I have come across a statement I don't understand while reading the following post :https://almostsuremath.com/2010/05/03/girsanov-transformations/
Given a probability measure $\mathbb{P}$, we have an equivalent measure $\mathbb{Q}$, and so we can write $\mathbb{Q}=U\cdot \mathbb{P}$, for a strictly positive random variable $U$. Indeed, this $U$ is just the Radon Nikodym derivative $d\mathbb{Q}/d\mathbb{P}$. So using this we can define the martingale $$U_t = E[d\mathbb{Q}/d\mathbb{P}|\mathscr{F}_t].$$ Then from Lemma 1 below, we have that $$\mathbb{E}_{\mathbb{Q}} [d\mathbb{P}/d\mathbb{Q}|\mathscr{F}_t] =U_t^{-1}.$$ Thus $U_t^{-1}$ is a uniformly integrable martingale with respect to $\mathbb{Q}$ and so if we choose a cadlag version of $U^{-1}$ then by martingale convergence theorem , it will convergeto the limit $\mathbb{E}_{\mathbb{Q}}[d\mathbb{P}/d\mathbb{Q}|\mathscr{F}_\infty].$
However, I cannot figure out why $\sup_t U_t^{-1}$ will be finite as well. What result allows us to conclude this?
