I'm reading Stochastic Processes by Richard F. Bass and got stuck by Exercise 1.3 for a while.
Let $\{\mathcal{F}_t\}$ be a filtration satisfying the usual conditions (complete and right-continuous) and let $\mathcal{B}[0,t]$ be the Borel $\sigma$-field on $[0,t]$. A real-valued stochastic process $X$ is progressively measurable if for each $t\ge 0$, the map $[0,t]\times \Omega\colon (s,\omega)\mapsto X(s,\omega)$ is $(\mathcal{B}[0,t]\otimes\mathcal{F}_t)$-measurable.
Show that if $X$ is adapted to $\{\mathcal{F}_t\}$ and has right-continuous paths, then $X$ is progressively measurable.
My attempt. For each $n\in\mathbb{N}$, define on $[0,\infty)\times \Omega$ a function $$Y^{(n)}(s,w)=\sum_{k=0}^\infty X((k+1)/2^n,\omega)1_{[k/2^n,(k+1)/2^n)}(s).$$
Let $t\ge 0$ be given. It is clear that the map $[0,t]\times \Omega\colon (s,\omega)\mapsto Y^{(n)}(s,\omega)$ is $(\mathcal{B}[0,t]\otimes \mathcal{F}_{t+2^{-n}})$-measurable. Let $\hat{k}(n)$ be the unique integer such that $$\frac{\hat{k}(n)}{2^n}\le s < \frac{\hat{k}(n)+1}{2^n}.$$ Then $$Y^{(n)}(s,\omega)=X((\hat{k}(n)+1)/2^n,\omega)\underset{n\to\infty}{\longrightarrow}X(s,\omega)$$ since $(\hat{k}(n)+1)/2^n\downarrow s$ by construction and $X$ has right-continuous paths. Hence, the map $[0,t]\times \Omega\colon (s,\omega)\mapsto X(s,\omega)$ is $\bigcap_{n\in\mathbb{N}}(\mathcal{B}[0,t]\otimes \mathcal{F}_{t+2^{-n}})$-measurable. The proof would be finished if I can show $$\bigcap_{n\in\mathbb{N}}(\mathcal{B}[0,t]\otimes \mathcal{F}_{t+2^{-n}})\subseteq \mathcal{B}[0,t]\otimes \mathcal{F}_{t}.$$
I am stuck here. I haven't used the usual conditions, especially the right-continuity of $\{\mathcal{F}_t\}$. Is this the right direction?
Thank you!
Best Answer
The following is a longer version of a standard argument (see e.g. Karatzas and Shreve, Proposition 1.13). Define $$Y^{(n)}(\omega,s)=\begin{cases}X_{t(k+1)2^{-n}}(\omega)& s \in (tk2^{-n},t(k+1)2^{-n}],\,k=\{0,1,...,2^n-1\}\\ X_0(\omega)&s=0\end{cases}$$ We write more explicitly $$Y^{(n)}(\omega,s)=X_0(\omega)\mathbf{1}_{\{0\}}(s)+\sum_{1\leq k<2^n}X_{t(k+1)2^{-n}}(\omega)\mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s)$$ (i) To see that this is $\mathscr{F}_t\otimes \mathscr{B}[0,t]$-measurable, first note that all $\omega \mapsto X_{t(k+1)2^{-n}}(\omega)$ are $\mathscr{F}_t$-measurable functions, since $X$ is adapted, and $s \mapsto \mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s)$ are $\mathscr{B}[0,t]$-measurable functions. Also for any $A,B \in \mathscr{B}(\mathbb{R})$: $$\begin{aligned}\{(\omega,s):X_{t(k+1)2^{-n}}(\omega)\in A\}&=\{\omega: X_{t(k+1)2^{-n}}(\omega)\in A\}\times [0,t]\in \mathscr{F}_t\otimes \mathscr{B}[0,t]\\ \{(\omega,s):\mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s)\in B\}&=\Omega \times \{s:\mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s) \in B\}\in \mathscr{F}_t\otimes \mathscr{B}[0,t]\end{aligned}$$ So the functions $(\omega,s)\mapsto Y^{(n)}(\omega, s)$ are linear combinations of $\mathscr{F}_t\otimes \mathscr{B}[0,t]$-measurable functions, and are thus measurable.
(ii) The sequence $Y^{(n)}(s,\omega)$ approaches $X_s(\omega)$ from the right and the right continuity guarantees that the limit is $X_s(\omega)$. The limit of a sequence of measurable functions is measurable whenever it exists; we then have that $X_s(\omega)$ is also $\mathscr{F}_t\otimes \mathscr{B}[0,t]$-measurable. But this means that $X_s(\omega)$ is $\mathscr{F}_t$-progressively measurable.