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Suppose you have a random variable $X_1:\Omega \to \mathbb{R}$, which is $\mathcal{F}$ measurable and a random variable $X_2:\Omega \to \mathbb{R}$, which is $\mathcal{F}_t$ measurable with $\mathcal{F}_t\subset \mathcal{F}$, where $(\mathcal{F}_t)$ is a filtration with the usual condition and $\mathcal{F}_t\subset \mathcal{F}$ for all $t$. Now suppose we have $X_1=X_2$ a.s. Is $X_1$ then also $\mathcal{F}_t$ measurable? I think it is true, using that $\mathcal{F}_0$ contains all the null set of $\mathcal{F}$, but I'm not able to wirte down a rigorous proof. If someone could explain / prove why this holds (if it is indeed true), that would be very helpful.

hulik

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Since $\mathcal{F}_t$ contains all null sets of $\mathcal{F}$, then $\{X_1\neq X_2\}\in\mathcal{F}_t$. For $A\in\mathcal{B}(\mathbb{R})$ we have that $$ \begin{align*} X_1^{-1}(A)&=\{X_1\in A,X_1=X_2\}\cup \{X_1\in A,X_1\neq X_2\}\\ &=\{X_2\in A,X_1=X_2\}\cup \{X_1\in A,X_1\neq X_2\}. \end{align*} $$ Can you finish the proof now?

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  • $\begingroup$ ah! since $X_2$ is $\mathcal{F}_t$ measurable the first set is also $\mathcal{F}_t$. The second set is a set of measure zero, hence by usual condition contained in $\mathcal{F}_0\subset\mathcal{F}_t$, am I right? $\endgroup$
    – user20869
    Nov 20, 2012 at 15:56
  • $\begingroup$ @hulik: Exactly! $\endgroup$
    – Stefan Hansen
    Nov 20, 2012 at 16:05
  • $\begingroup$ sorry, but why is the set $\{X_2\in A,X_1=X_2\}\in \mathcal{F}_t$? Of course the first part is, but why is $\{X_1=X_2\}\in \mathcal{F}_t$? $\endgroup$
    – user20869
    Nov 20, 2012 at 17:08
  • $\begingroup$ Because $\{X_1\neq X_2\}=\{X_1=X_2\}^c\in\mathcal{F}_t$ $\endgroup$
    – Stefan Hansen
    Nov 20, 2012 at 17:08

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