Let $(\Omega, (\mathcal F\_t), \mathbb P)$ denote the usual Wiener space where $\Omega = C[0,\infty)$, etc., and where $(W\_t)\_{t \geq 0}$ denotes the Wiener process.
Let $Z \in L^1(\mathbb P)$ with $Z > 0$, $\mathbb P$-almost surely. Let $Z\_t = \mathbb E [ Z | \mathcal F\_t ]$ denote the density process, which is a positive UI martingale.


By e.g. [Kallenberg, 2002, Lemma 18.21], there exists a process $\theta\_t$ such that
$Z\_t = \exp \left( \int\_0^t \theta\_s \ d W\_s - \frac 1 2 \int\_0^t \theta\_s^2 \ ds \right)$ for all $t \geq 0$.


Let $\mathbb Q$ be defined, as usual, by $d \mathbb Q/d \mathbb P = Z$. If $\mathbb E^{\mathbb Q} \left[ \int\_0^{\infty} \theta\_t^2 \ d t \right] <\infty$, it is not hard to show (see below) that the relative entropy
$\mathcal H(\mathbb Q;\mathbb P) := \mathbb E^{\mathbb Q} \left[ \ln Z \right]$ is finite and is in fact given by
$\mathcal H(\mathbb Q;\mathbb P) = \frac 1 2 \mathbb E^{\mathbb Q} \left[ \int\_0^{\infty} \theta\_t^2 \ d t \right]$.


I do not succeed in showing the reverse direction, i.e. that finite relative entropy implies square integrability of $(\theta\_t)$, (unless $Z$ is bounded from above and below by a constant). I expect that there exists a counterexample to the general statement. 



> 
> Can you provide such a counterexample?
> 
> 
> 


**Proof of $\mathcal H(\mathbb Q;\mathbb P) = \frac 1 2 \mathbb E^{\mathbb Q} \left[ \int\_0^{\infty} \theta\_t^2 \ d t \right]$ if $\mathbb E^{\mathbb Q} \left[ \int\_0^{\infty} \theta\_t^2 \ d t \right] <\infty$:**


\begin{align\*} \mathbb E^{\mathbb Q} | \ln Z | & = \mathbb E^{\mathbb Q} \left| \int\_0^{\infty} \theta\_s \ d W\_s - \frac 1 2 \int\_0^{\infty} \theta\_s^2 \ d s \right| \\
& \leq \left( \mathbb E^{\mathbb Q} \left[\int\_0^{\infty} \theta\_s^2 \ ds \right] \right)^{1/2} + \frac 1 2 \mathbb E^{\mathbb Q} \left[ \int\_0^{\infty} \theta\_s^2 \ d s \right] < \infty.\end{align\*}
In particular, $\ln Z$ is $\mathbb Q$-integrable, so that $\mathcal H(\mathbb Q;\mathbb P) = \mathbb E^{\mathbb Q} \ln Z < \infty$.
Furthermore,
\begin{align\*}
\mathcal H(\mathbb Q;\mathbb P) - \frac 1 2 \mathbb E^{\mathbb Q} \left[\int\_0^{\infty} \theta\_t^2 \ d t \right]& = \mathbb E^{\mathbb Q} \left[ \ln Z - \frac 1 2 \int\_0^{\infty} \theta\_t^2 \ d t \right] \\
& = \mathbb E^{\mathbb Q} \left[ \int\_0^{\infty} \theta\_t \ d W\_t - \int\_0^{\infty} \theta\_t^2 \ d t \right] \\
& = \mathbb E^{\mathbb Q} \left[\int\_0^{\infty} \theta\_t \ d W\_t^{\mathbb Q} \right] = 0,
\end{align\*}
where the expectation of the stochastic integral with respect to $W^{\mathbb Q} = W - \int\_0^{\cdot} \theta\_t \ d t$ vanishes since, by $\mathbb Q$-square integrability of $\theta$, this is in fact the expectation of a martingale that is null at zero.



Example of Girsanov change of density with finite relative entropy, but with infinite integral over squared changed drift
I can now answer my own question. There exists no such counterexample. I can show that finite relative entropy in this setting implies $\mathbb E^{\mathbb Q} \int\_0^{\infty} \theta\_t^2 \ d t < \infty$. This can be proven by a localization argument using stopping times $(\tau\_n)$, such that $Z^{\tau\_n} \geq \frac 1 n$ and $[Z]^{\tau\_n} \leq n^2$.


