Proof of Slutsky's Theorem

Throughout, $\mathbb{E}'$ is expectation on $(\Omega',\mathcal{F}',P')$; $X_{n}+Y_{n}$, $Y_{n}X_{n}$, $X_{n}+c$ and $cX_{n}$ are random variables on $(\Omega,\mathcal{F},P)$, and $X+c$, $cX$ on $(\Omega',\mathcal{F}',P')$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii)); $d_{\mathbb{R}}(x,y)=|x-y|$ (The Distance on the Real Numbers (i), The Distance on the Real Numbers Is a Map with Values the Absolute Differences (i)), so the hypothesis on $(Y_{n})$ says $P(|Y_{n}-c|>\varepsilon)\to0$ for every $\varepsilon>0$ (Convergence in Probability (i)); and Convergence in Distribution through Expectations of Bounded Continuous and of Bounded Lipschitz Functions is applied with every $(\Omega_{n},\mathcal{F}_{n},P_{n})$ equal to $(\Omega,\mathcal{F},P)$ and with $(\Omega',\mathcal{F}',P')$ in place of $(\Omega,\mathcal{F},P)$.

Step 1 (reduction). By Convergence in Distribution through Expectations of Bounded Continuous and of Bounded Lipschitz Functions (ii), it suffices to show, for every bounded Lipschitz $f:\mathbb{R}\to\mathbb{R}$, that $\mathbb{E}f(X_{n}+Y_{n})\to\mathbb{E}'f(X+c)$ and $\mathbb{E}f(Y_{n}X_{n})\to\mathbb{E}'f(cX)$. Let $|f|\le M$ and let $L$ be a Lipschitz constant; the random variables $f(\cdot)$ below are bounded by $M$, hence integrable (Bounded Measurable Functions on a Finite Measure Space Are Integrable (i)).

Step 2 (the shifted and scaled sequences). Let $g(x)=x+c$ (resp. $g(x)=cx$). Since $|(x+c)-(y+c)|=|x-y|$ and $|cx-cy|=|c|\,|x-y|$ (The Distance on the Real Numbers Is a Map with Values the Absolute Differences (i), Rules of Arithmetic and Order in an Ordered Field (x)), $g$ is Lipschitz with constant $1$ (resp. $|c|$), so $f\circ g$ is bounded by $M$ and Lipschitz with constant $L$ (resp. $L|c|$), in particular continuous (A Lipschitz Map is Uniformly Continuous and Continuous (ii)). As $f(X_{n}+c)=(f\circ g)(X_{n})$ and $f(X+c)=(f\circ g)(X)$ (resp. $f(cX_{n})=(f\circ g)(X_{n})$ and $f(cX)=(f\circ g)(X)$), Convergence in Distribution through Expectations of Bounded Continuous and of Bounded Lipschitz Functions (ii) applied to $X_{n}\xrightarrow{d}X$ and the function $f\circ g$, gives

$$\mathbb{E}f(X_{n}+c)\to\mathbb{E}'f(X+c),\qquad\mathbb{E}f(cX_{n})\to\mathbb{E}'f(cX).$$

Step 3 (proof of (i)). For $\varepsilon>0$ let $D_{n}=\{|Y_{n}-c|>\varepsilon\}$. Off $D_{n}$ the Lipschitz bound gives $|f(X_{n}+Y_{n})-f(X_{n}+c)|\le L|Y_{n}-c|\le L\varepsilon$, and on $D_{n}$ this difference is at most $2M$ because $|f|\le M$. Hence $|f(X_{n}+Y_{n})-f(X_{n}+c)|\le L\varepsilon+2M\,\mathbf{1}_{D_{n}}$, and by Rules for Expectations, Variances and Covariances (v), Rules for Expectations, Variances and Covariances (iv), Rules for Expectations, Variances and Covariances (iii), Rules for Expectations, Variances and Covariances (ii) and Rules for Expectations, Variances and Covariances (i)

$$\bigl|\mathbb{E}f(X_{n}+Y_{n})-\mathbb{E}f(X_{n}+c)\bigr|\le L\varepsilon+2M\,P(D_{n}).$$

Now let $\delta>0$. Choose first $\varepsilon>0$ with $L\varepsilon<\delta/2$, and then, since $P(D_{n})\to0$, an $N_{1}\in\mathbb{N}$ with $2M\,P(D_{n})<\delta/2$ for every $n\ge N_{1}$; thus $|\mathbb{E}f(X_{n}+Y_{n})-\mathbb{E}f(X_{n}+c)|<\delta$ for $n\ge N_{1}$. By Step 2 there is $N_{2}\in\mathbb{N}$ with $|\mathbb{E}f(X_{n}+c)-\mathbb{E}'f(X+c)|<\delta$ for $n\ge N_{2}$, so

$$\bigl|\mathbb{E}f(X_{n}+Y_{n})-\mathbb{E}'f(X+c)\bigr|<2\delta\qquad(n\ge\max(N_{1},N_{2})).$$

As $\delta>0$ was arbitrary, $\mathbb{E}f(X_{n}+Y_{n})\to\mathbb{E}'f(X+c)$.

Step 4 (the $X_{n}$ are uniformly unlikely to be large). Let $\eta>0$. By Properties of Distribution Functions (iii) and Properties of Distribution Functions (iv) there are reals $m_{1},m_{2}>0$ such that the distribution function $F_{X}$ satisfies $F_{X}(t)<\eta/2$ for $t\le-m_{1}$ and $F_{X}(t)>1-\eta/2$ for $t\ge m_{2}$. Put $K=m_{1}+m_{2}+1>0$; since $-K\le-m_{1}$ and $K-1\ge m_{2}$, we have $F_{X}(-K)<\eta/2$ and $F_{X}(K-1)>1-\eta/2$. Pointwise on $\Omega'$,

$$\mathbf{1}_{\{|X|\ge K\}}\le\mathbf{1}_{\{X\le-K\}}+1-\mathbf{1}_{\{X\le K-1\}},$$

since the right side is nonnegative, and if $|X|\ge K$ then either $X\le-K$, where the right side is at least $1$, or $X\ge K>K-1$, where it equals $1$. Taking expectations on $(\Omega',\mathcal{F}',P')$ (Rules for Expectations, Variances and Covariances (i), Rules for Expectations, Variances and Covariances (ii), Rules for Expectations, Variances and Covariances (iii), Rules for Expectations, Variances and Covariances (iv)) gives

$$P'(|X|\ge K)\le F_{X}(-K)+1-F_{X}(K-1)<\eta .$$

The set $C_{K}=\{x\in\mathbb{R}:|x|\ge K\}$ is closed: its complement is the open ball $B_{d_{\mathbb{R}}}(0,K)$ (The Distance on the Real Numbers Is a Map with Values the Absolute Differences (i)), which is open (Open Ball in a Metric Space is Open (i)) and so belongs to the topology of open sets of $(\mathbb{R},d_{\mathbb{R}})$ (The Open Subsets of a Metric Space Form a Topology (i)). Hence Portmanteau Theorem on a Metric Space (iii), applied to the laws, $P_{X_{n}}\Rightarrow P_{X}$ (Convergence in Distribution of Random Elements of a Metric Space (i)), gives

$$\limsup_{n\to\infty}P(|X_{n}|\ge K)\le P'(|X|\ge K)<\eta ,$$

and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (v), with the margin $\eta-\limsup_{n}P(|X_{n}|\ge K)>0$ (the limit superior as in Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (i)), shows that $P(|X_{n}|\ge K)<\eta$ for all large $n$.

Step 5 (proof of (ii)). Let $\eta>0$, and take $K$ as in Step 4. Choose next $\varepsilon>0$ with $LK\varepsilon<\eta$, and let $D_{n}=\{|Y_{n}-c|>\varepsilon\}$. Off $D_{n}\cup\{|X_{n}|\ge K\}$ we have $|Y_{n}X_{n}-cX_{n}|=|Y_{n}-c|\,|X_{n}|\le\varepsilon K$ (Rules of Arithmetic and Order in an Ordered Field (x)), so $|f(Y_{n}X_{n})-f(cX_{n})|\le LK\varepsilon<\eta$ there; hence, as in Step 3,

$$|f(Y_{n}X_{n})-f(cX_{n})|\le\eta+2M\bigl(\mathbf{1}_{D_{n}}+\mathbf{1}_{\{|X_{n}|\ge K\}}\bigr),\qquad\bigl|\mathbb{E}f(Y_{n}X_{n})-\mathbb{E}f(cX_{n})\bigr|\le\eta+2M\,P(D_{n})+2M\,P(|X_{n}|\ge K).$$

Finally choose $N\in\mathbb{N}$ with $P(D_{n})<\eta$ and $P(|X_{n}|\ge K)<\eta$ for every $n\ge N$ (as $P(D_{n})\to0$, and by Step 4). Then $|\mathbb{E}f(Y_{n}X_{n})-\mathbb{E}f(cX_{n})|\le(1+4M)\eta$ for $n\ge N$, and with Step 2, as in Step 3, $|\mathbb{E}f(Y_{n}X_{n})-\mathbb{E}'f(cX)|<(2+4M)\eta$ for all large $n$. As $\eta>0$ was arbitrary, $\mathbb{E}f(Y_{n}X_{n})\to\mathbb{E}'f(cX)$.