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If $\sigma=0$ there is no randomness: the spot follows a single deterministic path. That is, the measure consists of a point mass at that path. Any equivalent measure can again only give a point mass at that same path, with the same drift. So in this case we must have $\mu = r$ to have an equivalent martingale measure. This is arbitrage free, but there ...

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I think that you are a bit confused: the support of the Black-Scholes model is $(0,+\infty)$, that is to say the underlying asset price is non-negative, like a stock. The Vasicek model has an OU process whose support is $(-\infty,+\infty)$, that is to say the underlying can be negative. Therefore all equivalent measures (of which the martingale is one) ...

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As per your comments, this is the Kunita Watanabe decomposition. See the post at http://math.stackexchange.com/questions/413103/kunita-watanabe-decomposition and the presentation http://www.eurandom.nl/events/workshops/2011/ISI_MRM/Presentation/Vanmaele.pdf

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