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?