# Continuous Sigma-martingales and local martingales are equivalent sometimes?

I was reading through this paper, and they mentioned in the beginning-most portion of it that $\sigma$-martingales and local-martingales are equivalent if they are continuous.

1. Why must they be continuous (ie: I know local implies sigma martingale in general (straightforward by one of their characterization) but what is an example of a local-martingale which is not a sigma martingale)?
2. Why are these equivalent for continuous semi-martingales?

Unfortunately the link is broken ...

Sigma-martingales are covered in Philip Protters' book: Stochastic Integral and Differential Equations. In the second edition (Vers 2.1) Chapter IV.9 treats this topic.

Corollary 2 of this chapter establishes that a local martingale is a sigma martingale.

Theorem 91 establishes that a sigma martingale which is a special semimartingale is a local martingale. Since continuous semimartingales are special, it follows that any continuous sigma-martingale is a local martingale.

The famous Emery example is given directly before Theorem 34 - it is a badly behaving stochastic integral and the standard example of a sigma martingale which is not a local martingale (but with jumps): consider $$X=U \, \mathbf{1}_{t \ge T}$$ where $$P(U=1)=P(U=-1)=1/2$$, and $$T$$ is exponential with intensity $$1$$. Then $$X$$ is a martingale (in its own filtration). Now choose $$H_t=\frac 1 t \, \mathbf{1}_{t >0}$$. Then the stochastic integral $$Z=H \cdot X$$ is not a local martingale because $$E[|Z_\tau|]= \infty$$ for any stopping time $$\tau$$ with $$P(\tau >0)>0$$.

As a partial answer (that is an answer to $1$ only): the process:

$M_t:=\int |Cos(t)|dB_t$, where $B_t$ is a Brownian Motion is a $\sigma$-martingale but not a local martingale.

• @ CSA : I'm not sure about this $M_t$ is indeed a martingale and a fortiori a local martingale becasue the expectation of its quadaratic variation is finite at all time. Best regards Sep 4, 2015 at 9:38