I'm reading something that defines the following class of functions:

Let $\mathcal V$ be the class of functions $f(t,\omega):[0,\infty) \times \Omega \rightarrow \mathbb{R}$ such that:

  1. $(t,\omega) \rightarrow f(t,\omega)$ is $\mathcal B \times \mathcal F$ measurable, where $\mathcal B$ are the Borel sets on $[0,\infty)$.

  2. $f(t,\omega)$ is $\mathcal F_t$ adapted.

  3. $E\bigg[\int_S^Tf(t,\omega)^2dt\bigg]$ is finite

My question -- aren't 1 and 2 the same thing?


1 Answer 1


No they are not. For an example of a non progressively measurable process which is adapted have look at example 1.38 here

An even simpler example here

The converse is true, being adapted means in 2. that $f(t,\omega)$ seen only as a function of $\omega$ are measurable with respect to the sigma algebra $\mathcal{F}_t$. Taking a process $f$ that is progressively measurable. As $f({t}\times A_t)^{-1} \in \mathcal{B}(\mathbb{R})$ is true for any $A_t \in \mathcal{F}_t$ by hypothesis of progressive measurability identifying ${t}\otimes \mathcal{F}_t$ to $\mathcal{F}_t$ means that $f_t$ is $\mathcal{F}_t$-measurable which is exactly meaning that it is adapted.

Best regards


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