# Stochastic Exponential: $dZ=-\lambda Z dM + dL$ to $dZ=-\lambda Z dM + Zd\tilde{L}$ while $\tilde{L}$ is still orthogonal to $M$

I have a question concerning the paper http://www.researchgate.net/publication/228648002_No_arbitrage_and_the_growth_optimal_portfolio, Lemma 6.3, which is based on http://www.math.ethz.ch/~mschweiz/Files/martdens.pdf Prop. 5. Actually this proposition does not state what is used, but I think the following is done:

Say, we have a local Martingale $M$, a predictable process $\lambda$, a local martingale $L$ which is orthogonal to $M$ in the sense that $ML$ is a local martingale such that

$$Z=1-\int Z \lambda dM+L$$

is a strictly positive local martingale. Can you always rewrite $Z$ to $Z=\mathcal{E}(-\lambda\cdot M + \tilde{L})$ with $d\tilde{L}_t=\frac{1}{Z_{t-}}dL_t$ so $\tilde{L}$ is still orthogonal to $M$?

One gets

$$d(M\tilde{L})=\tilde{L}dM+\frac{M}{Z}dL+\frac{1}{Z}d[M,L]$$

Now, if $\tilde{L}$, $\frac{M}{Z}$ and $\frac{1}{Z}$ are well behaved, e.g. locally bounded everything is fine. Otherwise the integrals might not be local martingales. But that is not obvious, is it? I've seen the same trick a couple of times, always without going into details.

Thank you!

-