Is a predictable process adapted?

Solution 1:

There is also the concept of an optional process:

Definition 1: The optional $\sigma$-algebra $O$ is generated by all adapted càdlàg processes (continue à droite et limit à gauche/continuous from the right and limit from the left). A stochastic process $X=(X_t)_{t\geq 0}$ is called optional, if $X$ is measurable w.r.t. $O\subseteq \mathcal{P}(\Omega\times \mathbb{R}_{\geq 0})$.

It is possible to show that $P\subseteq O$ ($P$ denotes the predictable $\sigma$-algebra). Thus, every predictable process is optional. For optional process, we have the following result:

Theorem 2: If $X$ is optional, then $X$ is $\mathcal{F}\otimes\mathcal{B}(\mathbb{R}_{\geq 0})-\mathcal{B}(\mathbb{R})$ measurable and the random variable $$X_T 1_{T<\infty}:\Omega\rightarrow \mathbb{R}$$ $$\omega\mapsto X_{T(\omega)}1_{T(\omega)<\infty}$$ is $\mathcal{F}_T-\mathcal{B}(\mathbb{R})$-measurable for any stopping time $T$.

Applying this to the stopping time $T\equiv t$ for $t\geq 0$ gives $X_t=X_T 1_{T<\infty}$ is $\mathcal{F}_T$-measurable. Since $\mathcal{F}_T=\mathcal{F}_t$, we see that a predictable process is adapted to the filtration $(\mathcal{F}_t)_{t\geq 0}$

Regarding your second question, we have the following statement:

Theorem 3: Let $T$ be a stopping time and $X$ a predictable process. Then $X_T 1_{\{T<\infty\}}$ is $\mathcal{F}_{T -}$ measurable.

Applying this to the stopping time $T\equiv t$ yields $X_t=X_t 1_{t<\infty}$ is $\mathcal{F}_{t-}$ measurable for all $t\geq 0$ and a predictable process $X$.

The proof for theorem 3 can be found in the first edition of "Foundations of modern probability theory" by Olav Kallenberg in lemma 22.3.