled it to be safe.. $\endgroup$ 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.

  • $\begingroup$ Great counterexample! Thanks! $\endgroup$
    – MrFranzén
    Jun 11, 2020 at 15:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .