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For an article I am writing, I would like to know that two somewhat different looking conditions are in fact equivalent. Here is the setting. $X$ is a compact (and first countable) metric space and $\mu$ is a Radon probability measure on $X$. That is: $\mu$ is a measure on the $\sigma$-algebra of Borel serts of $X$ (the $\sigma$-algebra generated by the open sets), has total measure one, and is inner regular, that is the measure of any Borel set $B$ is the sup of the measures of the compact subsets of $B$.

Now let $\{x_n\}$ be a sequence in $X$. Let's say that this sequence is "$\mu$-uniformly-distributed-A" if for any open subset $O$ of $X$ $$ \mu(O) =\lim_{N \to \infty} { \#(O,N) \over N}$$ where $\#(O,N)$ is the number of $x_k$ in $O$ for $k = 1, \ldots, N$. (Or, in other words, the measure of $O$ is the "average number of the $x_n$ that are in $O$").

On the other hand, let's say that the sequence is "$\mu$-uniformly-distributed-B" if for any continuous real valued function $f : X \to R$, $$ \int f(x) \, d\mu = \lim_{N \to \infty} {1\over N}\sum_{k = 1}^N f(x_k)$$ (in other words the integral of $f$ is the "average value of $f$ on the $x_n$"). (Note that if we assume this equality not for all continuous functions but rather for all the characteristic functions of open sets, then it reduces to the definition of "$\mu$-uniformly-distributed-A".) So, as you have no doubt guessed, what I want to know is if "$\mu$-uniformly-distributed-A" and "$\mu$-uniformly-distributed-B" are in fact always equivalent.

It is well-known that for the special case where $X = [0,1]$ and $\mu$ is Lebesque measure the two are equivalent---see for example Theorem B of section 3.5 of Volume 2 of Knuth's "Art of Computer Programming"---but I don't see how to generalize the argument there. So does anyone know if this equivalence always does hold, and if so can they direct me to a proof in the literature.

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Minor quibble: every metric space is first countable and any compact metric space is second countable, so you needn't specify "(and first countable)". –  Pete L. Clark May 18 '11 at 3:48
    
Ouch ! I see where I went wrong now. For the case of Lebesgue measure on [0,1], the classic condition for a sequence to be uniformly distributed is: for any sub-interval O = (a,b) the ratio #(O,N/)N should approach b-a. It follows that the same is true for a FINITE union of intervals, but as Michael Renardy points out, it does NOT follow for an arbitrary open set O, even though O is a countable disjoint union of open intervals (its components). I'll have to rethink the correct definition of the ``A'' condition for the general case---presumably what Gerald Edgar says below. Thanks to all ! –  Dick Palais May 18 '11 at 4:12
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A common way to rephrase your condition A in a form that is equivalent to B, is to restrict to a particular class of open sets (that generates the Borel $\sigma$-algebra: when your space is $[0,1]$ with an absolutely continuous measure, the intervals are fine; when your space is a manifold with an a.c.m., you can take the interiors of connected domains with smooth boundary. What's important is that the characteristic functions are reasonnably approximable by continuous functions. –  Benoît Kloeckner May 18 '11 at 7:24
    
@Pete L. Clark: Thanks Pete, of course you are correct. I started out writing the question for the more general case of a 1st countable, locally compact Hausdorff space, and then for simplicity switched to metric---and forgot to remove the 1st countable. :-( –  Dick Palais May 18 '11 at 16:11
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5 Answers 5

They are not equivalent. Suppose $X = [0,1]$, $\mu$ is a unit mass at 0, and $x_n = 1/n$. This sequence is $\mu$-uniformly-distributed-B, because for any continuous $f$, $\int f(x) \, d\mu = f(0) = \lim_{N \to \infty} \frac{1}{N} \sum_{k=1}^N f(1/k)$.

However, it is not $\mu$-uniformly-distributed-A: take $O = (0,1]$

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Nice example. It indicates that in what I am doing I will need to restrict the class of measures I am dealing with. (I think I have to assume the boundaries of open sets should be null-sets.) –  Dick Palais May 18 '11 at 4:31
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Actually, for Lebesgue measure on the unit interval, the definition of "uniformly distributed A" does not seem to make sense. Consider any sequence $x_n$ and define $O$ to be the union of all intervals centered at $x_n$ with length $\epsilon_n$. Then all $x_n$ are in $O$ by construction, but we can make $\sum \epsilon_n$ as small as we want!

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A VERY good point ! And it tipped me off about where I had gone wrong in reinterpreting the B-condition. See my remark to the question Thanks. –  Dick Palais May 18 '11 at 4:19
    
@Michael Renardy: Note that if the x_n are uniformly distributed (in any sense) they are dense, so your open set O is an open dense set, so its boundary is its complement. But since O has small measure (< 1) this means its boundary has positive measure. What has been pointed out by others here (and is proved in Chapter 1 of Billingsley's book) is that in my definition of Uniformly-Distributed-A, I should have restricted to open sets whose boundary have measure zero, and then Uniformly-Distributed-A and Uniformly-Distributed-B would indeed be equivalent. Thanks again for your nice example. –  Dick Palais May 22 '11 at 7:35
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A sequence is $\mu-$ uniformly distributed -B iff the limit relation A holds for each Borel set $M$ whose boundary has $\mu-$ measure $0.$

Edit

In the meantime I found the following books which have proofs of this theorem:

"Uniform distribution of sequences" by L. Kuipers and H. Niederreiter, Wiley 1974

and P. Billingsley, "Convergence of probability measures", Wiley 1999.

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Yes, Billingsley's book has just what I needed. Many thanks. –  Dick Palais May 20 '11 at 16:46
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See discussion of "weak convergence" or "narrow convergence" of measures. Let $\mu_n$ be the measure with mass $1/n$ at each of $x_1,\dots,x_n$. Your condition B is what can be taken as the definition for: $\mu_n$ converges narrowly to $\mu$. The equivalent condition in terms of open sets is not your condition A, but rather $$ \mu(O) \le \liminf_{N \to \infty} \frac{\#(O,N)}{N} $$ for all open sets $O$.

Reference: Gilman & Jerison, Rings of Continuous Functions.

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On the other hand, A implies B. If A holds for open sets, it also holds for closed sets (simply consider complements on both sides of the equation). Every continuous function can be approximated uniformly by a linear combination of characteristic functions of open and closed sets (just divide up the range of the function into a disjoint union of open and closed intervals).

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