# Regarding proof of converse to Girsanovs theorem

This is regarding an argument from Arbitrage Theory by Thomas Björk - Theorem 11.6, but is attempted self contained.

Consider a Wiener process W on probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in[0,T]},P)$. Assume that the filtration is generated by W, and that $\mathcal{F}=\mathcal{F}_T$.

Assume Q is a probability measure absolutely continuous with respect to P on $\mathcal{F}_T$. Denote $L_T$ the Radon-Nikodym derivative and define $L_t = E[L_T \lvert \mathcal{F}_t]$ a martingale.

By the martingale representation theorem we can find $g_t$ such that $$dL_t = g_t dW_t$$ If we define $\phi_t = \frac{1}{L_t} g_t$ this should suffice as a Girsanov transformation, showing that in the filtration of the Wiener process any absolutely continuous transformation is of a Girsanov type.

The text says: "There remains a small problem namely when $L_t=0$ but also this can be handled". How does one handle that problem?

• The set of $\omega\in\Omega$ where $L_t=0$ is a set that must have $Q$-measure zero. What $W_t'$ is doing there doesn't matter. – Pp.. Jan 3 '15 at 14:43
• So you simply define $\phi_t = 0$ on $\{\omega \in \Omega : L_t(\omega)=0\}$? – Henrik Jan 3 '15 at 15:55
• I guess that works. The $Q$-measure doesn't see what you do in there. – Pp.. Jan 3 '15 at 16:01