Show that a process is Brownian motion 
For a Brownian motion $(B_t)_{t \geq 0}$ consider
  $$
X_t=\int_0^t\sin\left(\frac{1}{B_s}\right)\;ds+B_t.
$$
  I want to show that $(X_t)_t$ is a Brownian motion. 

Clearly, one can check the definition of BM and invoke the fact that $\int_{t_i}^{t_{i+1}}\sin(B_s^{-1})\;ds$ is measurable wrt. $\sigma(\{B_s,t_i\leq s\leq t_{i+1}\})$ together with the Markov property. However, this might be a bit tedious. 
I was wondering whether I can show that $(X_t)_t$ is a martingale and therefore conclude with Levy's characterization.
 A: Let $(B_t)_{t \geq 0}$ be a Brownian motion on a probabiliy space $(\Omega,\mathcal{A},\mathbb{P})$. The process
$$X_t := \int_0^t  \sin \left( \frac{1}{B_s} \right) \,ds + B_t$$
is a Brownian motion with respect to some probability measure $\mathbb{Q}$ on $(\Omega,\mathcal{A})$ (this is a direct consequence of Girsaonov's theorem); as far as I can see, it is, however, not a Brownian motion on the original probability space.
To show that $(X_t)_{t \geq 0}$ is not a Brownian motion on $(\Omega,\mathcal{A},\mathbb{P})$ it clearly suffices to show that $(X_t)_{t \geq 0}$ is not a $\mathbb{P}$-martingale. Since $(B_t)_{t \geq 0}$ is a martingale, we know that $(X_t)_{t \geq 0}$ is a martingale iff
$$M_t := \int_0^t \sin \left( \frac{1}{B_s} \right) \, ds$$
is a martingale. Denote by $(\mathcal{F}_t)_{t \geq 0}$ the canonical filtration of $(B_t)_{t \geq 0}$. If we fix $s \leq t$, then it follows from the independence and stationarity of the increments of $(B_t)_{t \geq 0}$ that
$$\begin{align*} \mathbb{E}(M_t \mid \mathcal{F}_s) &= \int_0^s \sin(B_r^{-1}) \, dr + \int_s^t \mathbb{E}(\sin(B_r^{-1}) \mid \mathcal{F}_s) \, dr \\ &= M_s+ f(t-s,B_s) \end{align*}$$
where
$$f(u,x) := \int_0^{u} \mathbb{E} \left( \sin \left[ \frac{1}{x+ B_r} \right] \right) \, dr.$$
Thus, $(M_t)_{t \geq 0}$ is a martingale iff
$$f(t-s,B_s) = 0 \quad \text{$\mathbb{P}$-almost surely for all $s \leq t$}. \tag{1}$$
Clearly, we can choose $\delta_1,\delta_2>0$ such that
$$ \forall \left| x- \frac{\pi}{2} \right| \leq \delta_1, |y| \leq \delta_2: \quad \frac{1}{x+y} \in \left[ \frac{\pi}{4}, \frac{3\pi}{4} \right].$$
Moreover, the continuity of the sample paths of Brownian motion implies that there exists $u>0$ such that
$$\mathbb{P}(|B_r| \leq \delta_2) \geq \frac{3}{4} \quad \text{for all $r \in [0,u]$.}$$
This implies
$$\mathbb{E}\left( 1_{|B_r| \leq \delta_2} \sin \left[ \frac{1}{x+B_r} \right] \right) \, dr \geq \sin \left( \frac{\pi}{4} \right) \frac{3}{4}$$
for all $|x-\pi/2|\leq \delta_1$ and $r \in [0,u]$. On the other hand,
$$\left| \mathbb{E}\left( 1_{|B_r| > \delta_2} \sin \left[ \frac{1}{x+B_r} \right] \right) \right| \leq \mathbb{P}(|B_r|>\delta_2) = 1- \mathbb{P}(|B_r| \leq \delta_2)\leq \frac{1}{4}$$
for all $r \in [0,u]$. Hence,
$$f(u,x) \geq \int_0^u \left[ \sin \left( \frac{\pi}{4} \right) \frac{3}{4} - \frac{1}{4} \right] \, dr > \frac{u}{4} >0$$
for any $|x-\pi/2| \leq \delta_1$. Since $\mathbb{P}(|B_s-\pi/2| \leq \delta_1)>0$ for any $s>0$, this implies that $(1)$ does not hold true for $t :=s+u$.
