I see some posts that are related to this one, e.g.

Borel Measures: Atoms (Summary)

I have a sort of particular question: I have one professor saying the following is true, while another says it's false. I think it's true. Neither have a justification haha

Suppose we take a Borel probability measure on a compact Euclidean space. Then can we approximate it by uniform atomic measures in the weak*-topology of Borel probability measures on X?

Specifically, if $\mu$ is a Borel probability measure on the compact Euclidean $X$, then does there exist a sequence of (not necessarily distinct!) points $( x_i )$ in X so that $$\frac{1}{n} \sum_{i = 1}^n \delta_{x_i} \xrightarrow{weak^*} \mu$$

where $\delta_{x_i}$ is the Dirac measure at $x_i$? The reason I think it's true is that we know this measure has a density by the Lebesgue Density Theorem.

  • $\begingroup$ Sorry, I don't follow! The relevant map(s) are those of $C(X)$, since we're in the weak*-topology. Calling those atomic measures $\mu_n$, weak* convergence means that for every $f \in C(X)$, we have $\int_Xf d\mu_n \rightarrow \int_X f d\mu$. I made a slight edit above to elucidate this. $\endgroup$ Apr 1, 2016 at 1:18
  • $\begingroup$ My impression of the necessary conditions here, btw, are the compactness, existence of a Borel-measurable density and separability. So can a Borel probability measure with density on a separable compact Hausdorff space always be approximated as so? $\endgroup$ Apr 1, 2016 at 1:33
  • $\begingroup$ Er, you might need 1st-countability (and thus 2nd) as well, would have to think back through a lot of theorems haha $\endgroup$ Apr 1, 2016 at 1:40

2 Answers 2


Here's one way to do it without getting your hands too dirty. It's not constructive.

Let ${\cal A}$ be the union of a countable dense set in $C_0({\mathbb R}^n)$, and indicators of all balls of integer radius centered at the origin (not necessary, but helpful later).

Let $\{X_1,X_2,\dots \}$ be an IID $\mu$-distributed sequence.

Let $\mu_n = \frac{1}{n} \sum_{j\le n} \delta_{X_j}.$

Then $\mu_n$ is a random probability measure. Observe that for any bounded and measurable function,

$$\int f d\mu_n= \frac {1}{n} \sum_{j\le n } f(X_j).$$

Then by LLN

$$ \lim_{n\to\infty} \int f d\mu_n = \int f d\mu ,~\forall f \in {\cal A}\quad (*),$$

a.s. (here is where we use the countability of ${\cal A}$).

Let $\{\bar \mu_n\}$ be a realization of $\{\mu_n\}$ (or equivalently, $\{X_1,\dots\}$) for which $(*)$ holds. It remains to show that $\{\bar \mu_n\}$ converges weakly.

Let $f\in C_b({\mathbb R}^d)$, by a standard procedure, for every $M$, there exists $f_M\in C_0({\mathbb R}^d)$ such that $f_M$ coincides with $f$ on $\{x:|x|<M\}$ and $|f_M |\le |f|$. Fix $\epsilon>0$ and choose an integer $M$ such that $\mu(\{x:|x|>M\})<\epsilon$

There exists a continuous $g_M\in {\cal A}$ such that $\|f_M-g_M\|<\epsilon$.

We then have

$$ \int f d \bar \mu_n = \int g_M d \bar \mu_n + \int (f_M -g_M) d \bar \mu_n + \int (f- f_M) d \bar \mu_n.$$

The first integral on the RHS converges to $\int g_M d\mu$ by construction. The second is bounded in absolute value by $\epsilon$. The third is bounded by $2\|f\|\mu_n (\{x:|x|>M\})$, which by assumption (recall the indicator functions in ${\cal A}$) converges to $2\|f\|\mu(\{x:|x|>M\})=2\|f\|\epsilon$.


$$ \limsup_{n\to\infty} |\int f d\bar \mu_n - \int g_M d\mu| \le \epsilon (1+2\|f\|).$$

But $$\begin{align*} |\int g_M d\mu - \int f d \mu| &\le \int |g_M -f_M| d \mu + \int |f_M -f| d\mu\\ & \le \epsilon + 2\|f\|\mu (\{x:|x|>M\})\le \epsilon (1+2\|f\|)\end{align*}.$$

The result follows.

  • $\begingroup$ Oh sweet, I like this proof actually! I think it's a hands-dirty proof, but it's not the one I was thinking. I was thinking to approximate atomic measures on an arbitrary sequence of finite sets (not necessarily a nested sequence) with a nested one, then use those, then diagonalize - er, so pretty dirty as well if the details get put down. Thanks again! $\endgroup$ Apr 1, 2016 at 23:57

Here is a constructive and short proof.

Let $0=y_0<y_1<\dots <y_n=1$ such that $y_k=k/n$ and $I_k=[y_{k-1},y_k)$ for $k\in[n-1]$ and $I_n=[y_{n-1},y_n]$; $\{I_k\}_k$ is a partition of the $[0,1]$ interval. Then, define $Z_k=F^{-1}_\mu(I_k)$ for $k\in[n]$, where $F_\mu$ is the distribution function related to $\mu$. Since $F_\mu$ is a distribution function, one can easily show that $\{Z_k\}_k$ is a partition of $\mathbb{R}$, where $Z_k$ is an interval of the form $(z_{k-1}=-\infty, z_k)$ for $k=1$, $[z_{k-1},z_k)$ for $k\in\{2,\dots ,n-1\}$ and $[z_{k-1},z_k=\infty)$ for $k=n$. Let $$F_{\mu_n}=\frac{n}{n-1}\sum_{k=1}^{n-1}y_{k}\chi_{Z_{k+1}}$$ One can easily show that $F_{\mu_n}$ is a distribution function to which we can relate a probability measure $\mu_n$ of the form $$\mu_n=\frac{1}{n-1}\sum_{k=1}^{n-1}\delta_{z_k}$$ Now, by construction $|F_\mu(x)-F_{\mu_n}(x)|\leq 1/n$ for all $x\in\mathbb{R}$. Therefore $\mu_n\rightarrow \mu$ in distribution. Thus, the result follows from Pormanteau's theorem, which establish that, for probability measures, convergence in the weak*-topolgy is equivalent to convergence in distribution.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .