Proof of $O_P(1) o_P(1) = o_P(1)$

Question

Consider $U_n=O_P(1)$ and $X_n=o_P(1)$. Show that $U_nX_n=o_P(1)$.

Since $o_P(1)$ is equivalent to convergence in probability, it suffices to show that $P(|U_nX_n|>\varepsilon) \to 0$. However, I am not quite sure if my proof is right:

$$P(|U_nX_n|>\varepsilon) = P(|U_nX_n|>\varepsilon, |U_n| \leq M) + P(|U_nX_n|>\varepsilon, |U_n|>M).$$

For the first term on the right-hand side it holds:

$$P(|U_nX_n|>\varepsilon, |U_n| \leq M) \leq P(|U_n| \leq M) < \epsilon.$$

From the definition of $O_P(1)$ we know that for each $\epsilon$ such an $M$ exists.

For the second term, it holds that:

$$P(|U_nX_n|>\varepsilon, |U_n| > M) \leq P(|MX_n|> \varepsilon) = P\left(|X_n|> \frac \varepsilon M\right) \to 0.$$

Hence,

$P(|U_nX_n|>\varepsilon) < \epsilon$.

Is this correct?

Answer 1

It seems that you reversed the $\leq$ and $\gt$: it should be $$ P(|U_nX_n|>\varepsilon, |U_n| \color{red}{\gt} M) \leq P(|U_n| \color{red}{\gt} M) < \epsilon $$ and $$P(|U_nX_n|>\varepsilon, |U_n|\color{red}{\leq} M) \leq P(|MX_n|> \varepsilon) = P\left(|X_n|> \frac \varepsilon M\right) \to 0.$$