Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $K=(K_t, t\ge 0)$ be a process adapted to a filtration $(\mathcal F_t, t\ge 0)$ such that the map $t \mapsto K_t (\omega)$ is continuous for all $\omega \in \Omega$. We define the process $V=(V_t, t\ge 0)$ by $$ V_t := \int_0^t K_s \mathrm d s \quad \forall t \ge 0. $$
Theorem $V$ is adapted to $(\mathcal F_t, t\ge 0)$.
My questions
- Could you have a check on my below attempt?
- In my proof, I use the fact the everywhere pointwise limit of a sequence of measurable functions is again measurable. It seems the function $V_t$ defined by $V_t := \int_0^t K_s \mathrm d s$ is not necessarily measurable if the sample paths are continuous only almost everywhere. Could you confirm if my understanding is fine?
Proof Fix $t>0$. We will prove that $V_t$ is $\mathcal F_t$-measurable. We fix an increasing sequence $0=t_0^n<\cdots<t_{p_n}^n=t$ of subdivisions of $[0, t]$ whose mesh tends to $0$. Let $$ X_n := \sum_{i=0}^{p_n - 1} K_{t_i} (t_{i+1} - t_i) \quad \forall n \ge 1. $$
Then $X_n \overset{n \to \infty}{\longrightarrow} V_t$ pointwise everywhere. Because $X_n$ is $\mathcal F_t$-measurable for all $n$. So $V_t$ is $\mathcal F_t$-measurable.