# 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?

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?

Proof Fix $$t>0$$. We will prove that $$V_t$$ is $$\mathcal F_t$$-measurable. We fix an increasing sequence $$0=t_0^n<\cdots of subdivisions of $$[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.

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.
• So the measurability of $K$ depends on the fact that $K(\cdot , \omega)$ is continuous for all $\omega \in \Omega$. Could you elaborate on the case where $K(\cdot , \omega)$ is continuous for almost all $\omega \in \Omega$? Feb 7 at 21:15