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I'm trudging through Protter, and as far as I can tell, when he defines the stochastic integral of simple predictable processes, he fails to mention why it is adapted.

Simple predictable definition: $H$ 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 $H = Z_0\mathbf{1}_{\{0\}} + \sum_{k = 1}^{n-1} Z_k\mathbf{1}_{(T_k,T_{k+1}]}$ and $Z_k$ is $\mathcal{F}_{T_{k}}$ measurable. Define $\mathcal{S} = \{H: H \text{ is simple predictable}\}$ We can put a norm $||\cdot||_u$ on $\mathcal{S}$ by $||H||_u = \sup_{t \geq 0}||H_t||_{\infty}$.

Define $L^0 = \{Z: Z \text{ is } \mathcal{F}\text{-measurable}\}$ to be the set of random variables with the topology induced by convergence in probability.

Stochastic integral definition: Let $X$ be a process. Define the stochastic integral $\int_0^tH_s dX_s = Z_0X_0 + \sum_{k = 1}^{n-1} Z_k(X^{T_{k+1}} - X^{T_k})$.

Semimartingale definition: Let $X$ be a process. Then $X$ is a semimartingale if $X$ is cadlag, adapted and for each $t \geq 0$, the map $ \mathcal{S}_u \rightarrow L^0$ given by $H \mapsto \int_0^t H_s dX_s$ is continuous.

I know that simple predictable processes are adapted, cadlag. I know that if $X$ is adapted and cadlag then so is $X^T$. I know that if $X$ is cadlag, then $Z_k(X^{T_{k+1}} - X^{T_k})$ is cadlag. However, I dont know if $X$ is cadlag, adapted implies $Z_k(X^{T_{k+1}} - X^{T_k})$ is adapted. That $Z_k$ variable is throwing me off. The reason I want to know is because in protter, there is a theorem.

Definition: Define $\mathbb{D} = \{\text{cadlag adapted processes}\}$

Theorem: If $X$ is a semimartingale, then the map $\mathcal{S} \rightarrow \mathbb{D}$ given by $H \mapsto \int_0^t H_s dX_s$ is continuous with respect to some topologies blah blah blah.

Here protter never mentioned why such a map would be well defined. I don't know if the integral is adapted because $X$ is cadlag, adapted or if because $X$ satisfies the continuity condition of being a semimartingale. but I cannot figure out why $Z_k(X^{T_{k+1}} - X^{T_k})$ is adapted.

Could anybody expain why? The $Z_k$ is causing me trouble in showing this.

I tried splitting up $\Omega = \{t \leq T_k\} \cap \{T_k < t \leq T_{k+1}\} \cap \{T_{k+1}<t \}$, but could not show that for a borel set B, $\{Z_k(X^{T_{k+1}} - X^{T_k}) \in B\}\cap \{T_k < t \leq T_{k+1}\} \in \mathcal{F}_t$

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Edit: I think I got it:

Rewrite \begin{align} (Z_k(X^{T_{k+1}}-X^{T_{k}}))_t &= \mathbf{1}_{t < T_k} (Z_k(X^{T_{k+1}}-X^{T_{k}}))_t + \mathbf{1}_{t \geq T_k} (Z_k(X^{T_{k+1}}-X^{T_{k}})_t \\ &= \mathbf{1}_{t \geq T_k} Z_k(X_t^{T_{k+1}}-X_{T_{k}}). \end{align}

$X^{T_k}$ and $X^{T_{k+1}}$ are adapted, i.e. we only have to show that $\mathbf{1}_{t>T_k} Z_k$ is $\mathcal{F}_t$-measurable. Hence we have to show that for a given Borel set $B$ \begin{align} A := \lbrace \mathbf{1}_{t \geq T_k} Z_k \in B \rbrace \in \mathcal{F}_t. \end{align}

First, note that \begin{align} A_1 := \lbrace T_k \leq t \rbrace \cap \lbrace \mathbf{1}_{t \geq T_k} Z_k \in B \rbrace = \lbrace T_k \leq t \rbrace \cap \lbrace Z_k \in B \rbrace \in \mathcal{F}_t \end{align} as $Z_k$ is $\mathcal{F}_{T_k}$-measurable, i.e. $\lbrace Z_k \in B \rbrace \in \mathcal{F}_{T_k}$.

Second, consider $A_2 := \lbrace T_k > t \rbrace \cap \lbrace \mathbf{1}_{t \geq T_k} Z_k \in B \rbrace$.

1) If $0 \notin B$, $A_2 = \emptyset$.

2) If $0 \in B$, $A_2 = \lbrace T_k > t \rbrace$.

In both cases we obtain $A_2 \in \mathcal{F}_t$ and hence $A = A_1 \cup A_2 \in \mathcal{F}_t$.

Remark: If $T_i$ are fixed numbers instead of stopping times it is clear as we can use $s \leq t \implies \mathcal{F}_s \subseteq \mathcal{F}_t$.

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    $\begingroup$ Wow, after applying the staring method at your solution for a bit, it clicked. That was a clever thing to rewrite it like that, I'll try to remember that. Thank you for your help. $\endgroup$ – Ceeerson Oct 10 '18 at 16:24
  • $\begingroup$ Yes, decomposing 1 using the indicator-fct. is a classic :) $\endgroup$ – Stockfish Oct 10 '18 at 16:28

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