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.


1 Answer 1


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

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

contradicting the assumption.

$(\impliedby)$ : 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.


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