$X_n$ is bounded in probability and $Y_n$ converges to 0 in probability then $X_nY_n$ congerges to probablity with 0 I want to show : $X_n$ is bounded in probability and $Y_n \rightarrow 0$ in probability  then $X_nY_n \rightarrow 0 $ in probablity. 
I know the following definitions that is 
Definition 2.17 : We say that $X_n$ is bounded in probability if $X_n = O_P (1)$,
i.e. if for every $\epsilon$ > 0, there exist $M$ and $N$ such that $P(|Xn| < M) >
1 − \epsilon $ for $n > N$.
So I want to show that for every
$\epsilon$ > 0 and $\epsilon '$ > 0 there exist $M>0 $ and $n_o $ such that 
$P(|X_n| <M) > 1 -\epsilon/2$
$P(|X_n| <\epsilon /M) > 1 -\epsilon'/2$
for every $n>n_0$
Then I want to show that 
$P(|X_n| <M$ and $|Y_n| <\epsilon /M) > 1-\epsilon' $

$P(|X_nY_n| >\epsilon) \leq P(|X_n| \leq M, |X_nY_n| >\epsilon)+P(|X_n| > M, |X_nY_n| >\epsilon)\leq P(|Y_n| >\frac {\epsilon} M)+[1-P(|X_n|<M)]$. For $n >N$ the second term is less than $\epsilon$ and the first term tends to $0$ as $n \to \infty$.
