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My class notes on stochastic calculus say that the if $(\alpha_s(\omega))_{s\in \mathbb{R_+}}$ is progressive then $\int_0^t \alpha_s ds$ is a pathwise continuous process?

How does the joint measurability of the process $\alpha$ imply the above continuity? How could I rigorously show this?

Davide Giraudo
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user3503589
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    You may need boundness, but if for a fixed $\hat\omega$ the map $\alpha_s(\hat\omega)$ is measurable in $s$, then it's Lebesgue integral is an absolute continuous function in $t$. – SBF Dec 03 '14 at 07:08
  • @Ilya But how could I rigorously show that? – user3503589 Dec 03 '14 at 21:40
  • First, assume that that sample space has just one element. Can you show it there? – SBF Dec 04 '14 at 06:53
  • @Ilya If I fix $\omega$, then $\alpha$ is just a deterministic map $\mathbb{R}+ \to \mathbb{R}$. I am bit confused because $\alpha$ could be discontinuous and and we could have a situation where the integral $\int_0^t \alpha_s ds$ might not be continuous. $\alpha^{-1}(A) \in \mathcal{B(R+)}$ where $A \in \mathcal{B(R)}$ , allows for alpha to be discontinuous and therefore we could have discontinuous path. Something must be wrong in my reasoning. But I can't see it. I would really appreciate any help – user3503589 Dec 04 '14 at 09:11
  • Since $\Omega = {\omega}$ is a simple special case, you need to prove the statement at least for that case. If the statement does not hold there, of course in general it neither holds. Now, discontinuity is not an issue in most of the cases since integration w.r.t. Lebesgue measure is "smoothing". E.g. see conditions for continuity here. More clear now? – SBF Dec 04 '14 at 09:15
  • @Ilya Thank you I think I understand now. Sorry for the late reply, I somehow missed the notification – user3503589 Dec 13 '14 at 11:54

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