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In my lecture notes they use the fact, that every supermartingale $(M_t)$ for which the map $t\mapsto E[M_t]$ is constant is already a martingale. Unfortunately I can't prove it. Some help would be appreciated.

By definition of a supermartingale we have: $E[M_0]\ge E[M_s]\ge E[M_t]$ for $t\ge s\ge 0$. I also know that $M_s\ge E[M_t|\mathcal{F}_s]$. If I would take expectation I would get an equality by assumption. However I do not see how this helps here to prove $M_s= E[M_t|\mathcal{F}_s]$

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up vote 5 down vote accepted

Fix $t\geq s$. Then $M_s-E[M_t\mid M_s]$ is a non-negative random variable. Its expectation is $$E[M_s]-E[E[M_t\mid M_s]]=E[M_s]-E[M_t]=0$$ by assumption.

A non-negative random variable $X$ with $0$ expectation is $0$ almost everywhere, since $$P(X\geq 2^{-n})\leq 2^nE X=0.$$

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A very nice proof! Thanks a lot! – user20869 Aug 1 '12 at 10:15

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