6
$\begingroup$

I'm beginning to study about stochastic processes, and currently focusing on stopping times and hitting times. The textbook I'm using is "Stochastic Integration Theory" by Medvegyev (and Karatzas & Shreve as a second reference), and in some of the theorems the following measurable projection theorem is used.

If the space $(\Omega,\mathcal{A},\mathbb{P})$ is complete and $$U \in \mathcal{B}(\mathbb{R}^n) \otimes \mathcal{A},$$ then $$\text{proj}_\Omega (U) := \{x: \exists t\text{ such that }(t,x) \in U\} \in \mathcal{A}.$$

On the authors homepage there is a note containing a proof as well as many definitions such as Suslin (also called analytic) sets and auxiliary lemmas, however I find the material to be lacking in rigor and it is missing some assumptions. Therefore I am looking for a textbook in which the measurable projection is covered in detail. I've looked at the textbooks by Kechris, and Srivastava without finding what I was looking for.

$\endgroup$
4
  • $\begingroup$ Dellacherie and Meyer, Probabilities and Potential, Chapter III. $\endgroup$
    – zhoraster
    Jan 8, 2016 at 7:24
  • $\begingroup$ I didn't find this in Chapter III, but should certainly be inside this book. All such facts are there. $\endgroup$
    – zhoraster
    Jan 8, 2016 at 7:40
  • $\begingroup$ Yeah I just looked at Chapter III, but to be honest, the notation and organization is a little messy so it's hard to find the actual theorem. $\endgroup$
    – Olorun
    Jan 8, 2016 at 7:54
  • $\begingroup$ @Olorun Agreed with not best place... Anyways,,, suggestion. Take it or leave it. I had got the same helpful hint on Math.SE and Chemistry.SE both, so thought of passing it on... $\endgroup$
    – Shailesh
    Jan 13, 2016 at 4:40

1 Answer 1

4
$\begingroup$

Theorem 13 in Chapter III of the first volume of Dellacherie & Meyer (cited by @zhoraster; see the foot of page 43 in the English translation) tells you that the projection onto $\Omega$ of $U$ is $\mathcal A$-analytic. As such, this projection is $\mathcal A$-measurable, because $(\Omega,\mathcal A,\Bbb P)$ is complete; see no. III-33 at the top of p. 58 of D. & M.

$\endgroup$

You must log in to answer this question.

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