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I am stuck with the following problem : let $M_t$ be a right continuous martingale. Show that the map $t \to M_t$ is right continuous from $\mathbb{R}^{+}$ to $L^1$.

I know that, fixed $\omega$, the map $f: t \to M_t(\omega)$ from $\mathbb{R}^{+}$ to $\mathbb{R}$ is right continuous and therefore I thought that it could be a good idea to prove $g: t \to M_t$ is right-continuous by seeing it as a composition of right-continuous functions. In particular, we could define $h : M_t \to M_t(\omega)$ and consider $g= h^{-1} \circ f$. Now, if we prove that $h^{-1}$, if it exists, is right continuous, we are done. Is it feasible ?

Fred G.
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    Another idea could be the following : proving right continuity on $L^1$ should mean $\lim_{s \to t^{+}} \mathbb{E} \left[ |M_t-M_s \right]=0$. What do you think ? – Fred G. Nov 06 '15 at 12:41
  • Ops, there is a typo : $|M_t - M_s | $. – Fred G. Nov 06 '15 at 12:41

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