How can we deduce uniqueness for SDEs by Girsanov's theorem? Let $\mu\in L^{\infty}(\mathbb{R};\mathbb{R})$ be a bounded deterministic function. Then my understanding is that by using Girsanov's theorem, we can deduce uniqueness (in law) for the following stochastic equation 
$$dX_t = dB_t + \mu(X_t)dt, \quad X_0\equiv 0.\quad(\star)$$ 
What we do is set $Y_t$ as the solution to $$dY_t =dW_t, \quad Y_0\equiv 0$$ where $W$ is an $(\mathcal{F}_t, P)$-BM. That is, $Y$ uniquely solves the driftless equation.
Then we define $$M_t:=\int_0^t\mu(W_s)dW_s$$ so $M$ is a $(\mathcal{F}_t, P)$-martingale. Then we define the probability measure $Q$ by $$\frac{dQ}{dP}\Big|_{\mathcal{F}_t}=\mathcal{E}(M)_t$$
Where $\mathcal{E}$ denotes the stochastic exponential of $M$, which is a martingale since $\mu$ was taken bounded.  
Then by Girsanov's theorem, we have that $$\tilde{W}_t:=W_t-\langle W,M\rangle_t=W_t-\int_0^t\mu(W_s)ds$$ is a $(Q, \mathcal{F}_t)$-BM.
We then simply observe that $dY_t=d\tilde{W}_t+\mu(Y_t)dt$. So that $(Y, \tilde{W})$ with $(Q,\mathcal{F}_t)$ solves SDE $(\star)$.

What I don't understand is how the above construction shows me that $(\star)$ has a unique solution. Any help at all would be much appreciated!
 A: There are two different notions of uniqueness for stochastic equations; in distribution and pathwise. For your equation there is actually uniqueness in both senses! 
Let us begin with distributional uniqueness: Assume we have a solution 
$$
X_t = \int_0^t \mu(X_s)ds + B_t
$$
and define 
$$
Z_t = \exp\{ -\int_0^t \mu(X_s) dB_s - \frac{1}{2} \int_0^t \mu^2(X_s)ds\} .
$$ 
By the Novikov condition, since $\mu$ is bounded, $Z$ is a proper martingale and if we define $dQ := Z_T dP$ then $X$ is a Brownian motion w.r.t. $Q$. Since $X$ satisfies the above SDE we can rewrite
$$
Z_t = \exp\{ -\int_0^t \mu(X_s) dX_s + \frac{1}{2} \int_0^t \mu^2(X_s)ds\} .
$$
Let $A$ be a measurable subset of $\mathbb{R}^d$. We get
\begin{align*}
P(X_t \in A) & = E_P[1_A(X_t)]  = E_Q [1_A(X_t) Z_T^{-1}] \\
& = E_Q[ 1_A(X_t)  \exp\{ \int_0^T \mu(X_s) dX_s - \frac{1}{2} \int_0^T \mu^2(X_s)ds\} ] \\
& = E_P[ 1_A(B_t)  \exp\{ \int_0^T \mu(B_s) dB_s - \frac{1}{2} \int_0^T \mu^2(B_s)ds\} ] ,\\
\end{align*}
so that the distribution of $X$ is completely determined by $\mu$ which proves the claim.
To see pathwise uniqueness, assume we have two solutions $X^1$ and $X^2$ defined on the same probability space. We have
$$
(X_t^2 - X_t^1)^+ = \int_0^t 1_{(X^2_s > X^1_s)} (\mu(X_s^2) - \mu(X_s^1)) ds
$$
so that 
\begin{align*}
X_t^1 \vee X_t^2 & = X_t^1 + (X_t^2  - X_t^1)^+ \\
& = \int_0^t \mu(X_s^1)ds + B_t + \int_0^t 1_{(X^2_s > X^1_s)} (\mu(X_s^2) - \mu(X_s^1)) ds \\
& = \int_0^t \mu(X_s^1) + 1_{(X^2_s > X^1_s)} (\mu(X_s^2) - \mu(X_s^1))ds + B_t \\
& = \int_0^t \mu(X_s^1 \vee X_s^2) ds + B_t
\end{align*}
so that also $X^1 \vee X^2$ is also a solution of the same equation. Then, from uniqueness in distribution $X^1$ and $X^1 \vee X^2$ have the same distribution. This is only true if $X^1 = X^1 \vee X^2 = X^2, P-a.s.$ which is exactly pathwise uniqueness.
Notice in the above that proof of distributional uniqueness is true in any dimension, but for pathwise uniqueness the proof relies heavily on $d=1$. Actually, there is also pathwise uniqueness when $d > 1$ (even when $d= \infty$) but the proof is much harder.
