I'm looking for an example of a stochastic process $X$ that is $\mathbb{F}$-adapted, but not progressively measurable. One example I found is the following:

$(\Omega, \mathfrak{A}) = (\mathbb{R^+}, \mathfrak{B}(\mathbb{R^+}))$, $\mathcal{F}_t = \sigma({x}:x \in \mathbb{R}^+)$ for all $t \geq 0$;

$X(t,\omega) := \cases{0, \mbox{if}~  \omega \neq t,\\ 1, \mbox{if}~ \omega = t}$.

I don't see why this process is adapted, but not progressively measurable.

For $X_t$ we have

$X_t (\omega) = \cases{0, \mbox{if}~ \omega \neq t,\\ 1, \mbox{if}~ \omega = t}.$


$X_t^{-1}(\{0\}) = \emptyset$

$X_t^{-1}(\{1\}) = \{t\}$

and since $\emptyset,\{t\} \in \mathcal{F}_t$ for all $t \geq 0$, $X_t$ is $\mathcal{F}_t$-measurable and therefore adapted. (correct so far?)

But can someone please explain, how I can see, that this process is not progressively measurable?

Or does someone have another easy understandable example?


1 Answer 1


I find the following example easier to understand:

Consider any probability space $(\Omega,\mathcal{A},\mathbb{P})$. Set $\mathcal{F}_t := \mathcal{A}$ for all $t \geq 0$ and let $A \subseteq [0,T]$ be a set which is not Borel-measurable. The process

$$X_t(\omega) := 1_A(t)$$

takes only values in $\{0,1\}$ and for each fixed $t \geq 0$, we have either $X_t = 1$ or $X_t = 0$. This implies in particular that $\omega \mapsto X_t(\omega)$ is measurable (it is a constant!). On the other hand, we have

$$\{(t,\omega) \in [0,T] \times \Omega; X_t(\omega) = 1\} = A \times \Omega \notin \mathcal{B}(\mathbb{R}) \otimes \mathcal{A}$$

and this means that the mapping

$$X: [0,T] \times \Omega \to \mathbb{R}$$

is not measurable, i.e. $(X_t)_t$ is not progressively measurable.

  • 1
    $\begingroup$ For $(X_t)_{t\geq 0}$ being a stochastic process, don't we need that $(t,\omega )\mapsto X_t(\omega )$ being $\mathcal B([0,\infty ))\otimes \mathcal F$ measurable ? If not, do you know an example of process s.t. $(t,\omega )\mapsto X_t(\omega )$ being $\mathcal B([0,\infty ))\otimes \mathcal F$ measurable but not $\mathcal B[0,t]\otimes \mathcal F$ measurable for all $t$ ? $\endgroup$
    – John
    Jul 28, 2019 at 12:06
  • 2
    $\begingroup$ @John Being a stochastic process just means that $\omega \mapsto X_t(\omega)$ is measurable; no joint measurability is required (...perhaps there are authors which define it a different way, but in general no such assumption is made). If $[0,\infty) \times \Omega \ni (t,\omega) \mapsto X_t(\omega)$ is $\mathcal{B}([0,\infty)) \otimes \mathcal{F}$-measurable for some $\sigma$-algebra $\mathcal{F}$, then $[0,T] \times \Omega \ni (t,\omega) \mapsto X_t(\omega)$ is $\mathcal{B}([0,T]) \otimes \mathcal{F}$-measurabl for any $T>0$; [...] $\endgroup$
    – saz
    Jul 28, 2019 at 12:56
  • 2
    $\begingroup$ this is a direct consequence of the fact that the mapping $$[0,T] \times \Omega \ni (t,\omega) \mapsto (t,\omega) \in [0,\infty) \times \Omega$$ is measurable. $\endgroup$
    – saz
    Jul 28, 2019 at 12:57
  • $\begingroup$ So at the end, being progressively measurable is equivalent than to be $\mathcal B([0,\infty ))\otimes\mathcal F$ measurable, right ? $\endgroup$
    – John
    Aug 25, 2019 at 11:20

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