# What is "progressive" about progressively measurable processes?

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.

• with respect to time Jan 18, 2021 at 2:29

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.
• How does the fact that $t \mapsto \int_{0}^{t}X_sds$ is progressive if X is, follow from Fubini's theorem? Oct 25, 2022 at 6:12
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.