# Vague convergence of dirac measure

Task. Let $$x_n$$ be a series in $$\mathbb R^d$$. When does $$\mu_n = \delta_{x_n}$$ converge a) vaguely and when does it converge in b) variation distance?

Regarding a). We want to have

$$\int f \, d \mu_n = f(x_n) \rightarrow \int f \, d \mu$$

for all $$f: \mathbb R^d \rightarrow \mathbb R$$ continuous with a compact support, where $$\mu$$ is some probability measure.

But I don't really know how to continue from here. My conjecture is that such a $$\mu$$ exists if and only if $$x_n$$ converges against some $$x$$. The direction "$$\Leftarrow$$" is trivial because $$f$$ is continuous. But how about the other direction?

Regarding b). So I am not certain if I am allowed to use the formula from our lecture "for the discrete case" or not. The dirac measure is discrete but how about the convergence measure $$\mu$$? However, if I am allowed to use that formula, the following is my approach. We want:

$$d_{\text{TV}}(\mu_n,\mu)= \frac 1 2 \sum_{x \in \mathbb R^d} |\mu_n(\{x\})-\mu(\{x\})| \rightarrow 0$$

Which we can write as:

$$|1-\mu(\{x_n\})| + \sum_{x \in \mathbb R^d, x\neq x_n}|\mu(\{x\})| \rightarrow 0$$

Sice each term in this is $$\geq 0$$, each of them must converge to $$0$$. But that means that $$\mu$$ is 1 on $${x_n}$$ and $$0$$ everywhere else, meaning it is a dirac measure of some $$x$$ with $$x_n \rightarrow x$$ uniformly.

a) $$\mu_n$$ converges if and only if either

1. $$(x_n)_{n\in\mathbb{N}}$$ converges to some point $$x_* \in \mathbb{R}^d$$, or
2. $$\|x_n\| \to \infty$$ as $$n \to \infty$$.

By identifying $$(x_n)_{n\in\mathbb{N}}$$ as a sequence in the one-point compactification $$\mathbb{S}^d = \mathbb{R}^d \cup \{\infty\}$$, these two conditions can be combined into a single one:

Condition. $$(x_n)_{n\in\mathbb{N}}$$ converges in $$\mathbb{S}^d$$.

$$( \implies )$$ : Assume $$(\mu_n)_{n\in\mathbb{N}}$$ converges. Since $$\mathbb{S}^d$$ is compact, $$(x_n)_{n\in\mathbb{N}}$$ always have limit points, it suffices to show that it has a unique limit point.

For the sake of proof, assume otherwise that $$(x_n)_{n\in\mathbb{N}}$$ has more than one limit point. Then we can choose two subsequences $$(x_{n})_{n\in I}$$ and $$(x_{n})_{n\in J}$$ and $$a, b \in \mathbb{S}^d$$ with $$a \neq b$$ such that $$(x_n)_{n\in I} \to a$$ and $$(x_n)_{n\in J} \to b$$. Since $$a$$ and $$b$$ are distinct, at least one of them is not $$\infty$$ and we may assume $$a \neq \infty$$. Then we can choose $$f \in C_c(\mathbb{R}^d)$$ such that $$f(a) = 1$$ and $$f(b) = 0$$ (if $$b \neq \infty$$). Then

$$\left( \int f \, \mathrm{d}\mu_n \right)_{n\in I} \to 1 \qquad\text{but}\qquad \left( \int f \, \mathrm{d}\mu_n \right)_{n\in J} \to 0,$$

$$(\impliedby)$$ : Assume $$(x_n)_{n\in\mathbb{N}}$$ converges in $$\mathbb{S}^d$$, and let $$a$$ be the limit. Then for any $$f \in C_c(\mathbb{R})$$,
\begin{align*} \lim_{n\to\infty} \int_{\mathbb{R}^d} f \, \mathrm{d}\mu_n = \lim_{n\to\infty} f(x_n) = \begin{cases} f(a), & a \neq \infty \\ 0, & a = \infty \end{cases} \end{align*}
Hence, $$\mu_n \to \delta_{a}$$ vaguely if $$a \neq \infty$$ and $$\mu_n \to 0$$ vaguely if $$a = \infty$$.
b) If $$\mu_n \to \mu$$ in total variation distance, then $$|\mu_n(\mathbb{R}^d) - \mu(\mathbb{R}^d)| \to 0$$ and hence $$\mu$$ is a probability measure. Moreover, $$|1 - \mu(\{x_n\})| \to 0$$ and hence $$\mu(\{x_n\}) \to 1$$. Using this, it is easy to prove that $$(x_n)$$ is eventually constant. (Hint. Assume otherwise and show that this implies $$\mu$$ has total mass strictly greater than one.) Then $$\mu$$ is the Dirac mass concentrated at that point.
Conversely, if $$(x_n)$$ is eventually constant with the eventual value of $$a$$, then clearly $$\mu_n$$ converges to $$\delta_a$$ in total variation distance.