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I would like to know if a discontinuous local martingale with paths of finite variations almost surely is a martingale. I feel that it should be the case but can't find a straightforward argument.

As noted by Did in its comments below, the continuous case follows easily from the widely known fact that continuous local martingales have paths of almost surely infinite variations. So I modified the question to make it more specifically about discontinuous local martingales.

Best regards.

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Continuous local martingales of finite variation are constant. Do you mean finite quadratic variation? –  Did Aug 9 '13 at 9:20
    
@ did : Hi, yes they are, but what about the discontinuous case ? –  TheBridge Aug 9 '13 at 9:21
    
So you mean finite variation and the local martingales of interest are not continuous? –  Did Aug 9 '13 at 9:25
    
Indeed, otherwise the conclusion holds as you noted. –  TheBridge Aug 9 '13 at 9:38
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