A question related to Novikov's condition

Question

The well-known 'Novikov condition' says:

Let $ L = (L_t)_{t \geq 0} $ be a continuous local martingale null at 0 and $ Z = \exp(L - \frac{1}{2} \langle L \rangle) $ its stochastic exponential.

If

$ E[\exp(\frac{1}{2} \langle L \rangle_\infty)] \ < + \infty $,

then $ Z $ is a (uniformly integrable) martingale on $ [0, +\infty] $.

Now to my question:

Is it also true that in this case $ Z_\infty > 0, P-a.s. $?

I'm interested in this question, because $ Z_\infty > 0 $ would ensure that the measure $ Q $ defined by $ Q[A] := E_P [Z_\infty 1_A] $ is equivalent to $ P$.

Thanks for your help! Regards, Si

Answer 1

Yes because $\mathbb E [L_{\infty},L_{\infty}] < \infty$ will make L an $\mathbb L^2$ bounded martingale, so $L_{\infty}$ is finite (and the limit of $L_t$.)