Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(\mathcal F_t)_t$ a filtration. In all example I can see, we always have that $$\mathbb E\int_0^T H_s^2ds=\int_0^T \mathbb E[H_s^2]ds,$$ when $s\mapsto H_s$ is a.s. continuous and $\omega \mapsto H_s(\omega )$ adapted. So is it true in general that if $s\mapsto H_s$ a.s. continuous and $\omega \mapsto H_s(\omega )$ adapted, then $$\mathbb E\int_0^T H_s^2ds=\int_0^T \mathbb E[H_s^2]ds \ \ ?$$

I know that since $H_s\geq 0$ for all $\omega $ and all $s$, it's enough to prove that $(\omega ,s)\mapsto H_s(\omega )$ is measurable. I would say that it's true, but I have difficulties to prove it. But maybe it's not true ? but I can't provide a counter-example.

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    $\begingroup$ If $H$ is jointy measurable (in $s,\omega$), this is always true thanks to Tonelli's theorem. $\endgroup$ – zhoraster May 3 at 12:16

$H_s(\omega) =\lim_{n \to \infty} H_{\frac {[ns]} {n}}(\omega)$ a.s. so $H$ is jointly measurable (w.r.t. the completion of $P$). Hence the equation is true.

$H_{\frac {[ns]} {n}}(\omega)=\sum_k H_{\frac k n} (\omega) I_{k \leq ns <k+1}$. Each term here is a product of a measurable function of $s$ and a measurable function of $\omega$. This makes each term, hence the sum jointly measurable. Since limit of measurable functions is measurable it follows that $H_s(\omega)$ is jointly measurable.

  • $\begingroup$ thank you, but I'm not sure to completely understand your argument... I agree that $H_{\frac{[ns]}{n}}$ is measurable for all $n$, but I don't get why $(s,\omega )$ is jointly measurable (in $\Omega \times [0,t], \mathcal F\otimes \mathcal B([0,T], \mathbb P\otimes \lambda )$ where $\lambda $ is the lebesgue measure. $\endgroup$ – user659895 May 3 at 9:03
  • $\begingroup$ @user659895 I have included a proof now. $\endgroup$ – Kavi Rama Murthy May 3 at 9:11

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