Let $ \xi \_t(\omega), t\in[0,\infty)$, be a random process and let $ \xi \_t(\omega)\in \{\mathfrak F\_t\}$ be some filtration. Even if $ \xi \_t(\omega) $ is $ \mathfrak F\_t $ measurable then $\int\_0^t\xi \_s(\omega)L(ds)$ is not necessarily $ \mathfrak F\_t $ measurable. I am looking for a process that will demonstrate this statement. Maybe I should use Heaviside function? Thanks in advance for any tips.



An example of a measurable random process with non-measurable integral
The following example is a variant of an example of Sierpinski that Martin Väth told me; it depends on the Continuum Hypothesis. Let $<^\*$ be a well-ordering of $[0,1]$. By CH, $\{s: s<^\*t\}$ is always at most countable. Let $M\subset [0,1]$ and define $k: [0,1]\times [0,1]\to \mathbb{R}$ by $k(s,\omega)=1$ if $\omega\in M$ and $\omega <^\* s$, and $k(s,\omega)=0$ otherwise. Then $k$ is partially measurable: for each $s$, $k(s,\omega)$ is non-zero at most countably often, and for each $\omega\in M$, $k(s,\omega)\neq1$ at most countably often while $k(s,\omega)=0$ if $\omega\notin M$. The latter implies that $\int\_0^1 k(s,\omega)\,ds=1$ resp. $=0$ if $\omega\in M$ resp. $\omega\notin M$. Hence $\omega \mapsto \int\_0^1 k(s,\omega)\,ds$ is nothing but the indicator function $\chi\_M$ of $M$. Therefore this function is not measurable if $M$ was not measurable. 


