Proof*. We define the following sequences of functions \begin{align*} m_k(\bb x) &= \inf\{ h(\bb y)+ k \|\bb x-\bb y\|: \bb y\in\R^d\}, \quad k\in\mathbb{N}\\ M_k(\bb x) &= \sup\{ h(\bb y) - k \|\bb x-\bb y\|: \bb y\in\R^d\}, \quad k\in\mathbb{N}. \end{align*} Since (i) $m_k(\bb x)\leq m_{k+1}(\bb x)$, (ii) $h_{k+1}(\bb x)\leq h_k(\bb x)$, (iii) $m_k(\bb x)\leq h(\bb x)$ ($h(\bb x)=\{ h(\bb y)+ k \|\bb x-\bb y\|: \bb y\in\R^d\}$ for $\bb y=\bb x$), and similarly (iv) $h\leq M_k$, we have \begin{align*} m_1(\bb x) &\leq m_2(\bb x) \leq \cdots \leq h(\bb x) \leq \cdots\leq M_2(\bb x)\leq M_1(\bb x). \end{align*}

We note the following.

1. Since monotonic limits of bounded sequences necessarily converge, the limits $\lim_{k\to\infty} m_{k}(\bb x)$ and $\lim_{k\to\infty} M_k(\bb x)$ are well defined and \[\lim_{k\to\infty} m_k(\bb x) \leq h(\bb x)\leq \lim_{k\to\infty} M(\bb x).\]
2. The functions $m_k,M_k$ are continuous since \begin{align*} m_k(\bb x) &= \inf\{ h(\bb y)+ k \|\bb x-\bb y\|: \bb y\in\R^d\} \\ &\leq \inf\{ h(\bb y)+ k \|\bb z-\bb y\|: \bb y\in\R^d\} + \|\bb y - \bb z\|\\ &= m_k(\bb z)+k\|\bb y - \bb z\|, \end{align*} implying that \[ |m_k(\bb x)-m_k(\bb z)| \leq k\|\bb y - \bb z\|\] (and similarly in the case of $M_k$).
3. Since $h$ is bounded, $m_k$ and $M_k$ are bounded as well ($m_k$ is an infimum of a bounded function plus a non-negative term and similarly for $M_k$).
4. If $\bb x$ is a continuity point of $h$, then \[\lim_{k\to\infty} m_k(\bb x) = h(\bb x)= \lim_{k\to\infty} M(\bb x).\] This follows from the following argument. For $\epsilon>0$, select $\delta$ such that $\|\bb x-\bb y\|<\delta$ implies $|h(\bb x)-h(\bb y)|<\epsilon$. Then if $r>(h(\bb x)-\inf_{\bb w} h(\bb w))/\delta$, we have for an arbitrary $\epsilon>0$ \begin{align*} \lim_k m_k(\bb x) &\geq m_r(\bb x)\\ &=\min\left\{\inf_{\bb y:\|\bb x-\bb y\|<\delta} h(\bb y) - r\|\bb x-\bb y\|,\,\, \inf_{\bb y:\|\bb x-\bb y\|\geq \delta} h(\bb y) - r\|\bb x-\bb y\| \right\} \\ &\geq \min\left\{ h(\bb x)-\epsilon, \inf_{\bb w} h(\bb w) + (h(\bb x) - \inf_{\bb w} h(\bb w))\frac{\delta}{\delta} \right\} = h(\bb x)-\epsilon, \end{align*} and similarly in the case of $M_k$.
5. Observation 4 above, together with Equation (*) implies \[\E\lim_k m_k(\bb X)=\E(h(\bb X))=\E\lim_k M_k(\bb X).\]

Observation 5 above, together with monotone convergence theorem (Proposition F.3.4) give \begin{align*} \E m_k(\bb X) &\nearrow \E \lim_k m_k(\bb X) = \E(h(\bb X)),\\ \E M_k(\bb X) &\searrow \E \lim_k M_k(\bb X)=\E(h(\bb X)), \end{align*} implying that for all $\epsilon>0$, there exists $r$ such that whenever $k>r$, we have $\E (m_k(\bb X)-M_k(\bb X)) < \epsilon$.

Proof of The Portmanteau Theorem*. Statement 4 implies statement 3 since continuous functions are measurable. Statement 3 implies statement 2 since continuous function on a compact set (in $\R^d$ a compact set is a closed and bounded set) are bounded. It remains to show that $(4)\Rightarrow (1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4)$.

$(4) \Rightarrow (1)$: For every $\bb z$ that is a continuity point of $F_{\bb X}$ we construct a function $h_{\bb z}$ such that $\E(h_{\bb z}({\bb X}^{(n)}))\to\E(h_{\bb z}(\bb X))$ implies $F_{{\bb X}^{(n)}}(\bb z)\to F_{\bb X}(\bb z)$. This will prove the claim that ${\bb X}^{(n)}\tood \bb X$. Specifically, we define $h_{\bb z}(\bb x)=1$ if $\bb x\leq \bb z$ (the inequality holds for all components of the two vectors) and $h_{\bb z}(\bb x)=0$ otherwise. Since $\E h_{\bb z}(\bb x)=F_{\bb x}(\bb z)$, the convergence $\E(h_{\bb z}({\bb X}^{(n)}))\to\E(h_{\bb z}(\bb X))$ implies $F_{{\bb X}^{(n)}}(\bb z)\to F_{\bb X}(\bb z)$. The functions $h_{\bb z}(\bb x)$ satisfy the conditions of (4): they are bounded and measurable, and since \[ \lim_{\epsilon\to 0} F_{\bb X}(\bb z+\epsilon\cdot (1,\cdots,1)) - F_{\bb X}(\bb z-\epsilon\cdot(1,\cdots,1)) =0 \] (recall that $\bb z$ is a continuity point of $F_{\bb X}$), the set of points at which $h_{\bb z}$ is discontinuous has probability 0 (under the distribution of $\bb X$).

$(1) \Rightarrow (2)$: We assume ${\bb X}^{(n)}\tood \bb X$ and consider a continuous function $h$ that is zero outside a bounded and closed set $C$. Since $C$ is compact, and continuous functions on a compact set are uniformly continuous (Proposition B.3.5), $h$ is uniformly continuous. Given $\epsilon>0$, we can find $\delta>0$ such that $\|\bb x-\bb y\| < \delta$ implies $\|h(\bb x)-h(\bb y)\| < \epsilon$. Since $C$ is bounded we can obtain a finite partition $Q$ of $C$ into rectangular cells of the form $\{\bb x: \bb a < \bb x\leq \bb b\}$ such that the distance between points within each cell is less than $\delta$. We can also assume, without loss of generality, that the boundaries between the cells have probability zero under $\bb X$ (since there can only be a countable number of regions with positive probability).

Based on this partition, we can define a simple function (function that takes on finitely many values) whose values are constant on the partition cells \begin{align*} h'(\bb x) = \sum_{i=1}^r a_i I_{A_i} (\bb x) \qquad \text{where} \qquad a_i=h\left(\max_{\bb x\in A_i}(x_1),\ldots,\max_{\bb x\in A_i}(x_d)\right). \end{align*} In fact, we can write the function $h'$ in the following form \[h'(\bb x)=\sum_{i=1}^m \beta_i I_{(-\bb\infty,{\bb b}_i]}(\bb x)\] using the method described in the proof of Proposition 8.2.3 (see also Figure 8.1). Note that by construction of the partition, each point ${\bb b}_i$ is a continuity point of $F_{\bb X}$ and therefore ${\bb X}^{(n)}\tood \bb X$ implies \begin{align} \lim_{n\to\infty} \E(h'({\bb X}^{(n)})) = \lim_{n\to\infty} \sum_{i=1}^m \beta_i F_{{\bb X}^{(n)}}(\bb x) = \sum_{i=1}^m \beta_i F_{{\bb X}}(\bb x) = \E(h'({\bb X})). \tag{**} \end{align} We finally note that \begin{align*} &|\E(h({\bb X}^{(n)}))-\E(h(\bb X))| \\ &= |\E(h({\bb X}^{(n)})) +\E(h'({\bb X}^{(n)}))-\E(h'({\bb X}^{(n)}))+\E(h'({\bb X}))-\E(h'({\bb X})) -\E(h(\bb X)) |\\ &\leq |\E(h({\bb X}^{(n)}))-\E(h'({\bb X}^{(n)}))| + |\E(h'({\bb X}^{(n)})) -\E(h'({\bb X}))| + |\E(h'({\bb X}))-\E(h(\bb X)) |\\ &\leq \epsilon + |\E(h'({\bb X}^{(n)})) -\E(h'({\bb X}))| + \epsilon \\ &\to 2\epsilon \qquad \text{ as } n\to\infty. \end{align*} Above, the first inequality follows from the triangle inequality, the second inequality follows from the uniform convergence of $h$ and the construction of the partition $Q$ and $h'$, and the convergence follows from (**). Since this holds for all $\epsilon>0$, we have $|\E(h({\bb X}^{(n)}))-\E(h(\bb X))|\to 0$ or $\E(h({\bb X}^{(n)}))\to\E(h(\bb X))$.

$(2) \Rightarrow (3)$: For an arbitrary continuous function $h$ such that $\|h(\bb x)\|\leq M$ and an arbitrary $\epsilon>0$ we will show that $|\E(h({\bb X}^{(n)}))-\E(h(\bb X))|\to \epsilon$. Since this holds for all $\epsilon>0$, this implies $\E(h({\bb X}^{(n)}))\to\E(h(\bb X))$.

It is possible to find $\alpha$ such that $\P(\|\bb X\|\geq \alpha) < \epsilon/(2M)$, and a continuous function $0\leq h'(\bb x)\leq 1$ such that $h'(\bb x)=0$ if $\|\bb x\|\geq \alpha+1$ and $h'(\bb x)=1$ if $\|\bb x\|\leq \alpha$. It follows that $\E(h'(\bb X))\geq 1-\epsilon/(2M)$ and \begin{align*} &|\E(h({\bb X}^{(n)}))-\E(h(\bb X))| \\ &= |\E(h({\bb X}^{(n)}))-\E(h(\bb X)) + \E(h({\bb X}^{(n)})h'({\bb X}^{(n)})) \\ &\quad -\E(h({\bb X}^{(n)})h'({\bb X}^{(n)}))+ \E(h({\bb X})h'({\bb X})) - \E(h({\bb X})h'({\bb X}))|\\ &\leq |\E(h({\bb X}^{(n)}))-\E(h({\bb X}^{(n)})h'({\bb X}^{(n)}))| +|\E(h({\bb X}^{(n)})h'({\bb X}^{(n)}))-\E(h({\bb X})h'({\bb X}))| \\ &\quad +|\E(h({\bb X})h'({\bb X})) - \E(h(\bb X))| \\ &\to |\E(h({\bb X}^{(n)}))-\E(h({\bb X}^{(n)})h'({\bb X}^{(n)}))| +0 +|\E(h({\bb X})h'({\bb X})) - \E(h(\bb X))|\\ &\to \epsilon/2+\epsilon/2=\epsilon. \end{align*} The first inequality above follows from the triangle inequality. The first convergence follows from statement (2) and the fact that $hh'$ is a continuous function that vanishes outside a bounded and closed set. The second convergence follows from \begin{align*} |\E(h({\bb X}^{(n)}))-\E(h({\bb X}^{(n)})h'({\bb X}^{(n)}))| &\leq \E(|h({\bb X}^{(n)})|\cdot | 1- h'({\bb X}^{(n)})|) \\ &\leq M \E(| 1- h'({\bb X}^{(n)})|) \\ &= M \E(1- h'({\bb X}^{(n)})) \\ &= M (1- \E h'({\bb X}^{(n)})) \\ &\to M (1- \E h'({\bb X})) \\ &\leq M \epsilon/(2M)=\epsilon/2 \end{align*} (with a similar bound applying to the term $|\E(h({\bb X}))-\E(h({\bb X})h'({\bb X}))|$).

$(3) \Rightarrow (4)$: Let $h$ be a bounded measurable function, for which $\P(\bb X\in \{\bb x: h\text{ is continuous at }\bb x\})=1$. Given $\epsilon>0$, we use the previous lemma to obtain continuous and bounded functions $m,M$ such that $m\leq h\leq M$ and \begin{align*} \E(h(\bb X)) - \epsilon &\leq \E m(\bb X) \\ &= \lim \E m({\bb X}^{(n)}) \\ &\leq \liminf \E( h ({\bb X}^{(n)}))\\ &\leq \limsup \E(h({\bb X}^{(n)})) \\ &\leq \lim \E M({\bb X}^{(n)}) \\ &= \E(M(\bb X))\\ &\leq \E(h(\bb X))+\epsilon. \end{align*} Since $\epsilon$ is arbitrary, $\liminf \E(h({\bb X}^{(n)}))=\limsup \E(h({\bb X}^{(n)}))$, and $\E(h({\bb X}^{(n)}))\to \E(h(\bb X))$.