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I have the following, seemingly simple question:

Consider a stochastic process $(X_t)$ satisfying $X_t\le X_s$ a.s. for all $t\le s.$ My question is: Does there exist a modification $\tilde{X}$ of $X$, which almost surely has increasing sample paths $t\mapsto\tilde{X}_t(\omega)$?

I assume such a modification exists, but I did not manage to proof it.

Thanks in advance!

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    $\begingroup$ Does $\tilde X_t(\omega)=\mathop{ess\ sup}_{s\le t}X_s(\omega)$ work? $\endgroup$ – Anthony Quas Jun 1 '14 at 21:11
  • $\begingroup$ How do you show measurability? However the idea seems to work (sup over rational times, countability of times with essential jumps, etc.) $\endgroup$ – fedja Jun 2 '14 at 1:33
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Yes, this works. As suggested by fedja, focus for a moment on the rational times. The function (on $\mathbb Q$) $f_{\omega}(t)=X_t(\omega)$ is almost surely increasing. Set $\widetilde{X_t}=0$ for all $t\in\mathbb R$ on the exceptional set. Now if $t\in\mathbb R$ is arbitrary, then $f_{\omega}(t-)\le X_t(\omega) \le f_{\omega}(t+)$ a.s. because only countably many times are involved (of course, the null set will depend on $t$). Set $\widetilde{X_t}(\omega)=f_{\omega}(t+)$ (say) on the null set where this fails.

All paths are increasing now because the required inequalities hold if at least one rational time is involved.

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