Slutsky's Theorem

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If $X_n\rightarrow X$ in distribution, and $Y_n\rightarrow0$ in distribution, then
(1) $X_n+Y_n\rightarrow X$ in distribution;
(2) $X_nY_n\rightarrow 0$ in distribution.


$\bullet$ Proof.

(1) Recall that $Y_n\overset{d}{\rightarrow}c\implies Y_n\overset{P}{\rightarrow}c$ for constant $c$.  Let $$f\in C_K=\{\mbox{functions are continuous in a compact set and vanish outside.}\}$$ Then $|f|\leq M<\infty$ and $f$ is uniformly continuous, i.e. $$\forall\,\varepsilon>0,\,\exists\,\delta>0\mbox{ s.t. }|x-y|\leq\delta\implies|f(x)-f(y)|\leq\varepsilon.$$Consider $$\begin{array}{l}|\mathscr{E}\{f(X_n+Y_n)\}-\mathscr{E}\{f(X_n)\}|\\
\quad\leq\mathscr{E}\{|f(X_n+Y_n)-f(X_n)|\}\\
\quad= \mathscr{E}\{|f(X_n+Y_n)-f(X_n)|I[|f(X_n+Y_n)-f(X_n)|\leq\varepsilon]\}\\
\quad\quad+\mathscr{E}\{|f(X_n+Y_n)-f(X_n)|I[|f(X_n+Y_n)-f(X_n)|>\varepsilon]\}\\
\quad\leq\varepsilon\mathscr{P}\{|f(X_n+Y_n)-f(X_n)|\leq\varepsilon\}\\
\quad\quad+2M\mathscr{P}\{|f(X_n+Y_n)-f(X_n)|>\varepsilon\}\;\mbox{by }|f|\leq M\\
\quad\leq\varepsilon+2M\mathscr{P}\{|Y_n|>\delta\}\quad\mbox{by uniformly continuity}\\
\quad\rightarrow\varepsilon\quad\mbox{by }Y_n\overset{P}{\rightarrow}0.\end{array}$$Thus we have $$\underset{n\rightarrow\infty}{\lim}\mathscr{E}\{f(X_n+Y_n)\}=\underset{n\rightarrow\infty}{\lim}\mathscr{E}\{f(X_n)\}=\mathscr{E}\{f(X)\}$$since $X_n\overset{d}{\rightarrow}X\iff\mathscr{E}\{f(X_n)\}\rightarrow\mathscr{E}\{f(X)\}$ if $f\in C_K$.  By the same reason, we also have $$\mathscr{E}\{f(X_n+Y_n)\}\rightarrow\mathscr{E}\{f(X)\}\iff X_n+Y_n\overset{d}{\rightarrow}X.$$

(2) Given $\varepsilon>0$, choose $A_0$ such that $A_0$ is the continuous point of $F_X$ and  is large enough so that $$\underset{n\rightarrow\infty}{\lim}\mathscr{P}\{|X_n|>A_0\}=\mathscr{P}\{|X|>A_0\}<\varepsilon.$$  This also means that there exists $n_0(\varepsilon)$ large such that $\mathscr{P}\{|X_n|>A_0\}<\varepsilon$ for $n>n_0(\varepsilon)$.  Choose $A\geq A_0$ and $A$ is free from $\varepsilon$, we have $\mathscr{P}\{|X_n|>A\}<\varepsilon$ for large $n$. Then $$\begin{array}{rl}\mathscr{P}\{|X_nY_n|>\varepsilon\}
& = \mathscr{P}\{|X_nY_n|>\varepsilon,\,|X_n|>A\}+\mathscr{P}\{|X_nY_n|>\varepsilon,\,|X_n|\leq A\}\\
&\leq\mathscr{P}\{|X_n|>A\}+\mathscr{P}\{|Y_n|>\frac{\varepsilon}{A}\}\\
&\leq\varepsilon+\mathscr{P}\{|Y_n|>\frac{\varepsilon}{A}\}\\
&\rightarrow\varepsilon\quad\mbox{by }Y_n\overset{P}{\rightarrow}0.\end{array}$$ We have $$X_nY_n\overset{P}{\rightarrow}0\implies X_nY_n\overset{d}{\rightarrow}0.$$

$\Box$