Varied Type of Slutsky's Theorem (1): Converge in Probability

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If $X_n\rightarrow X$ and $Y_n\rightarrow Y$ both in probability, then
(1) $X_n\pm Y_n\rightarrow X\pm Y$ in probability;
(2) $X_nY_n\rightarrow XY$ in probability.


$\bullet$ Proof.

(1) For every $\varepsilon>0$, consider the probability $$\begin{array}{rl}\mathscr{P}\left\{|(X_n+Y_n)-(X+Y)|>\varepsilon\right\}&\leq\mathscr{P}\left\{|X_n-X|+|Y_n-Y|>\varepsilon\right\}\\ &=\mathscr{P}\left\{|X_n-X|+|Y_n-Y|>\varepsilon,|Y_n-Y|\leq\frac{\varepsilon}{2}\right\}\\ &\quad+\mathscr{P}\left\{|X_n-X|+|Y_n-Y|>\varepsilon,|Y_n-Y|>\frac{\varepsilon}{2}\right\}\\ &\leq\mathscr{P}\left\{|X_n-X|>\frac{\varepsilon}{2}\right\}+\mathscr{P}\left\{|Y_n-Y|>\frac{\varepsilon}{2}\right\}\\ &\rightarrow0+0\mbox{  as }n\rightarrow\infty.\end{array}$$Since $\varepsilon$ is arbitrary, we have $X_n+Y_n\rightarrow X+Y$ in probability.  Since $|Y_n-Y|=|Y-Y_n|$, we also have $X_n-Y_n\rightarrow X-Y$ in probability.

(2) For every $\varepsilon>0$, consider the probability $$\begin{array}{rl}\mathscr{P}\left\{|X_nY_n-XY|>\varepsilon\right\}&=\mathscr{P}\left\{|X_nY_n-XY_n+XY_n-XY|>\varepsilon\right\}\\ &\leq\mathscr{P}\left\{|Y_n||X_n-X|+|X||Y_n-Y|>\varepsilon\right\}\\ &=\mathscr{P}\left\{|Y_n||X_n-X|+|X||Y_n-Y|>\varepsilon,|X||Y_n-Y|\leq\frac{\varepsilon}{2}\right\}\\ &\quad+\mathscr{P}\left\{|Y_n||X_n-X|+|X||Y_n-Y|>\varepsilon,|X||Y_n-Y|>\frac{\varepsilon}{2}\right\}\\ &\leq\mathscr{P}\left\{|Y_n||X_n-X|>\frac{\varepsilon}{2}\right\}+\mathscr{P}\left\{|X||Y_n-Y|>\frac{\varepsilon}{2}\right\}.\quad(\star)\end{array}$$Thus, to show $X_nY_n\overset{p}{\rightarrow}XY$ is equivalent to show $\star\rightarrow0$.  For the first part of $\star$ and for this $\varepsilon$,we choose a sequence $A_n(\varepsilon)$ such that $\mathscr{P}\{|Y_n|>A_n(\varepsilon)\}<\varepsilon$.  Then for $A>\max_n{A_n(\varepsilon)}$, $$\begin{array}{rl}\mathscr{P}\left\{|Y_n||X_n-X|>\frac{\varepsilon}{2}\right\}&=\mathscr{P}\left\{|Y_n||X_n-X|>\frac{\varepsilon}{2},|Y_n|>A\right\}\\ &\quad+\mathscr{P}\left\{|Y_n||X_n-X|>\frac{\varepsilon}{2},|Y_n|\leq A\right\}\\ &\leq\mathscr{P}\left\{|Y_n|>A\right\}+\mathscr{P}\left\{|X_n-X|>\frac{\varepsilon}{2A}\right\}\\ &\rightarrow0+0\mbox{  as }n\rightarrow\infty. \end{array}$$Similarly, for the second part of $\star$, we choose $A$ such that $\mathscr{P}\{|X|>A\}<\varepsilon$.  Then we have $$\mathscr{P}\left\{|X||Y_n-Y|>\frac{\varepsilon}{2}\right\}\rightarrow0.$$Thus, $$\star\rightarrow0\implies X_nY_n\overset{p}{\rightarrow}XY.$$

$\Box$