Is Girsanov the only way to change measure?

Question

Let $\mathbb{P}$ and $\mathbb{Q}$ be equivalent measures and let $Z = \frac{d\mathbb{Q}}{d\mathbb{P}}$.

Let $L_t = \mathbb{E}^\mathbb{P}[Z \mid \mathcal{F}_t]$. This is a martingale under $\mathbb{P}$. Assume $\mathcal{F}_t$ is a filtration generated by a $\mathbb{P}$-Brownian motion $W_t$.

Then $dL_t = \varphi_t L_t \, dW_t$ for some process $\varphi_t$, by the martingale representation theorem.

If we assume $Z \in \mathcal{F}_T$ for some $T$, then we get a Girsanov transformation.

Am I missing some important conditions, or is it really the case that all change of measures arise this way?

Answer 1

Your argument is correct; in fact, this is often referred to as a mild converse to Girsanov's theorem (see, for instance, Theorem 11.6 in Bjork's Arbitrage Theory in Continuous Time).

Of note, the result hinges on the assumption that $\mathcal{F}_t = \sigma (W_s\, : s \leq t)$, and one cannot expect the result to be true for any filtration. In particular, this assumption is required when applying the martingale representation theorem. The Girsanov transform is defined in terms of $W_t$ and, in a general filtered space, you may expect to find many more processes, not necessarily determined by $W_t$.