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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?

user2379888
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1 Answers1

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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.]

John Dawkins
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  • 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. – user2379888 Jan 21 '19 at 02:34