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Let $\mathcal{F}_t$ be the $\sigma$-algebra generated by singletons on $\Omega=[0,1]$. And let \begin{align} X_t(\omega)=\mathbb{1}_{\{\omega=t\}} \qquad \text{for} \qquad 0 \leq \omega, t \leq 1. \end{align} Despite of that $X_t$ is $\mathcal{F}_t$-measurable for each $0 \leq t < \infty$. I do not see why X in $[0,1] \times \Omega$ is not $\mathcal{B}([0,1]) \otimes \mathcal{F}_1$-measurable.

In other words, why the set $\{(t,\omega) \mid t = \omega \}$ does not belong to the product $\sigma$-algebra $\mathcal{B}([0,1]) \otimes \mathcal{F}_1$.

What do you think?

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    $\begingroup$ You say "let $\mathcal{F}_t$ be the $\sigma$-algebra generated by singletons on $\Omega=[0,1]$," and yet you don't make it depend on any variable $t$. Then you introduce $\mathcal{F}_1$ -- what is that? The same as $\mathcal{F}_t$? I assume $\mathcal{B}$ denotes the Borel sets. $\endgroup$
    – ForgotALot
    Commented Mar 7, 2016 at 6:11

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First, we point out that $\sigma\{\{x\}:x \in [0,1] \} = \sigma\{A \subset [0,1]: A \text{ is countable or $A^c$ is countable}\}$.

We construct a process $X:[0,1]\times \Omega \to \mathbb{R}$ which is $\mathcal{B}([0,1])\otimes \mathcal{F}$-measurable and $\mathcal{F}_t$ adapted but not progressive measurable.

Take $X_t(\omega)=\mathbb{1}_{\{(t,\omega):t=\omega\}}$, then $X$ is $\mathcal{B}([0,1])\otimes \mathcal{F}$-measurable. Also $X$ is adapted since for each $t \in [0,1]$, $\{\omega: X_t(\omega)=1\}=\{t\} \in \mathcal{F}_t$ and $\{\omega: X_t(\omega)=0\}=[0,1] \setminus \{t\} = \{t\}^c \in \mathcal{F}_t$.

We need to show $\mathcal{B}([0,1])\otimes \mathcal{F}_1$-measurable. It can be shown that, \begin{align} \mathcal{B}([0,1])\otimes \mathcal{F} = \mathcal{G} = \big\{ \bigcup_{n \in \mathbb{N} } (A_n \times C_n): A_n \in \mathcal{B}([0,1]), C_n \in \mathcal{F}_1 \bigg\}, \end{align} and that $\mathcal{G}$ is a $\sigma$-algebra.

Now, we show that the diagonal $D= \{(t,\omega):X_t(\omega)=1\} = \{(t,t):t \in [0,1] \}$ is not in $\mathcal{G}$.

Assume $D \in \mathcal{G}$. Then we can find $A_n \in \mathcal{B}([0,1]), C_n \in \mathcal{F}_1$ such that $D = \bigcup_{n \in \mathbb{N}}(A_n \times C_n)$.

There are two possibilities:

$i)$ there is a set $C_n$ which is uncountable

$ii)$ all the $C_n$ are countable

•if $i)$ holds, then this set $C_n$ contains at least two distinct points $\omega_1, \omega_2$. Fix a $t \in A_n$. Then we find $(t,\omega_1), (t,\omega_2) \in A_n \times C_n$. However, $(t,\omega_1), (t,\omega_2) \in D$ is not possible.

• if $ii)$ holds, then $C = \cup_{n \in \mathbb{N}} C_n$ is countable. Hence $C^c$ is non-empty. Choosing $\omega \in C^c$, we see $ (\omega,\omega) \notin \cup_{n \in \mathbb{N}} \ A_n \times C_n$. But $(\omega,\omega)\in D$.

So, $D \notin \mathcal{G} = \mathcal{B}([0,1])\otimes \mathcal{F}_1$.

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  • $\begingroup$ iJup: It is, to this reader, not obvious that $\mathcal G$ is closed under the formation of complements. Could you provide some details here? $\endgroup$ Commented Mar 12, 2016 at 23:07

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