Let $X$ and $Y$ be two families of random variables, such that $X$ is uniformly-integrable and $Y \rightarrow c$ in probability to $c \in \mathbb{R}$. I have to show that for all $\epsilon >0, $ $$ P(|X_n(Y_n -c)| > \epsilon) \rightarrow 0~\text{in probability.}$$
I would really appreciate any hints or tips.
Let $M>0$. $|X_n(Y_n-c)|> \epsilon$ implies $|X_n| >M$ or $|Y_n-c| >\frac {\epsilon} M$. Uniform integbrability implies that given $\eta >0$ we can find $M$ such that $P(|X_n| >M)<\eta$ for all $n$. Hence $P(|X_n(Y_n-c)|> \epsilon) <\eta+ P(|Y_n-c| >\frac {\epsilon} M$. The second term tends to $0$ so the proof is complete.