0
$\begingroup$

In Baldi's book, Stochastic Calculus, Proposition 2.1 states that a right continuous adapted process, taking values in a topological space $E$ will be progressively measurable.

The proof starts out in the standard way, producing a sequence of elementary processes, $X^{(n)}$, with pointwise convergence to $X$ which are progressively measurable. It concludes that as the limit of such processes, it is also progressively measurable.

But is it true that this should hold in a topological space? Results that I am aware of on convergence of measurable functions to a measurable function are all on spaces which are at least metrizable. In fact, there appear to be counterexamples when you do not have a metric.

Does the additional structure here (right continuity? adaptedness?) allow us to circumvent having a metric?

Or am I just parsing the book incorrectly?

$\endgroup$
0
$\begingroup$

If the topological space $E$ is a separable metric space, then the Borel $\sigma$-field is generated by the open balls, and Baldi's assertion is correct.

Also, the discussion here: When are pointwise limits of measurable functions measurable? is instructive. [N.B.: By a theorem of Tikhonov, a second countable regular space is metrizable.]

$\endgroup$
  • $\begingroup$ Thanks for the link, and I agree with your assertion. The passage in Baldi, "Let us assume from now on that the state space $E$ is a topological space with its Borel $\sigma$-algebra..." Just curious if there's any other way out of this. $\endgroup$ – user2379888 Jan 21 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.