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Let $b: [0,T] \times \Omega \times \mathbb{R} \rightarrow \mathbb{R}$ with the properties:

  1. For all $ x \in \mathbb{R}$ the process $(t,\omega) \mapsto b(t,\omega, x)$ is progressively measurable.

  2. There exits $ C > 0 $ such that for all $\omega, t, x_1, x_2 $ we have $$ | b(t,\omega, x_1) - b(t,\omega, x_2) | \leq C | x_1 - x_2 |. $$

Let $X = (X_t)_{t \in [0,T]}$ be progressively measurable. I want to show that the process $$ (t, \omega) \mapsto b(t,\omega, X_t(\omega) ) $$ is progressively measurable. This should follow from 1. and 2. but I was not able to prove it. I hope someone can help me out.

Edit: I actually need a more general result. Let $\mathcal{P}_2(\mathbb{R})$ be the space of measures on $\mathbb{R}$ with finite second moment. Equip this space with the Wasserstein metric $W^2.$ Then, $\mathcal{P}_2(\mathbb{R})$ is a Polish space.

Let $b: [0,T] \times \Omega \times \mathbb{R} \rightarrow \mathbb{R}$ with the properties:

  1. For all $ x \in \mathbb{R}, \mu \in \mathcal{P}_2(\mathbb{R})$ the process $(t,\omega) \mapsto b(t,\omega, x, \mu)$ is progressively measurable.

  2. There exits $ C > 0 $ such that for all $\omega, t, x_1, x_2 , \mu_1, \mu_2$ we have $$ | b(t,\omega, x_1, \mu_1) - b(t,\omega, x_2, \mu_2) | \leq C \big( | x_1 - x_2 | + W^2(\mu_1, \mu_2) \big). $$

Let $X = (X_t)_{t \in [0,T]}$ be progressively measurable. Let $P(X_t)$ be the push forward measure of $X_t$. I want to show that the process $$ (t, \omega) \mapsto b(t,\omega, X_t(\omega), P(X_t) ) $$ is progressively measurable.

I tried to imitate the proof of the easier case. However, the problem seems to be that I cannot split $\mathcal{P}_2(\mathbb{R})$ into disjoint sets of "small radius". Do you have an idea how to deal with this?

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  • $\begingroup$ may I know whether $t\to P(X_t)$ is Borel measurable under the above condition? $\endgroup$
    – John
    Jul 30, 2020 at 22:37

1 Answer 1

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Consider $b_n(t,\omega,x):=b(t,\omega,k/n)$ for $x\in[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,\ldots$. The corresponding process $$ X^{(n)}_t(\omega):=b_n(t,\omega,X_t(\omega))=\sum_k b(t,\omega,k/n)1_{[k/n,(k+1)/n)}(X_t(\omega)) $$ is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence: $$ \lim_n\sup_{0\le t\le T}|X^{(n)}_t(\omega)-X_t(\omega)|=0. $$ It follows that $X$ is progressive as well.

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  • $\begingroup$ Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, \omega).$ Is this correct? $\endgroup$
    – White
    Aug 7, 2018 at 9:32
  • $\begingroup$ No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see. $\endgroup$ Aug 7, 2018 at 13:20
  • $\begingroup$ Thanks a lot, I posted a more general version. Maybe you could also help me out with this one. $\endgroup$
    – White
    Aug 7, 2018 at 14:16
  • $\begingroup$ For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof. $\endgroup$
    – White
    Aug 7, 2018 at 15:29

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