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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.

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  • $\begingroup$ I think your question doesn't really have to do with filtrations and stochastic processes. You are given a mapping $K: (s,\omega)\mapsto \xi_s(\omega)$ on $[0,t]\times \Omega$, and you seem to be assuming ("measurable random process") that $K$ is measurable for the product $\sigma$-algebra of $[0,t]\times \Omega$. But then it is embodied in the proof of Fubini's theorem that $\omega\mapsto \int_0^t K(s.\omega)\,ds$ is measurable. Maybe I'm missing something? $\endgroup$
    – Dirk Werner
    Commented Dec 2, 2018 at 22:40
  • $\begingroup$ @Dirk We don't know if the process will be joint measurable. So everything is not obvious. $\endgroup$
    – Emerald
    Commented Dec 3, 2018 at 14:27
  • $\begingroup$ Thanks for the clarification, but ... may I suggest to use precise wording in a question? For the record, in stochastic analysis a process $(\xi_s)_s$ is called measurable if $(s,\omega)\mapsto \xi_s(\omega)$ is measurable for the product $\sigma$-algebra. Incidentally, $\xi_s(\omega)$ is a number, not a process. Returning to what you wanted to ask, it seems to me now that the question is: If $s\mapsto K(s,\omega)$ is measurable (and integrable) for each $\omega$ and $\omega\mapsto K(s,\omega)$ is measurable for each $s$, need $\omega\mapsto \int_0^t K(s,\omega)\,ds$ be measurable? $\endgroup$
    – Dirk Werner
    Commented Dec 3, 2018 at 22:34
  • $\begingroup$ @Dirk, I mean, if $ω ↦K(s,ω) $is measurable (and integrable) for each $ ω$, need $ω↦∫^t_0 K(s,ω)ds$ be measurable? I'm looking for a counterexample. $\endgroup$
    – Emerald
    Commented Dec 4, 2018 at 6:41

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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.

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