Since $|Z_tY_t|\leq C Z_t$ and $Z$ is uniformly integrable, the continuous local martingale $ZY$ is locally in Doob's class and hence is a true martingale. For $A\in\mathcal F_s$ and $s\leq t$, <Bayes formula for conditional expectation> gives
$$
\widetilde{\mathbb E}[\mathbf1_A Y_t]
=\mathbb E[\mathbf1_AZ_\infty Y_t]
=\mathbb E[\mathbf1_AZ_tY_t]
=\mathbb E[\mathbf1_AZ_sY_s]
=\widetilde{\mathbb E}[\mathbf1_A Y_s].
$$
The bounded process $Y$ is integrable under $\widetilde{\mathbb P}$, so this is exactly the martingale property.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 202 / 2 / b / Solution

If $d\widetilde{\mathbb P}=Z\,d\mathbb P$ and $Z_t=\mathbb E[Z\mid\mathcal F_t]>0$, then
$$
\widetilde{\mathbb E}[X\mid\mathcal F_t]
=\frac{\mathbb E[ZX\mid\mathcal F_t]}{Z_t}.
$$


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