Progressive measurability implies adaptedness

Question

I've read that every progressively measurable process is also adapted, but I can't prove it using the definition of measurability. Can anyone give me a proof of this result ?

Answer 1

It seems to me that this fact is a great example of something that is evident to someone experienced, but in fact confuses a great many students by being labeled as such with no explanation, myself included.

First, let's lay down some definitions. Let $(\Omega,\mathcal F, P)$ be a probability space and $\mathcal F_t$ a filtration on it.

The process $(X_t)_{t\geq 0}$ is

• adapted if $X_t:\Omega \to \mathbb R$ is $\mathcal F_t$ measurable for each $t\geq 0$
• progressively measurable if $X:[0,t]\times \Omega \to \mathbb R$ is $\mathcal B([0,t])\otimes \mathcal F_t$ measurable for each $t>0$

In terms of sets, this means

• Adapted: $\{\omega: X_t(\omega) \in B\}\in \mathcal F_t$ for all, $t\geq 0 , B\in \mathcal B(\mathbb R)$:
• Progressively measurable: $\{ (s,\omega): X_s(\omega) \in B \} \in \mathcal B([0,t])\otimes \mathcal F_t$ for all $t>0, B \in \mathcal B(\mathbb R)$

The key realization is that the set $\{\omega: X_t(\omega) \in B\}$ is, in fact, a cross-section (or slice) of the set $\{ (s,\omega): X_s(\omega)\in B\}$ at the point $t$, which yields the implication progressively measurable $\Rightarrow$ adapted.

The measurability of cross-sections is typically introduced in measure theory courses/books when introducing product measures and can be shown by a standard measure theoretic argument.

Answer 2

On this basic issue, many textbooks make a basic mistake, they state that: for a measurable subset $E$ defined in a product measure space $(X\times Y,\mathcal{A}\times \mathcal{B},\mu\otimes \nu)$, the $x$-sections of $E$, $\{y\in Y:(x,y)\in E\}$ are measurable for all $x\in X$. In fact, this is not the case in general. Fubini's theorem tells us that these sections are measurable only for almost all $x\in X$ (for a proof, see Section 20.1 of Royden's Real Analysis).

Here is a correct answer of your question:

Question. Let $X=(X_t)_{t\geq 0}$ be a progressively measurable stochastic process on $(\Omega,\mathscr{F},(\mathscr{F}_t),\mathbb{P})$, i.e., for each $t\geq 0$, restricted on $\Omega\times [0,t]$, $X$ is $\mathscr{B}([0,t])\times \mathscr{F}_t$-measurable. Then $X$ is adapted w.r.t. $(\mathscr{F}_t)$.

Proof. Let $t\geq 0$ be arbitrary, note that \begin{equation} (\Omega,\mathscr{F}_t)\stackrel{f}{\longrightarrow}([0,t]\times \Omega,\mathscr{B}([0,t])\times \mathscr{F}_t)\stackrel{g}{\longrightarrow}(\mathbb{R},\mathscr{B}(\mathbb{R})), \end{equation} \begin{equation} \omega\mapsto (t,\omega)\mapsto X_t(\omega)=X(t,\omega). \end{equation} By the progressive measurability of $X$, $g$ is measurable; to prove $g\circ f$ is measurable, it suffices to show $f$ is measurable.

It's well known that the collection of subsets of $[0,t]\times \Omega$: \begin{equation} \mathscr{G}=\{(a,b]\times A:0\leq a\leq b\leq t,A\in \mathscr{F}_t\} \end{equation} generates the product $\sigma$-algebra $\mathscr{B}([0,t])\times \mathscr{F}_t$; moreover, $\mathscr{G}$ is a $\pi$ system. Now we consider the collection of subsets of $[0,t]\times\Omega$: \begin{equation} \mathscr{C}=\{E\in \mathscr{B}([0,t])\times \mathscr{F}_t:f^{-1}(E)\in \mathscr{F}_t\}. \end{equation} We can easily verify that $\mathscr{C}$ is a Dynkin system and $\mathscr{G}\subseteq \mathscr C$. By Sierpiński-Dynkin's $\pi$-$\lambda$ theorem, we obtain: $\mathscr{C}= \sigma(\mathscr{G})=\mathscr{B}([0,t])\times \mathscr{F}_t$, that is to say, $f$ is measurable. The proof is complete.