Before the proof of Girsanov's theorem, we were proving the following result in class:-
Lemma- Let $Q$ and $P$ be mutually locally equivalent probability measures on $(C[0,\infty),\mathcal{B}(C[0,\infty))$, and let $\alpha_t:= \frac{dQ_t}{dP_t}$. Here, $Q_t$ denotes $Q$ restricted to the sigma algebra $\mathcal{F}_t$, the sigma-algebra generated by the coordinate process upto time $t$. Then process $Y_t$ is a $Q$ - local martingale iff $\alpha_t Y_t$ is a $P$ - local martingale.
Suppose $Y$ is a $Q$ - local martingale. Let $T_n$ denote the usual reducing sequence for $Y$. Using the definition of the Radon-Nikodym derivative, we were able to show that $\alpha_tY_{t\wedge T_n}$ is a martingale. To show that $\alpha_{t\wedge T_n}Y_{t\wedge T_n}$ is a martingale, we used the optional stopping theorem, which is where my confusion is. To use the optional stopping theorem, we need to show that the process $\alpha_{t\wedge T_n}Y_{t\wedge T_n}$ is uniformly integrable. $Y$ is bounded until time $T_n$ and hence it suffices to show that $\alpha_{t\wedge T_n}$ is uniformly integrable. I am stuck here. I tried to do the following:- Choose $K$ large enough so that $\mathbb{P}_P[\alpha_t>K]\leq(1/K)\mathbb{E}_P[\alpha_t]=(1/K)\mathbb{P}_Q[\alpha_t]=(1/K)<\delta$. If $P$ and $Q$ were equivalent instead of locally equivalent, we could now have said that $\mathbb{E}_P[\alpha_t;\alpha_t>K]=\mathbb{P}_Q[\alpha_t>K]<\epsilon$, thereby giving uniform integrability, but we do not have equivalence of $P$ and $Q$ here. I think that maybe the fact that we have to prove uniform integrability of $\alpha$ stopped at $T_n$ instead of the whole process must be used.