# When are simple predictable processes adapted?

I am trying to work through protters book and some supplementary notes on my own, but am stuck. I wish to know when a simple predictable process is adapted.

Our definition of simple predictable will be: $$X$$ is simple predictable if there exist bounded random variables $$Z_k$$, $$k = 0,1,\cdots,n$$ and finite stopping times $$0 = T_1 \leq T_2 \leq \cdots \leq T_n < \infty$$ such that $$X = Z_0\mathbf{1}_{\{0\}} + \sum_{k = 1}^{n-1} Z_k\mathbf{1}_{(T_k,T_{k+1}]}$$ and $$Z_k$$ is $$\mathcal{F}_{T_{k}}$$ measurable.

Since predictable processes in general are just those processes that are measurable with respect to the predictable sigma algebra generated by left continuous and adapted processes, predictable processes must therefore be adapted. However, starting with the definition above, I find it hard to see that the simple predictable processes are adapted.

I would like to show that these processes generate the predictible processes. I already know that processes of the form $$\mathbf{1}_{(S,T]}$$ are left continuous and adapted... adapted since for any $$t \geq 0$$, the function $$h_t(\omega) = \mathbf{1}_{(S,T]}(t,\omega)= \mathbf{1}_{\{S is $$\mathcal{F_t}$$ measurable if we assume that our filtration is right continuous.

I would like to know why something of the form $$Z_k\mathbf{1}_{(T_k,T_{k+1}]}$$ is adapted if $$Z_k$$ is $$\mathcal{F}_{T_k}$$ measurable like assumed in protter.

Thanks

Update:

For a simple example, let $$S \leq T$$ be finite stopping times, $$Z$$ a bounded $$\mathcal{F}_S$$ measurable random variable and $$t \geq 0$$. Put $$g_t(\omega) = Z(\omega)\mathbf{1}_{(S,T]}(t,\omega)$$. We want to show that $$g_t$$ is $$\mathcal{F}_t$$ measurable.

I understand that if $$0 \not \in B$$, then $$g_t^{-1}(B) = (Z^{-1}(B) \cap \{S \leq t\}) \cap \{S = t\}^c \cap \{T \geq t\} \in \mathcal{F}_t$$.

But I cannot figure out why $$g_t^{-1}(0) = \{Z = 0\} \cup \{S \geq t\} \cup \{T < t\}$$ must be in $$\mathcal{F_t}$$.

In any case, I think your $$X$$ is adapted for the following reason. The question is whether $$\{X_t\in B\}\in\mathcal F_t$$ for all Borel sets $$B$$ and $$t> 0$$. And we can rewrite $$\{X_t\in B\}$$ as \begin{align} \{X_t\in B\} & = \cup_{k=1}^n(\{Z_k\in B\} \cap \{t\in(T_k,T_{k+1}]\} )\\ & = \cup_{k=1}^n(\{Z_k\in B\} \cap \{T_k\le t\} \cap \{t=T_k\}^c \cap \{t\le T_{k+1}\} ) \end{align} where I take $$T_{n+1}$$ to be $$\infty$$ for convenience. Now, $$\{Z_k\in B\} \cap \{T_k\le t\}\in\mathcal F_t$$ by the assumption $$Z_k\in\mathcal F_{T_k}$$ and so are $$\{t=T_k\}^c$$ and $$\{t\le T_{k+1}\}$$ as they are stopping times. The conclusion follows.
• Hello, I agree when $0 \not \in B$, but I will respond to when $B = \{0\}$ in an update – Ceeerson Oct 7 '18 at 15:19
• Uh... what's special about $\{0\}$? It's just another Borel set. – AddSup Oct 8 '18 at 7:19
• I'm saying that the way you split up $X_t^{-1}(B)$ is incorrect if $0 \in B$. You can see for example in my update what $X_t^{-1}(B)$ looks like if $B = \{0\}$. – Ceeerson Oct 8 '18 at 14:24
• For $t>0$, $\cup_{k=1}^n \{ t\in(T_k,T_{k+1}] \}$ is a partition of $\Omega$. So $\{X_t\in B\}=\{X_t\in B\}\cap(\cup_{k=1}^n \{ t\in(T_k,T_{k+1}] \})=\cup_{k=1}^n(\{X_t\in B\}\cap\{t\in(T_k,T_{k+1}]\})$. But on $\{t\in(T_k,T_{k+1}]\}$, $X_t=Z_k$ so I changed $\{X_t\in B\}$ to $\{Z_k\in B\}$ to arrive at the above expression. (FYI, I focus on $t>0$ because $X_0$ is clearly measurable with respect to $\mathcal F_0$.) – AddSup Oct 8 '18 at 16:06
• If I took $n = 2, t = 2, T_1 \equiv 0, T_2 \equiv 1$, then $\{t \in (T_1, T_2]\} = \varnothing$ is not a partition of $\Omega$. Remember, in the definition that I supplied, the stopping times are finite. – Ceeerson Oct 8 '18 at 16:29