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Let $(\Omega, \mathscr{F}, P)$ be a measure space with filtration $\{\mathscr{F}_n\}_{n \in N}$. Let $A_n$ be adapted to the filtration $\mathscr{F}_{n}$, and $E|A_n - E[A_n| \mathscr{F}_{n-1}]|=0$, i.e. $A_n = E[A_n | \mathscr{F}_{n-1}]$ a.e -P. How do we get that $A_n$ is $\mathscr{F}_{n-1}$-measurable from this?

All we have here is that $A_n$ agrees with the $\mathscr{F}_{n-1}$-measurable function $Y$ that has $\int_C YdP = \int_C A_n dP$ for all $C \in \mathscr{F}_{n-1}$. How do we get that $A_n$ must be $\mathscr{F}_{n-1}$-measurable as well?

nomadicmathematician
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  • I don't see how $A_n = \mathbb{E}[A_n ~|~ \mathscr{F}{n-1}]$ makes sense? Did you mean $A{n-1} = \mathbb{E}[A_n ~|~ \mathscr{F}_{n-1}]$? – tortue Dec 05 '19 at 14:17
  • @TheoreticalEconomist That was a mistake. – nomadicmathematician Dec 05 '19 at 14:20
  • Ok. In which case, you can’t. I don’t even see how you conclude that $A_n = E[A_n \mid \mathscr F_n]$. – Theoretical Economist Dec 05 '19 at 14:22
  • I am a bit surprised by the negative answers. I thought that $E(X| \mathcal F)$ was $\mathcal F$ measurable by definition. – Giuseppe Negro Dec 05 '19 at 14:54
  • @GiuseppeNegro $\mathsf{E}[X\mid \mathcal{F}]$ is always $\mathcal{F}$-measurable (by definition). –  Dec 05 '19 at 17:15
  • @d.k.o.: Great, thank you for reassuring me. So, if $X=E(X|\mathcal F)$, obviously $X$ is $\mathcal F$-measurable (actually, this is a if and only if, I suppose). Then, where's the caveat with this question? It must be related with that "almost everywhere". – Giuseppe Negro Dec 05 '19 at 17:23
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    @GiuseppeNegro Actually, conditional expectations are uniquely defined up to $\mathsf{P}\mid_{\mathcal{F}}$-null sets. However, in order for $X$ to be a version of $\mathsf{E}[X\mid \mathcal{F}]$, it must be $\mathcal{F}$-measurable. –  Dec 05 '19 at 17:52
  • @d.k.o.: And this is an answer to the present question. – Giuseppe Negro Dec 05 '19 at 18:01

2 Answers2

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This is not true unless $\mathcal{G}$ contains all the $\mathsf{P}$-null sets in $\mathcal{F}$ (and $\mathcal{F}$ is $\mathsf{P}$-complete). For a counterexample take $\mathcal{G}=\{\emptyset,\Omega\}$ and $X=1_{A}$, where $\mathsf{P}(A)=0$.

If $\mathcal{G}$ satisfies the above-mentioned assumption, then for a version $Y$ of $\mathsf{E}[X\mid\mathcal{G}]$ and any Borel set $B$, $$ X^{-1}(B)=(X^{-1}(B)\cap N)\cup(Y^{-1}(B)\cap N^{c})\in\mathcal{G}, $$ where $N\equiv\{X\ne Y\}$, and thus $X$ is $\mathcal{G}$-measurable.

  • I actually got this question from a proof in Karatzas and Shreve. The link is https://math.stackexchange.com/q/3463699/135204. There clearly is something wrong with the proof there? – nomadicmathematician Dec 05 '19 at 14:22
  • Note that your statement above is slightly self-contradictory. In your case, your $A$ is very much measurable, so $(\Omega,\mathcal{F},\mathbb{P})$ might very well be complete, and $(\Omega, \mathcal{G},\mathbb{P})$ is trivially complete since $\mathbb{P}$ isn't the $0$-measure. Hence, the problem is much deeper: There might be $\mathcal{F}$-elements, which are $\mathbb{P}$-null sets that are invisible to your $\mathcal{G}$, completeness be damned! – WoolierThanThou Dec 05 '19 at 14:30
  • @nomadicmathematician Similar result (an increasing sequence is predictable if and only if it's natural) appears in Doob's "Classical Potential Theory and Its Probabilistic Counterpart" on page 483, where the author assumes that $\mathcal{F}_0$ contains the null sets. –  Dec 05 '19 at 15:21
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We can't, because the statement is not true.

Assume, that our probability space is $[0,1]$ with its Borel algebra $\mathcal{B}$ and the Lesbegue measure $m$. Let $\mathcal{F}=\{\emptyset,[0,1]\}$ be the trivial algebra. Then, for any $X\in L^1,$ we have $\mathbb{E}(X|\mathcal{F})=\mathbb{E}(X),$ which is almost surely equal to $1_{\mathbb{[0,1]}\setminus \mathbb{Q}}\cdot \mathbb{E}(X),$ which isn't constant and hence, isn't $\mathcal{F}$-measurable.

Note that $E(X)=0$ doesn't imply that $X=0$ almost surely unless $X$ is positive.

WoolierThanThou
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  • I actually got this question from a proof in Karatzas and Shreve. The link is https://math.stackexchange.com/q/3463699/135204. There clearly is something wrong with the proof there? – nomadicmathematician Dec 05 '19 at 14:23
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    There is an absolute value in that proof that your question is missing. However, it's still not true that if $X$ is measurable and $Y=X$ a.e. then $Y$ is measurable. However, the author probably means that there is an $\mathcal{F}_{n-1}$ version of $A$, which does follow from there. – WoolierThanThou Dec 05 '19 at 14:26