I am coming across a paper ( Proposition $1.1$ from <http://www.sciencedirect.com/science/article/pii/0304414987901840> ) that claims the following fact which I don't understand why:



> 
> On a probability space $(\Omega, \mathcal{F} , \mathbb{P})$ with a Brownian motion $(B\_t)\_{t \in [0,T]}$, let $(C\_t)\_{t \in [0,T]}$ be a bounded process and $(Y\_t)\_{t \in [0,T]}$ be a process defined by
>  $$ Y\_t = Y\_0 + B\_t + \int\_0^t C\_s \,ds,\quad 0\leq t \leq T,$$
>  where $Y\_0$ is a random variable independent of $(B\_t)\_{t \in [0,T]}$.
>  Then it claims that $(Y\_t)$ has a density function for every $t \in [0,T]$, by the Girsanov's theorem.
> 
> 
> 


Let $\mathbb{Q}$ be an equivalent measure to $\mathbb{P}$ defined by 
$$ \frac{d\mathbb{Q}}{d\mathbb{P}} = \exp \bigg\{ -\int\_0^T C\_s \, dB\_s - \frac{1}{2} \int\_0^T C^2\_s \,ds \bigg\}. $$ Then, by the Girsanov's theorem, $\{Y\_t - Y\_0 \}\_{t \in [0,T]}$ is a $\mathbb{Q}$- Brownian motion. This means that
$$ \mathbb{P} (Y\_t \leq y) = \mathbb{E}^{\mathbb{Q}} \bigg[ \frac{d\mathbb{P}}{d\mathbb{Q}} \mathbf{1}\_{ \{ Y\_t \leq y \} } \bigg] = \mathbb{E}^{\mathbb{Q}} \bigg[ \exp \bigg\{ \int\_0^T C\_s \, dB\_s + \frac{1}{2} \int\_0^T C^2\_s \,ds \bigg\} \mathbf{1}\_{(- \infty ,y]} (Y\_t) \bigg]. $$
How can we proceed from here? I don't know the distribution of $\exp \bigg\{ \int\_0^T C\_s \, dB\_s + \frac{1}{2} \int\_0^T C^2\_s \,ds \bigg\}$, nor do I know anything about the distribution of $Y\_0$. I am very confused. Any ideas?



Girsanov theorem and the density of a process
Short answer: it follows directly from the Radon-Nikodym theorem.


Longer answer: Let $\mu$ denote the law of $Y(t)$ under $\mathbb P\_{y\_0}$ and let $\nu$ denote the law of $Y(t)$ under $\mathbb Q\_{y\_0}$ for some initial condition $y\_0$.
From your application of Girsanov theorem, $\nu(A) = 0$ implies $\mu(A) = 0$. Therefore by the Radon-Nikodym theorem, $\mu$ admits a density $\rho(y)$ with respect to $\nu$. Of course, $\nu$ is just the Gaussian distribution centered at $y\_0$ with variance $t$, with density function $f(y)$ with respect to Lebesgue measure say. Now $\rho(y) f(y)$ is the density of $\nu$ with respect to Lebesgue measure. If $y\_0$ is not fixed but has a distribution $\gamma$, an analogous argument holds, basically replacing $y\_0$ above by $\gamma$, and noting that
$$ \nu(dy) = \frac 1 { \sqrt {2 \pi}} \left\{ \int\_{\mathbb R} \exp(-(y-\eta)^2/(2 t) ) \gamma(d\eta)\right\} \ d y.$$


