Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned [Martingale Convergence][1] here

>Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied
>- $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
>- $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.


By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.




  [1]: https://almostsuremath.com/2009/12/20/martingale-convergence/