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Given a continuous local martingale $X_t$ starting at 0, i.e. $X_0=0$, by the continuity of the paths, we can deduce that $\sup_{0\leq s\leq t}|X_s|<\infty$ P-a.s. Hence, $E[\sup_{0\leq s\leq t}|X_s|]<\infty$, which means, that the following holds: $$E[X_t\mid F_s]=E[\lim_{n\to \infty}X_{t\wedge T_n}\mid F_s]=\lim_{n\to\infty}E[X_{t\wedge T_n}\mid F_s]=\lim_{n\to\infty}X_{s\wedge T_n}=X_s$$.

The exchange of the limit with the expectation is justified by using dominated convergence theorem. Is this argumentation correct? I have the feeling that the claim is not true, but I cannot find the flaw in my argumentation.

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The flaw in your argumentation is that $\sup_{0 \leq s \leq t} |X_s| < \infty$ does not imply $\mathbb{E}(\sup_{0 \leq s \leq t} |X_s|)< \infty$. This means that the dominated convergence theorem is not applicable. For a given (local) martingale it is, in general, not true that $\mathbb{E}(\sup_{s \leq t} |X_s|) < \infty$, see this question, and therefore this flaw cannot be fixed.

In fact, there are known counterexamples to your assertion; see e.g. this question for such a counterexample.

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