Is $\int_0^t K_s \mathrm d s$ $\mathcal F_t$-measurable in case the sample paths of $(K_s, s\ge 0)$ are continuous only almost everywhere?

Question

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

1. Could you have a check on my below attempt?
2. 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?

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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.

Answer 1

Fix $t>0$. We need to prove that $K:[0,t]\times\Omega\rightarrow\mathbb{R}$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_{t}$-measurable. Then, that $V_{t}$ is $\mathcal{F}_{t}$-measurable follows from Fubini Theorem.

The joint-measurability of $K$ follows from the fact that $K$ is adapted to the filtration and that it has continuous sample paths. We prove this fact as follows:

For each $n$, define $K^{n}:[0,t]\times\Omega\rightarrow\mathbb{R}$ by $$ K^{n}(s,\omega)=K(0,\omega)1_{\{0\}\times\Omega}(s,\omega)+\sum_{k=0}^{2^{n}-1}K(\frac{t(k+1)}{2^{n}},\omega)1_{(\frac{tk}{2^{n}},\frac{t(k+1)}{2^{n}}]\times\Omega}(s,\omega). $$ Since $K(\cdot,\omega)$ is continuous for each $\omega\in\Omega$, it is routine to verify that $K^{n}\rightarrow K$ pointwisely. Note that for each $k$, $(s,\omega)\mapsto K(\frac{t(k+1)}{2^{n}},\omega)$ and $(s,\omega)\mapsto1_{(\frac{tk}{2^{n}},\frac{t(k+1)}{2^{n}}]\times\Omega}(s,\omega)$ are $\mathcal{B}([0,t])\otimes\mathcal{F}_{t}$-measurable. Hence, $K^{n}$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_{t}$-measurable. It follows that $K$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_{t}$-measurable.