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So I have just started to learn about stochastic processes, and I got to learn about progressively measurable processes.

Definition Let $(\Omega, \mathcal{F},P)$ be a probability space and $\{\mathcal{F_t}\}_{t\in[0,\infty)}$ be a filtration. A stochastic process $X:[0,\infty)\times \Omega \to \mathbb{R}$ is said to be $\mathcal{F}$-progressively measurable if, for any $T$, $X|_{[0,T]}$ is $\mathcal{B}([0,T])\otimes\mathcal{F}_T$-measurable.

So that is the definition, and I am failing to see how it is progressive. I would like to know why it was named in such a way.

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  • $\begingroup$ with respect to time $\endgroup$ Jan 18, 2021 at 2:29

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The word "progressive" refers to the progression (in time) of measurability conditions that a progressively measurable process satisfies: for each $t\ge 0$, the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal B([0,t])\otimes\mathcal F_t$-measurable. A progressive process is a little better than just adapted. One consequence is that if $X$ is progressive then $t\mapsto \int_0^t X_s ds$ is well defined and progressive (even predictable) provided $\int_0^t|X_s| ds<\infty$ for $t>0$, by Fubini's theorem.

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  • $\begingroup$ How does the fact that $t \mapsto \int_{0}^{t}X_sds$ is progressive if X is, follow from Fubini's theorem? $\endgroup$ Oct 25, 2022 at 6:12
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Let $I_t(\omega):=\int_0^t X_u(\omega)\,du$, assuming for simplicity that $X$ is bounded, so the integral is well defined. If $s\in[0,t]$ then $I_s$ is $\mathcal F_s$ measurable, by Fubini, because $(u,\omega)\mapsto X_u(\omega)$ is $\mathcal B([0,s])\otimes\mathcal F_s$ measurable. That is, $(I_s)_{s\ge 0}$ is adapted to $(\mathcal F_s)$. But also $s\mapsto I_s(\omega)$ is continuous, by analysis. It follows that $(I_s)$ is even predictable, and in particular progressive.

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