I have a question on [Lawler – Notes on the Bessel process](http://www.math.uchicago.edu/%7Elawler/bessel18new.pdf), on page 4. Let $X\_t$ be one-dimensional Brownian motion, and we want to use $N\_t$ as a measure-changing (local) martingale, defined as $$N\_t=\left(\frac{X\_t}{X\_0}\right)^a\exp\left(-\frac{a(a-1)}{2}\int\_0^t\frac{ds}{X\_s^2}\right),\quad t<T\_0,$$


up to the first visit to zero. It satisfies the SDE $$dN\_t=\frac{a}{X\_t}N\_tdX\_t,\quad N\_0=1.$$


Then it is supposed in the note that $0<\varepsilon<X\_0<R$, and $\tau=T\_{\varepsilon}\wedge T\_R$, as up to the first visit to $\varepsilon$ or $R$. Then it claims that $N\_s$ stopped by $\tau$, $$dN\_{t\wedge\tau}=\frac{a}{X\_{t\wedge\tau}}1(t<\tau)N\_{t\wedge\tau}dX\_t,$$


is uniformly bounded. **But why is this true?**


I have a rough thought that $N\_{t\wedge\tau}=N\_0+\int\_0^t\frac{a}{X\_{s\wedge\tau}}N\_{s\wedge\tau}dX\_{s\wedge\tau}\leq N\_0+\frac{a}{\varepsilon}\int\_0^tN\_{s\wedge\tau}dX\_{s\wedge\tau}$, a.s. as $t<\tau$. Then **if we have the stochastic version of Grönwall's inequality,** i.e. if we can read $N\_{t\wedge\tau}\leq N\_0\exp(\frac{a}{\varepsilon}X\_{t\wedge\tau})$, a.s., then it is uniformly bounded for sure. But I am quite pessimistic about whether this holds or not.



Uniform boundedness of this SDE? And possibly a stochastic Grönwall inequality?
On the time interval $[0,\tau]$, the exponential factor in the definition of $N\_t$ is bounded below by $0$ and above by $\exp(Kt)$ for some constant $K=K(\epsilon,R)\ge 0$. Therefore $0<N\_{t\wedge\tau}\le(R/\epsilon)^a\exp(K\cdot (t\wedge \tau))$. This is "uniform" provided you ignore the dependence on $t$.


