# Why are stochastic integrals adapted?

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

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 $$s \leq t \implies \mathcal{F}_s \subseteq \mathcal{F}_t$$.

• 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. – Ceeerson Oct 10 '18 at 16:24
• Yes, decomposing 1 using the indicator-fct. is a classic :) – Stockfish Oct 10 '18 at 16:28