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If I have the following expression:

$$E[M_{t\wedge n}|\mathcal{F}_s]$$

Where the set of random variables $M_{t\wedge n}$ is bounded in $L^2$, i.e.

$$\sup_{t} E[M^2_{t\wedge n}]<\infty$$

Hence they are uniformaly integrable. In fact $(M_t)$ is a local martingale. Now my question, why can I take the limit for $n\to \infty$ insight the expectation in the first expression to obtain

$$E[M_t|\mathcal{F}_s]$$ ?

share|cite|improve this question
What kind of convergence do you want? – Davide Giraudo Jun 4 '12 at 8:54
In fact I have the equality: $E[M_{n\wedge t}|\mathcal{F}_s]=M_{n\wedge s}$. The RHS converges a.s. to $M_s$ as $n\to \infty$. Now I would like to have $E[M_{n\wedge t}|\mathcal{F}_s] \to E[M_{t}|\mathcal{F}_s]$ a.s. – user20869 Jun 4 '12 at 9:02
I got stuck at this point, reading a proof in Revuz/Yor Proposition 1.23. Maybe this makes it easier to answer my question. – user20869 Jun 5 '12 at 6:44
I edited the question! I think I forgot something important – user20869 Jun 5 '12 at 13:59

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