the measurability of $\int_0^t X(s)ds$

Question

For a $F_t$ adapted process $X$, please prove that $\int_0^t X(s)ds$ is $F_t$ measurable. For simple processes $X$, the conclusion is obvious. I think we should use monotone class theorem to prove the case of general processes. However, I don't know the details. Thank you.

Answer 1

Hints:

1. Without loss of generality, we may assume $X(s) \geq 0$ for all $s \geq 0$ (otherwise consider the decomposition $X=X^+-X^-$ where $$X^+(s) := \max\{X(s),0\} \quad \text{and} \quad X^-(s) := \max\{-X(s),0\}$$ are both adapted processes).
2. Prove the claim if $X$ is a simple process.
3. Let $X \geq 0$ be an arbitrary adapted process. Then there exists a sequence $(X_n)_{n \in \mathbb{N}}$ of adapted simple processes such that $X_n \uparrow X$. By the monotone convergence theorem, $$\int_0^t X(s) \, ds = \sup_{n \in \mathbb{N}} \int_0^t X_n(s) \, ds.$$ Conclude that $\int_0^t X(s) \, ds$ is $\mathcal{F}_t$-measurable.

Alternative solution: For each fixed $\omega$, the mapping $s \mapsto X(s,\omega)$ is Riemann-integrable and

$$\int_0^t X(s,\omega) \, ds = \lim_{n \to \infty} \sum_{i=0}^n X \left( \frac{i}{n} t ,\omega \right) \frac{1}{n}.$$

Since $\sum_{i=0}^n X \left( \frac{i}{n} t ,\cdot\right) \frac{1}{n}$ is $\mathcal{F}_t$-measurable, it follows that $\int_0^t X(s,\cdot) \, ds$ is $\mathcal{F}_t$-measurable (as a pointwise limit of measurable functions).