I am studying strong Markov property and the following question came up.

Let $ Y_{ s}( \omega)$ be a bounded jointly measurable stochastic process and $ X_{ s}( \omega) $ be a standard Brownian motion for $ s \geq 0$, $\omega \in \Omega$ . Let $ P^ { x}$ be the probability associated with the Browman motion starting at $ x \in \mathbb R$ and $ E^{ x}$ be the corresponding expectation. Finally, let $\tau$ be a stopping time and $\mathcal F_{\tau}$ be the associated sigma-algebra.

I want to show:

$$ \phi (X _{\tau},\tau) \ \text{ is} \ \mathcal F_{\tau } \ \text{ measurable on }\{\tau < \infty \}\ \text{ where } \ \phi ( y, s):= E ^{ y} [ Y_{ s}]$$ for $ s \geq 0$ and $y \in \mathbb R$.

I know that $\tau$ and $X_{\tau} 1_{\tau < \infty}$ are $\mathcal F_{\tau}$ measurable and want to use this fact, but I cannot proceed on.

Any help is appreciated.

  • $\begingroup$ When you say that $Y$ is (jointly) measurable process, is it with respect to the filtration $\mathcal{F}$ ? $\endgroup$
    – TheBridge
    Sep 5, 2019 at 9:00
  • $\begingroup$ @TheBridge Yes, the big one $\endgroup$
    – Focus
    Sep 5, 2019 at 9:13
  • $\begingroup$ $\phi(X_{\tau},\tau) = E^{X_{\tau}}[Y_{\tau}]$ is the composition of $y\mapsto E^yY_{\tau}$ (which is non-random, so its measurability is not an issue) and $\omega \mapsto X_{\tau}(\omega)$. So you only have to show that $X_{\tau}$ is $\mathcal{F}_{\tau}$ measurable. $\endgroup$
    – Sayantan
    Sep 5, 2019 at 10:54
  • $\begingroup$ @Sayantan We are not considering such composition; we take expectaion of each section of $Y$ and then insert randomness again in two different places. I edited question to be more clear. $\endgroup$
    – Focus
    Sep 5, 2019 at 13:53

1 Answer 1


I came up with the following solution.

It suffices to show $\phi$ is jointly Borel m'ble.

Let a random variable $Y$ be called special if $Y(\omega)=\prod_{i=1} ^m f_i(\omega(t_i))$ for some $f_i \in C_b(\mathbb R)$ and $t_i \geq 0$. In other words, $Y=\prod _{i=1} ^m f_i \circ \pi_{t_i}$ where $\pi_{t_i}$ is projection onto time $t_i$ i.e., $\pi_{t_i} (\omega) := \omega(t_i)$.

$B \subseteq \Omega$ is a finite dimensional set if $B=\bigcap_{i=1}^m \pi_{t_i}^{-1}(A_i)$ for some $t_i \geq 0$ and open sets $A_i \subseteq \mathbb R$.


Consider the case $Y_s(\omega)=f(s)Y(\omega)$ for some $f \in C_b(\mathbb R)$ and $Y$ special.

Using DCT, it can be shown that $y \mapsto E^y(Y)$ is continuous since $Y$ is special .

Then $\phi(y,t)=f(s)E^y(Y)$ is continuous, hence measurable.


Consider the case $Y_s(\omega)= 1 _U Y$ for some open set $U \subset \mathbb R$.

It is a fact that $ 1_U$ is a monotone pointwise limit of continuous bounded functions. (generally holds in LCH spaces using Urysohn lemma, but can be explicitly constructed in metric spaces)

Then, using STEP1, $\phi(y,t)=1_U E^y(Y)$ is pointwise limit of measurable functions, hence measurable.


Consider the case $Y_s(\omega)=1_U(s) 1_B(\omega)$ for some open $U\subseteq \mathbb R$ and finite dimensional $B=\bigcap_{i=1}^m \pi_{t_i}^{-1}(A_i)$.

Claim: $1_B$ is monotone pointwise limit of some special random variables.

proof. \begin{align} 1_B &=1_{\bigcap_{i=1}^m \pi_{t_i}^{-1}(A_i)} = \prod_{i=1}^m 1_{A_i} \circ \pi_{t_i} \\&= \prod_{i=1}^m \lim_{n \to \infty} f_n^{(i)} \circ \pi_{t_i} = \lim_{n \to \infty} \prod_{i=1}^m f_n^{(i)} \circ \pi_{t_i} \end{align} for some $f_n^{(i)} \in C_b(\mathbb R)$ such that $f_n^{(i)} \uparrow 1_{A_i} $ as $n \to \infty$ for each $i$. ////

Then, using STEP2 and MCT, $\phi(y,t)=1_U(t) E^y 1_B$ is monotone pointwise limit of measurable functions, hence measurable.


Now, we can use the monotone class theorem. Note that $\mathcal P = \{U\times B: U \text{ open in } \mathbb R , B \text{ is finite dimensional}\}$ is a $\pi$-system generating the product $\sigma$-algebra on $\mathbb R \times \Omega$. Let $$\mathcal H:=\{Y_{s}(\omega): Y_{s}(\omega) \text{ is bounded jointly measurable and } \phi(y,s) = E^y(Y_s) \text{ is jointly Borel measurable}\}$$ Firstly, constant $1 \in \mathcal H$ is obvious. Secondly, $U\times B \in \mathcal P \implies 1_{U\times B} \in \mathcal H$ follows from STEP3. Lastly, $0\leq Y_n \uparrow Y $ for some $Y_n \in \mathcal H$ and $Y$ bounded implies $Y\in \mathcal H$ by MCT and the fact that pointwise limit of measurable is measurable.

Thus, by monotone class theorem, $\mathcal H$ contain all bounded $\sigma(\mathcal P)$ measurable function, which yields the result.


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