I have a question about the following version of the Girsanov theorem.
Girsanov's Theorem Let $\theta_t$ be an adapted process satiysfying the Novikov condition. Define a probability measure $\mathbb Q$ by $\frac{d\mathbb Q}{d\mathbb P}\Big|_{\mathcal F_t} = \mathcal E(\theta_t)$. Then the stochastic process $t \mapsto W_t - \int_0^t \theta_s ds$ is a standard Wiener process under the measure $\mathbb Q$.
From my understanding, the setting is as follows. We have a filtered probability space $(\Omega, (\mathcal F_t)_t, \mathbb P)$, some index set $T$ and some state space $S$ and the process $\theta$ is a function $\theta: \Omega \times T \to S$ such that each $\theta(\cdot,t) = \theta_t: \Omega\to S$ is measurable.
What I am confused about is the $t$-dependency of the density $\frac{d\mathbb Q}{d\mathbb P}\Big|_{\mathcal F_t}$. For each value of $t$, this defines $\mathbb Q|_{\mathcal F_t}$ via $\theta(\cdot,t)$ as $\mathbb Q(U) = \mathbb E_{\mathbb P}[1_U \mathcal E(\theta_t)] $ for $U\in \mathcal F_t$. But why is this consistent? Specifically, if $s>t$, then $U\in \mathcal F_t\subset \mathcal F_s$ and so $$\mathbb Q(U) = \mathbb E_{\mathbb P}[1_U \mathcal E(\theta_t)] = \mathbb E_{\mathbb P}[1_U \mathcal E(\theta_s)].$$ But this can only be true if $\theta_s|_U = \theta_t|_U$ for each $U\in \mathcal F_t$. Particularly for $U$ and $\Omega \setminus U$. Did I make a mistake when applying the definitions?