A Borel measure $\mu$ on $\mathbb{R}^k$ can be represented by a "distribution function" $F : \mathbb{R}^k \to \mathbb{R}$ via $F(x_1, \dots, x_k) = \mu ((-\infty, x_1] \times \dots \times (-\infty, x_k])$. Note that $\mu$ need not be a probability measure. If we have a sequence of Borel measures $\mu_n$ on $\mathbb{R}^k$ then $\mu_n \to \mu$ vaguely is defined by $\int g d\mu_n \to \int g d\mu$ for all continuous functions $g : \mathbb{R}^k \to \mathbb{R}$ with compact support. Denote by $F_n$ the distribution functions of $\mu_n$. In the one-dimensional case $k = 1$ the Helly-Bray theorem states that
$\mu_n \to \mu$ vaguely if and only if $F_n(x) \to F(x)$ at all continuity points $x \in \mathbb{R}$ of $F$.
Does this theorem generalize to $\mathbb{R}^k$? I have only seen a "Helly-Bray theorem" for weak(!) convergence (with bounded continuous test functions) of probability(!) measures on $\mathbb{R}^k$ (proven via the Portmanteau theorem). Weak convergence and vague convergence coincide for probability measures, so I suppose that this theorem should be generalizable to vague convergence of arbitrary Borel measures on $\mathbb{R}^k$.
Best Answer
The Helly-Bray theorem also holds for $\mathbb{R}^n$.
"$\Rightarrow$": Assume that $\mu_n \to \mu$ vaguely. By the Portmanteau theorem for vague convergence, $\mu_n(B) \to \mu(B)$ for all bounded $\mu$-continuity Borel sets $B \subseteq \mathbb{R}^n$. For $i = 1, \dots, n$ denote by $D_i \subseteq \mathbb{R}$ the set of continuity points of the marginal measure $\mu_i$ on $\mathbb{R}$. Then $D_i$ is countable and $C := D_1^c \times \dots \times D_n^c$ is dense in $\mathbb{R}^n$. For any point $u \in C$, the set $(-\infty, u]$ is a $\mu$-continuity set. Therefore, $u$ is a continuity point of $F$. Any rectangular box $(a, b]$ with $a, b \in C$ is a $\mu$-continuity set. Any corner $u$ of $(a, b]$ is contained in $C$. With this in mind, let $x$ be a continuity point of $F$. We can decompose $(-\infty, x]$ into a countable collection of boxes $(a^k, b^k]$ with $a^j, b^j \in C$. Since all these boxes $(a^j, b^j]$ are $\mu$-continuity sets, we get $F_n(x) = \sum_j \mu_n(a^j, b^j] \to \sum_j \mu(a^j, b^j] = F(x)$ by the bounded convergence theorem.
"$\Leftarrow$": Assume that $F_n(x) \to F(x)$ for all continuity points $x$ of $F$. For a box $(a, b]$ it holds $\mu(a, b] = \Delta^a_b F$ which is an alternating sum over values $F(x)$ with $x$ a corner of $(a, b]$. If $a, b \in C$ then all the corners of $(a, b]$ are contained in $C$ and since $F$ is continuous on $C$ we get $\mu_n(a, b] = \Delta^a_b F_n \to \Delta^a_b F = \mu(a, b]$. Let $g : \mathbb{R}^n \to \mathbb{R}$ be continuous with compact support. Then $\textrm{supp}(g) \subseteq (a, b]$ for some $a, b \in C$. Let $\varepsilon > 0$. Since $g$ is uniformly continuous on $(a, b]$ and $C$ is dense in $\mathbb{R}^n$ we can partition $(a, b]$ into finitely many boxes $(a^j, b^j]$, $j = 1, \dots, m$ with $a^j, b^j \in C$ such that $\sup_{x \in (a^j, b^j]} |g(x) - g(b^j)| < \varepsilon$ for all $j$. Decompose $\int g d\mu = \sum_j \int_{(a^j, b^j]} g d\mu$. We can approximate
$$\left|\int g d\mu - \sum_j g(b^j) \mu(a^j, b^j]\right| = \left|\sum_j \int_{(a^j, b^j]} (g(x) - g(b^j)) \mu(dx)\right| \\ \leq \sum_j \sup_{x \in (a^j, b^j]} |g(x) - g(b^j)| \mu(a^j, b^j] < \varepsilon \cdot \mu(a, b]$$
and similarly for all the $\mu_n$. It follows
$$\left|\int g d\mu_n - \int g d\mu\right| \leq \left| \int g d\mu_n - \sum_j g(b^j) \mu_n(a^j, b^j]\right| + \left| \sum_j g(b^j)(\mu_n(a^j, b^j] - \mu(a^j, b^j])\right| \\ + \left| \int g d\mu - \sum_j g(b^j) \mu(a^j, b^j] \right| \leq 2\varepsilon + \lVert g \rVert \sum_j |\mu_n(a^j, b^j] - \mu(a^j, b^j]|.$$
As $n \to \infty$, the right-hand side converges to $0$ (the sum is finite) and we get $\limsup_n |\int g d\mu_n - \int g d\mu| \leq 2 \varepsilon$. Since this is true for all $\varepsilon$, $\int g d\mu_n \to \int g d\mu$. Therefore, $\mu_n \to \mu$ vaguely.