distribution of $\{na\}$ when $a$ is irrational number (by $\{x\}$ I mean the fraction part of the real number $x$)
If $a$ is an irrational number and $n$ is a integral number, what is the distribution of $\{na\}$? I'm asking for some continuous function $f:[0,1]\to\mathbb R$ such $\int_{\alpha}^{\beta}f(x)\;dx$ gives the probability that $\{na\}$ falls between $\alpha$ and $\beta$.
When I calculated it for a bunch of irrational number, from $n=1$ to $10000$, I found that it's very close to uniform distribution. It's well known that {na} with a proper choose of $a$ could be arbitrary close to any real number in the $[0,1]$ interval. But this claim is more than that and wants the distribution to be uniform. 
I think that a quite simple simple proof may exist: If $a$ was rational, say $p/q$, a uniform discrete distribution have been existed. I mean if $n$ goes to infinity the number would fall into $[i/q,(i+1)/q]$ interval with probability $1/q$. Now If we could approximate $a$ with a rational $p/q$, with "sufficiently small" error, the same would happen for a. That is, {na}s would also fall into the $[i/q,(i+1)/q]$ with probability $1/q$. if $q$ goes to infinity the distribution would become continuous.
And at last ... I think $a = p/q + c/(q^2)$ where $c$ is smaller than or equal to one, is a sufficiently good approximation. Good in the sense that such an approximation causes a uniform distribution. 
 A: The distribution is known to be uniform (a result due to Weyl, I believe). An excellent reference for this (and much else) is Dym and McKean's book on harmonic analysis.
A: For rational $a$ the answer (explicit bound for the error term) is given by Ostrowski's theorem (Ostrowski A. Bemerkungen zur Theorie der Diophantischen Approximationen,-Abh.
Math. Sem Hamburg, 1922, 1, s. 77-98). It depends on the sum of partial quotients in continued fraction expansion of number $a$. For real number it is sufficient to take good rational approsimation (one of convegents). See also Khintchine A. Ya. Ein Satz uЁber KettenbruЁche, mit arithmetischen Anwendungen. — Mathematische Zeitschrift, 18: 1 (1923), 289–306.
A: For what it's worth, in the language of measures one can reformulate your statement as
$$\mu_n:=\frac{1}{n}\sum_{k=1}^n\delta_{k\alpha}\rightharpoonup\mathcal{L}^1$$
on the circle $\mathbb R/\mathbb Z$ (in the weak-* topology, meaning $\int f\,d\mu_n\to\int f\,d\mathcal{L}^1$ for any continuous $f$). Any limit measure $\mu_\infty$ has total mass $1$ and is invariant under the translation $\tau_\alpha$ (as $\|(\tau_\alpha)_*\mu_n-\mu_n\|\to 0$), so
$$\int e^{-2\pi i nx}\,d\mu_n=:\widehat{\mu_\infty}(n)=\widehat{(\tau_\alpha)_*\mu_\infty}(n)=e^{2\pi i n\alpha}\widehat{\mu_\infty}(n)$$
and thus
$\widehat{\mu_\infty}(n)=0$ for $n\neq 0$, $\widehat{\mu_\infty}(0)=1$. By uniqueness, $\mu_\infty=\mathcal{L}^1$. By compactness of measures we are done.
