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According to Rajendra Bhatia in his book Fourier Series, Weyl's Equidistribution Theorem states that if $x$ is an irrational number, then for every subinterval $[a, b]$ of $(0, 1)$ we have

$lim_{n \rightarrow \infty}$$\frac{1}{N}$$card${$k : 1 \leq k \leq N$, $\tilde{(kx)} \in [a, b]$} = $b - a$ where $\tilde{(kx)}$ is the fractional part of the number $kx$.

My question is what happens if we generalise to measurable subsets of $(0, 1)$?

Does $lim_{n \rightarrow \infty}$$\frac{1}{N}$$card${$k : 1 \leq k \leq N$, $\tilde{(kx)} \in A$} = $\mu(A)$ where $A$ is a measurable subset and $\mu$ the Lebesgue measure function?

Further, for non-measurable subsets $V$ is the sequence $\frac{1}{N}$$card${$k : 1 \leq k \leq N$, $\tilde{(kx)} \in V$} bounded above and below and if so, does it have the same set of sublimits for all irrational $x$?

After my last question that revealed I had momentarily forgotten all my undergraduate real analysis, I hope this one is worthy of mathoverflow... thanks...

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Sorry, this is way too much to hope for: ${\bf Z} x \bmod 1 = \lbrace k x \bmod 1 | x \in {\bf Z} \rbrace$ is countable, and thus of measure zero. Hence we could add it to or remove it from any measurable subset $A$, leaving $\mu(A)$ fixed but making the limit $1$ or $0$. Or, by having $A$ meet ${\bf Z} x \bmod 1$ in a set of multiples whose density in $[1,N]$ comes arbitrarily close to $0$ and $1$ as $N \rightarrow \infty$, we could prevent the proportion of multiples in $A$ from having any nontrivial upper lower bounds while keeping $A$ measurable. –  Noam D. Elkies Jan 26 '12 at 19:47
    
Thanks Noam... that is very clear now... –  George Lazou Jan 26 '12 at 20:21
    
Sorry if I am flogging a dead horse, but what happens if we require the limit to be equal for all irrational $x$? i.e. if $lim_{n \rightarrow \infty}$$\frac{1}{N}$$card${$k : 1 \leq k \leq N, \tilde{(kx)} \in A$} = $lim_{n \rightarrow \infty}$$\frac{1}{N}$$card${$k : 1 \leq k \leq N, \tilde{(ky)} \in A$} for all $x, y$ irrational, then does the limit equal $\mu(A)$? –  George Lazou Jan 26 '12 at 21:44
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@George: In that case it does. Lebesgue measure is invariant and ergodic for any irrational rotation, and in particular, Birkhoff's ergodic theorem implies that the limit frequency of visits to any measurable set is equal to the Lebesgue measure of that set for Lebesgue-a.e. starting point $x$. Thus if $A$ is measurable and $E$ has positive Lebesgue measure and the limiting frequency of visits exists and is equal for every $x\in E$, then that limit must equal $\mu(A)$. –  Vaughn Climenhaga Jan 26 '12 at 22:37
    
Related question: mathoverflow.net/questions/75777 –  Goldstern Jan 26 '12 at 23:37

2 Answers 2

up vote 2 down vote accepted

This is a very interesting question, which actually asks about the interplay between equidistribution (or harmonic analysis if you would like to call it that way) and ergodic theory.

As Vaughn mentioned, for any L^{p} function (p>=1), the pointwise ergodic theorem would imply that for Lebesgue almost every point $x$, the value of the ergdoic averages $$ \frac{1}{N}\sum_{n=0}^{N-1}f(x+n \alpha) \to \int fd\mu$$ where $mu$ is the Lebesgue measure. Hence one can take the characteristic function of your favourite measurable set, and get the "equidistribution" statement you would like, but only at the cost of convergence a.e.

Noam showed why a.e. is sharp here.

When one asks about continuous function (or more frequently in the field, takes a function from a suitable Sobolev space), one can get more information than the statement above (including for example estimation about the errors), see for example this inequality for your problem - http://en.wikipedia.org/wiki/Low-discrepancy_sequence , but this also comes at a cost, you usually have to limit the irrational number your dealing with to be Diophantine generic or so.

As you may know, Weyl's equidistribution was generalized to equidistribution of sequences of the form ${p(n)\alpha}$, where p(n) is some polynomial in $Z[x]$. Now one can ask about such "ergodic limits" for the sequence ${n^{2}\alpha}$, for general L^{p} function (continuous functions will be dealt easily by the equidistribution theorem). It was an open question for some time, but the main result here is by Bourgain, which showed that for any $p>1$, if the function is in $L^{p}$, then the averages will converge a.e. to the integral. This is the analogue of the usual pointwise ergodic theorem for the quadratic case. The interesting point here is that $p>1$ is sharp, there is today a counterexample for $p=1$ case.

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Noam Elkies pointed out that this generalization cannot be true because the set of values of this sequence has measure zero. However, another natural generalization is true: By definition, a sequence $(x_i)$ is equidistributed iff

$$(*)\qquad \frac1n \sum_{i=1}^n f(x_i) \ \to \ \int_0^1 f(x) \ dx$$

holds whenever $f$ is the characteristic function of an interval. It is a classical theorem that the following are equivalent:

  1. $(*)$ holds for all characteristic functions of an interval

  2. $(*)$ holds for all continuous functions

  3. $(*)$ holds for all Riemann-integrable functions

  4. $(*)$ holds whenever $f$ is one of the functions $e^{i\pi k x}$.

Condition 3 implies that uniformly distributed sequences can be used to approximate integrals. (Using the "discrepancy" of a sequence, the error can even be estimated in terms of the total variation of the function.)

It also means that the answer to the original question is yes for all sets whose characteristic function is Riemann-integrable; equivalently, for all sets whose topological boundary has measure zero.

Condition (4) means that only few functions have to be checked to conclude uniform distribution. For the sequence $x_k =k \alpha$, the criterion in (4) is very easy to verify, since the exponential sum appearing there is a finite geometric series.

(See also wikipedia's article on equidistribution)

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How are this "natural generalization" and "classical theorem" different from Weyl equidistribution itself? (For example, that's how Körner's Fourier Analysis states and proves Weyl's theorem if I remember right.) –  Noam D. Elkies Jan 27 '12 at 2:45
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I am not sure what you mean by "different". For me, (1) is the definition of equidistribution, (4) (or rather the equivalence between (1) and (4)) is Weyl's criterion, and (3) is the most useful consequence. They are all equivalent, so in this sense they are not different. I call the condition (3) a "generalization", because it is (apparently) a strengthening of (1) (though not strictly). –  Goldstern Jan 27 '12 at 3:25

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