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I am not sure in this proof how the monotone class theorem is applied.

I am familiar with the following version of the monotone class theorem. But I cannot tell what becomes $\mathscr{C}$ and $\mathscr{H}$ in this proof. So here $\mathscr{O}$ is the $\sigma$-algebra generated by all cadlag adapted processes. It seems like saying that the set of all processes that are $\mathscr{F} \otimes \mathbb{R}_+$ - measurable and meet (a) and (b) for all stopping times is obviously a vector lattice and is stable under pointwise convergence suggest that this set must be $\mathscr{H}$ in the monotone class theorem below. And the natural candidate for $\mathscr{C}$ would be the set of all $\{X \in B\}$ for all cadlag adapted processes $X$ and Borel sets $B$. But then I cannot show that this set is a $\pi$-class. How is the intersection also of this form? And I cannot see why iii) would be satisfied as well. Finally, in the notes here, we assume that processes take values in $\mathbb{R}^d$. So how can we extend this result on real processes to higher dimensions? I would greatly appreciate any help.

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    $\begingroup$ @TomookiYuasa I can't find an appropriate type for this argument. $\endgroup$ – nomadicmathematician Feb 15 at 18:19
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    $\begingroup$ @TomookiYuasa To apply the theorem you referred to, we need to let $\Phi$ be the set of all processes that are $\mathcal{F} \otimes \mathbb{R}_+$ measurable and meet (a) and (b) for all stopping times. Then, it satisfies (i) and (ii). But the conclusion gives that $\Phi$ contains all bounded $\mathcal{F} \otimes \mathbb{R}_+$ measurable processes. And this is not the conclusion we desire. $\endgroup$ – nomadicmathematician Feb 15 at 18:33
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    $\begingroup$ @TomookiYuasa Sorry I don't follow your argument. So we need to show that optional processes are both well measurable and satisfy (a) and (b) for all stopping times. The argument the book suggests is show that all cadlag adapted processes satisfy these conditions, then extend it to all optional processes by monotone class argument. I can't see how the type of monotone class theorem you referred to applies here. $\endgroup$ – nomadicmathematician Feb 15 at 18:59
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Lemma (Monotone class theorem for optional processes). Let ($d=1$) and $\Phi$ be a linear space of one-dimensional bounded ${\mathcal F} \otimes {\mathcal B}({\mathbb R}_{+})$-measurable processes satisfying the following two conditions:

(i) $\Phi$ contains all bounded, cadlag adapted processes;

(ii) if $\{\phi_{n}\}_{n \in {\mathbb N}}$ is a monotone increasing sequence of processes in $\Phi$ such that $\phi=\sup_{n \in {\mathbb N}}\phi_{n}$ is bounded, then $\phi \in \Phi$.

Then $\Phi$ contains all one-dimensional bounded optimal processes.

Proof of Lemma.

Step1. Define $$ {\mathcal O}':=\{B;B \subset \Omega \times {\mathbb R}_{+} \text{ such that } {\bf 1}_{B} \in \Phi\}. $$

Then ${\mathcal O}'$ satisfies the following properties:

(1) $\Omega \times {\mathbb R}_{+} \in {\mathcal O}'$;

(2) $A,B \in {\mathcal O}'$ with $A \subset B \Rightarrow B \setminus A \in {\mathcal O}'$;

(3) $\{A_{n}\}_{n \in {\mathbb N}} \subset {\mathcal O}'$ with $A_{n} \subset A_{n+1}$, $n \in {\mathbb N}$ $\Rightarrow \cup_{n}A_{n} \in {\mathcal O}'$.

Indeed, (1) holds by (i), (2) holds by (i) and the linearity of $\Phi$, and (3) holds by (ii). Thus ${\mathcal O}'$ is a d-system on $\Omega \times {\mathbb R}_{+}$ (cf. page 193, Williams (1991)).

Step2. Let $k \in {\mathbb N}$, $\{Y_{i}\}_{i=1}^{k}$ be cadlag adapted processes and $\{E_{i}\}_{i=1}^{k}$ be open sets in ${\mathbb R}^{1}$. Then $\cap_{i=1}^{k}Y_{i}^{-1}(E_{i}) \in {\mathcal O}'$. Indeed, let $i \in \{1,2,\ldots,k\}$ and define \begin{align*} \phi(x):= \left\{ \begin{array}{lL} 1, & x \leq 0 \\ 1-x, & 0 \leq x \leq 1 \\ 0, & 1 \leq x \end{array} \right., \quad \phi_{n}^{i}(x):=1-\phi(n\rho(x,E_{i}^{c})), \quad x \in {\mathbb R}^{1}, n \in {\mathbb N}, \end{align*} where $\rho(x,E_{i}^{c})=\inf\{|y-x|; y \in E_{i}^{c}\}$. Then $\{\phi_{n}^{i}\}_{n \in {\mathbb N}}$ is a sequence of real and bounded continuous functions such that $$ \uparrow\lim_{n \uparrow \infty}\phi_{n}^{i}(x)={\bf 1}_{E_{i}}(x), \quad x \in {\mathbb R}^{1}. $$ Thus for any $(\omega,t) \in \Omega \times {\mathbb R}_{+}$, we obtain $$ \uparrow\lim_{n \uparrow \infty}\prod_{i=1}^{k}\phi_{n}^{i}(Y_{i}(\omega,t)) =\prod_{i=1}^{k}{\bf 1}_{E_{i}}(Y_{i}(\omega,t)) =\prod_{i=1}^{k}{\bf 1}_{Y_{i}^{-1}(E_{i})}(\omega,t) ={\bf 1}_{\cap_{i=1}^{k}Y_{i}^{-1}(E_{i})}(\omega,t). $$ Hence ${\bf 1}_{\cap_{i=1}^{k}Y_{i}^{-1}(E_{i})} \in \Phi$ by (ii) since $\prod_{i=1}^{k}\phi_{n}^{i}(Y_{i}) \in \Phi$. This implies that $\cap_{i=1}^{k}Y_{i}^{-1}(E_{i}) \in {\mathcal O'}$.

Step3. Define \begin{align*} {\mathcal I}&:=\left\{\cap_{i=1}^{k}Y_{i}^{-1}(E_{i}); \{Y_{i}\}_{i=1}^{k} \text{ is cadlag adapted processes},\right. \\ &\hspace{3.4cm}\left.\{E_{i}\}_{i=1}^{k} \text{ is open sets in } {\mathbb R}^{1}, k \in {\mathbb N}\right\}. \end{align*} Then ${\mathcal I}$ is a $\pi$-system on $\Omega \times {\mathbb R}_{+}$ (i.e., $A,B \in {\mathcal I} \Rightarrow A \cap B \in {\mathcal I}$) and $\sigma({\mathcal I})={\mathcal O}$ (i.e., ${\mathcal O}$ is generated by ${\mathcal I}$). Indeed, $\sigma({\mathcal I}) \subset {\mathcal O}$ is obvious since ${\mathcal I} \subset {\mathcal O}$. Next, let $Y$ ba a cadlag adapted process and define $$ {\mathcal A}:=\left\{E \in {\mathcal B}({\mathbb R}^{1}); Y^{-1}(E) \in \sigma({\mathcal I})\right\}. $$ Then ${\mathcal A}={\mathcal B}({\mathbb R}^{1})$ since ${\mathcal A}$ is a $\sigma$-field ($\sigma$-algebra) on ${\mathbb R}^{1}$. Thus for any an open set $E$ in ${\mathbb R}^{1}$, $Y^{-1}(E) \in \sigma({\mathcal I})$ since $E \in {\mathcal B}({\mathbb R}^{1})={\mathcal A}$. This implies that $\sigma({\mathcal I}) \supset {\mathcal O}$.

Step4. ${\mathcal I} \subset {\mathcal O}'$ by Step2. Thus $d(I) \subset {\mathcal O}'$ by Step1. Hence ${\mathcal O}=\sigma({\mathcal I})=d({\mathcal I}) \subset {\mathcal O}'$ by Step3 (cf. page 193, Williams (1991)).

Step5. For any $A \in {\mathcal O}$, ${\bf 1}_{A} \in \Phi$ since $A \in {\mathcal O}'$ by Step4. This implies that for any a one-dimensional bounded optimal process is an element in $\Phi$ by the standard-machine argument (cf. page 56, Williams (1991)).

Thus this Lemma had been proved by Step5.

Proof of Proposition 1.21.

Step6. Set \begin{align*} \Phi&:=\left\{\phi; \phi \text{ is a one-dimensional bounded}\right. \\ &\hspace{1.4cm}\left.{\mathcal F} \otimes {\mathcal B}({\mathbb R}_{+})\text{-measurable process which satisfies (a) and (b).}\right\}. \end{align*} Then $\Phi$ is a linear space which satisfies the conditions (i) and (ii) in Lemma. Thus $\Phi$ contains all one-dimensional bounded optimal processes by Lemma. Indeed, let $\{\phi_{n}\}_{n \in {\mathbb N}}$ be a monotone increasing sequence of process in $\Phi$.

(a) For any $\omega \in \Omega$, $$ \phi(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)=\sup_{n \in {\mathbb N}}\phi_{n}(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)=\limsup_{n \uparrow \infty}\phi_{n}(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega). $$ Thus $\omega \mapsto \phi(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)$ is ${\mathcal F}_{T}$-measurable since $\omega \mapsto \phi_{n}(\omega,T(\omega))$ is ${\mathcal F}_{T}$-measurable (cf. page 31, Williams (1991)).

(b) For any $(\omega,t) \in \Omega \times {\mathbb R}_{+}$, $$ \phi(\omega,T(\omega) \wedge t)=\sup_{n \in {\mathbb N}}\phi_{n}(\omega,T(\omega) \wedge t)=\limsup_{n \uparrow \infty}\phi_{n}(\omega,T(\omega) \wedge t). $$ Thus $(\omega,t) \mapsto \phi(\omega,T(\omega) \wedge t)$ is ${\mathcal O}$-measurable since $(\omega,t) \mapsto \phi_{n}(\omega,T(\omega) \wedge t)$ is ${\mathcal O}$-measurable (cf. page 31, Williams (1991)).

Hence $\Phi$ contains all one-dimensional bounded optimal processes by Lemma.

Step7. Let ($d=1$) and $X$ be a one-dimensional optimal process. Let $n \in {\mathbb N}$ and set $$ X_{n}(\omega,t):=X(\omega,t) \wedge n, \quad (\omega,t) \in \Omega \times {\mathbb R}_{+}. $$ Then $X_{n}$ is a one-dimensional bounded optimal process since a function $x \mapsto x \wedge n$ is continuous (cf. page 30, 31, Williams (1991)). Thus $X_{n}$ satisfies (a) and (b) by Step6. This implies that $X$ satisfies (a) and (b). Indeed,

(a) For any $\omega \in \Omega$, $$ X(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)=\limsup_{n \uparrow \infty}X_{n}(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega). $$ Thus $\omega \mapsto X(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)$ is ${\mathcal F}_{T}$-measurable since $\omega \mapsto X_{n}(\omega,T(\omega))$ is ${\mathcal F}_{T}$-measurable (cf. page 31, Williams (1991)).

(b) For any $(\omega,t) \in \Omega \times {\mathbb R}_{+}$, $$ X(\omega,T(\omega) \wedge t)=\limsup_{n \uparrow \infty}X_{n}(\omega,T(\omega) \wedge t). $$ Thus $(\omega,t) \mapsto X(\omega,T(\omega) \wedge t)$ is ${\mathcal O}$-measurable since $(\omega,t) \mapsto X_{n}(\omega,T(\omega) \wedge t)$ is ${\mathcal O}$-measurable (cf. page 31, Williams (1991)).

Step8. Let $d \in {\mathbb N}$ and $X=(X^{1},X^{2},\ldots,X^{d})$ be a $d$-dimensional optimal process. Then for any $i \in \{1,2,\ldots,d\}$, $X^{i}$ satisfies the conditions (a) and (b) by Step7 since $X^{i}$ a one-dimensional optimal process. This implies that $X$ satisfies (a) and (b). Indeed,

(a) For any $E_{1},E_{2},\ldots,E_{d} \in {\mathcal B}({\mathbb R}^{1})$, \begin{align*} &\left\{\omega \in \Omega; X(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega) \in E_{1} \times E_{2} \times \cdots \times E_{d}\right\} \\ &=\cap_{i=1}^{d}\left\{\omega \in \Omega; X^{i}(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega) \in E_{i}\right\} \in {\mathcal F}_{T} \end{align*} since $\omega \mapsto X^{i}(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)$ is ${\mathcal F}_{T}$-measurable. Thus $\omega \mapsto X(\omega,T(\omega)){\bf 1}_{\{T<\infty\}}(\omega)$ is ${\mathcal F}_{T}$-measurable (cf. page 76, 30, Williams (1991)).

(b) For any $E_{1},E_{2},\ldots,E_{d} \in {\mathcal B}({\mathbb R}^{1})$, \begin{align*} &\left\{(\omega,t) \in \Omega \times {\mathbb R}_{+}; X(\omega,T(\omega) \wedge t) \in E_{1} \times E_{2} \times \cdots \times E_{d}\right\} \\ &=\cap_{i=1}^{d}\left\{(\omega,t) \in \Omega \times {\mathbb R}_{+}; X^{i}(\omega,T(\omega) \wedge t) \in E_{i}\right\} \in {\mathcal O} \end{align*} since $(\omega,t) \mapsto X^{i}(\omega,T(\omega) \wedge t)$ is ${\mathcal O}$-measurable. Thus $(\omega,t) \mapsto X(\omega,T(\omega) \wedge t)$ is ${\mathcal O}$-measurable (cf. page 76, 30, Williams (1991)).

Hence Proposition 1.21 had been proved by Step8.

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    $\begingroup$ Why does (b) hold by (i)? Doesn't (b) hold simply because $1_{B \backslash A} = 1_B - 1_A$, where $1_A, 1_B \in \Phi$, and $\Phi$ is linear? And in Step 3, I think you mean the sets $\cap_{i=1}^k Y_i^{-1}(E_i)$? So the proof concludes that $\Phi$ contains all bounded optional processes. How can we get rid of the boundedness restriction? Finally, how can we extend the result to $\mathbb{R}^d$-valued processes instead of just $\mathbb{R}$-valued? $\endgroup$ – nomadicmathematician Feb 16 at 7:33
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    $\begingroup$ One more question : why is the set of all measurable processes satisfying (a) and (b) in Proposition 1.21 stable under pointwise convergence? If $X_n \to X$ pointwise, then how do we show that $X_T 1_{T < \infty}$ is $\mathscr{F}_T$ -measurable and the stopped process $X^T$ is optional? In fact, I'm not sure how we get $(X_n)^T \to X^T$ and $(X_n)_T 1_{T < \infty} \to X_T 1_{T < \infty}$ from $X_n \to X$. $\endgroup$ – nomadicmathematician Feb 16 at 7:37
  • $\begingroup$ The proof was added and corrected. $\endgroup$ – 720773 Feb 16 at 11:21
  • $\begingroup$ To Question 1: Yes. We used the linearity of $\Phi$ to show ${\bf 1}_{B \setminus A}={\bf 1}_{B}-{\bf 1}_{A} \in \Phi$. To Question 2: One often use the approximation technique. To Question 3: We used a truncate technique in Step7 to rid of the boundedness restriction. To Question 4: We used a technique in Step8 to extend the result to general dimensions. To Question 5: These hold by a property of the measurability about pointwise convergence (or limit superior) (cf. page 31, Williams (1991)). $\endgroup$ – 720773 Feb 16 at 12:35
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    $\begingroup$ Excellent answer took me some time to go through them but seems flawless and resolves all the queries I had, except for $\mathcal{I}$ in Step 3 should be fixed to have elements of the form $\cap_{i=1}^k Y_i^{-1}(E_i)$? $\endgroup$ – nomadicmathematician Feb 16 at 13:27

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