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. $\endgroup$ – John Samples Apr 1 '16 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? $\endgroup$ – John Samples Apr 1 '16 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 $\endgroup$ – John Samples Apr 1 '16 at 1:40

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! $\endgroup$ – John Samples Apr 1 '16 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.