Let W be a Brownian motion, T a finite time horizon and $(\Omega, \Bbb{P}, \Bbb{F})$ a probability space with $\Bbb{F} = \Bbb{F}^W$ the natural augmented filtration of $W$. For any $\Bbb{F}$-adapted process $\eta$, I define an associated probability measure $\Bbb{Q}$ by setting $$\frac{d\Bbb{Q}}{d\Bbb{P}} \Big|_{\mathcal{F}_t} = \Sigma^\eta(t) := \exp({\int_0^t \eta(s) dW_s - \frac{1}{2} \int_0^t \lVert \eta(s) \rVert^2 ds}) = \mathcal{E} \Big( \int \eta dW \Big)_t$$ $\forall 0 \leq t \leq T$. Let $$\mathcal{B} := \{ \eta : \eta \mbox{ is a } \Bbb{F} \mbox{-adapted process s.t. } \Bbb{E}^\Bbb{P}[\Sigma^\eta(t)] = 1 \mbox{ and } \Bbb{E}^\Bbb{P}[(\Sigma^\eta(t))^2] < \infty \; \forall t \in [0,T] \}$$ I also define $$\mathcal{A} := \{ \Bbb{Q} : \frac{d\Bbb{Q}}{d\Bbb{P}} \in L^2(P)\} $$
I would like, for any element of $\Bbb{Q}$ of $\mathcal{A}$, to prove that there exists $\eta \in \mathcal{B}$ such that $\frac{d\Bbb{Q}}{d\Bbb{P}} \Big|_{\mathcal{F}_t} = \Sigma^\eta(t)$ $\forall 0 \leq t \leq T$.
I did the following :
Let $\Bbb{Q} \in \mathcal{A}$. A corollary of the Radon-Nikodym theorem is that for a given a probability measure $\Bbb{Q}$, the associated Radon-Nikodym derivative $ Z = \frac{d\Bbb{Q}}{d\Bbb{P}}$ exists if and only if $\Bbb{Q} \gg \Bbb{P}$. As a consequence, any element of $\mathcal{A}$ is absolutely continuous with respect to $\Bbb{P}$. Hence $Z$ is a strictly positive $\Bbb{P}$-martingale. $Z_0$ is $\mathcal{F_0}$ -measurable and $\Bbb{E}[Z_0] = 1$. But $\mathcal{F_0}$ is the trivial $\sigma$-algebra, hence $Z_0 \equiv 1$. By the martingale representation theorem, there exists $\vartheta \in L^2_{loc}(W)$ such that $$Z_t = \Bbb{E}[Z_t] + \int_\Bbb{R} \vartheta(s) dW_s = 1 + \int_\Bbb{R} \frac{\vartheta(s)}{Z_s}Z_s dW_s = 1 + \int_\Bbb{R} \psi(s)Z_sdW_s$$ where $\psi = \frac{\vartheta}{Z}$. This gives us $$dZ = Z \psi dW$$ with $Z_0 = 1$ and we know that the unique solution to this stochastic differential equation is the stochastic exponential, i.e $$Z_t = \mathcal{E} \Big( \int \psi dW \Big)_t = \Sigma^\psi(t)$$ But now I don't know how to prove that $\Sigma^\psi(t)$ is square-integrable with respect to $\Bbb{P}$ for all $t \in [0,T]$.
Thank you in advance for your help.