$\text { Consider the } \operatorname{SDE} d X_{t}=\operatorname{sign}\left(X_{t}\right) d t+d B_{t} \text { on } 0 \leq t \leq T, \text { where } \operatorname{sign}(x)=1\\ \text { for } x>0 \text { and } \operatorname{sign}(x)=-1 \text { for } x \leq 0 \text { . Show that it has a weak solution. }$
I am not sure if this is the way I am suppose to answer the question but this is how I did it:
$d x_{t}-\operatorname{sign}\left(x_{t}\right) d t=d B_{t}$
$\text { let } Y_{t}=\int_{0}^{t}-\operatorname{sign}\left(X_{s}\right) d s+\int_{0}^{t} d X_{s}$
We can see here that $\int_{0}^{t}\left|\operatorname{sign}\left(x_{s}\right)\right| d s<\infty$ as sign$(X_s)$ is bounded between -1 and 1.
We can also see that $\int_{0}^{t}\left(d x_{s}\right)^{2}=T<\infty$
Hence $Y_t$ is well defined and is a martingale as $[y, y]_{t}=\int_{0}^{t} \operatorname{sign}^{2}\left(X_{s}\right) d s=T \text { and }$ $\int_{0}^{t} d X_{S}=\int_{0}^{t} d t=T$
Therefore $[Y, Y]_{t}$ is a constant value which is a martingale. And $Y_t$ is Brownian Motion that satisfies the SDE $dX_t$