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Let $(\Omega,\mathcal F)$ be a measurable space and $X:\Omega\rightarrow\mathbb R^d$ a $\mathcal F$-measurable, bounded map. Let $(\mathbb Q_n)_{n\in\mathbb N}$ be a sequence of probability measures on $(\Omega,\mathcal F)$ and $\{\mathcal F_n\}_{n\in\mathbb N}$ a filtration of $\mathcal F$ such that $\mathcal F=\sigma(\mathcal F_n\mid n\in\mathbb N).$ Assume that $$\mathbb E_{\mathbb Q_n}[X\mid\mathcal F_n]=0$$ for all $n\in\mathbb N$. Does that imply that the sequence $(\mathbb Q_n)_{n\in\mathbb N}$ converges in the weak*-topology to a probability measure $\mathbb Q$ such that $$\mathbb E_{\mathbb Q}[X\mid\mathcal F]=0?$$ I guess that the Banach-Alaoglu Theorem is involved here, but I don't see how this precisely works, since the set of probability measures is in general not weak*-compact, as mentioned in the following link: Is the set of all probability measures weak*-closed?. If my question has a negative answer in general, can we still have a positive answer in the simpler case, where we assume that for every $n\in\mathbb N,$ $\mathcal F_n$ is generated by a countable partition?

After the comments of Michael Greinecker I realized several mistakes/typos...

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The result doesn't even hold when $(\mathcal{F}_n)$ is a constant sequence. Let $\mathbb{Q}_n$ be the probability measure on $[0,1]$ with density $$d(x)=\max\{2n-2n^2x,0\}$$ and let $X$ be the indicator function of the set $\{0\}$. Let $\mathcal{F_n}$ be the Borel $\sigma$-algebra. Then $\lim_{n\rightarrow\infty}\mathbb E_{\mathbb Q_n}[X\mid\mathcal F_n]=0$, but $E_\mathbb{Q}[X∣\mathcal{F}]=1$. Also, there exists an increasingly fine sequence $P_n$ of countable (or even finite) partitions of $[0,1]$ such that $\sigma(\bigcup_n P_n)$ is the Borel $\sigma$-algebra on $[0,1]$. If one assumes, which is possible wlog, that $\{0\}\in P_n$ for all $n$, one gets a counterexample to the weaker claim.

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  • $\begingroup$ Sorry, I just realizet that I have written something that I didn't want and changed it now... $\endgroup$ – Andy Teich Apr 4 '14 at 17:17
  • $\begingroup$ @AndyTeich My answer still applies. $\endgroup$ – Michael Greinecker Apr 4 '14 at 17:41
  • $\begingroup$ can there exist a weak*-converging subsequence? $\endgroup$ – Andy Teich Apr 5 '14 at 14:50
  • $\begingroup$ @AndyTeich The sequence of measures does converge weakly to the measure concentrated on $0$. The problem here is essentially that weak convergence does not preserve absolute continuity. The set $\{0\}$ only ceases to be a null-set in the limit. $\endgroup$ – Michael Greinecker Apr 5 '14 at 15:57

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