Uniformly distributed sequence in $\mathbb{R}$ We say that a sequence $(x_n)_{n=1}^\infty \subseteq \mathbb{R}$ is "uniformly distributed in $[a,b]$", with $a < b$, if $(x_n)_{n=1}^\infty \cap [a,b] \neq \varnothing$ and
$$\lim_{N \to \infty} \frac{\#\{n \leq N : x_n \in [c,d]\}}{\#\{n \leq N : x_n \in [a,b]\}} = \frac{d-c}{b-a}$$
for all $[c,d] \subseteq [a,b]$, with $c < d$, (here $\#S$ is the number of elements of a finite set $S$).
We say that a sequence $(x_n)_{n=1}^\infty \subseteq \mathbb{R}$ is "uniformly distributed in $\mathbb{R}$" if $(x_n)_{n=1}^\infty$ is uniformly distributed in any $[a,b]$, with $a < b$.
I have a construction for a sequence uniformly distributed in $\mathbb{R}$. Are there in the literature examples of uniformly distributed sequence in $\mathbb{R}$? References?
Thank you all for your help.
 A: Yes definitely. 
You can generate equidistributed sequences on $\bf R$ in the same fashion that you would do on $[0,1) \simeq {\bf R}/{\bf Z}$: by using an ergodic transformation preserving the Lebesgue measure. Note that the Lebesgue measure on $\bf R$ has infinite mass and thus the situation is slightly more subtle on $\bf R$ than on $[0,1)$.
The study of transformations preserving a measure of infinite mass is the topic of infinite measure theory. The standard reference is the book of J. Aaronson, an introduction to infinite measure theory edited by the AMS. The first example of a conservative ergodic transformation from $\bf R$ to $\bf R$ preserving the Lebesgue measure given in the book (and perhaps the simplest example) is the Boole transformation $$x \mapsto x - {1\over x}.$$
Applying the Hopf ratio ergodic theorem, we deduce that for almost all $x\in \bf R$, the sequence given by $x_{n+1} = x_n -{1\over x_n}$, $x_0=x$ is equidistributed according to your definition (which is almost the standard one).
Another way to generate an equidistributed sequence is to use a random walk on $\bf R$. So let $X_1$,...$X_k$... be an iid sequence of real valued random variables defined on some space $(\Omega, {\cal F}, P)$ which are both integrable and with zero expectation. We need a "non-arithmetic" assumption on the law of the $X_i$ because we don't want that the variables take values in a discrete subgroup of $\bf R$. $P_X = {1\over 1+\alpha}(\delta_{-\alpha}+\alpha \delta_1)$ with $\alpha$ positive irrational is ok.
Theorem the sequence $(S_n(\omega))$ is equidistributed on $\bf R$ for almost all $\omega$. 
I think that this result is stronger than the CLT. Maybe this can be deduced from the local CLT. Anyway, this follows from the conservativity and ergodicity of the lift of the shift on the skew-product $\Omega\times \bf R$, the same argument as above applies. In addition to the previous reference,  there are probably a few probability books where this is discussed but I can't remember any from the top of my head.
A: Let me guess ...
First: equally spaced points with difference $1$ going from $-1$ to $1$.
Then: equally spaced points with difference $1/2$ going from $-2$ to $2$.
Then: equally spaced points with difference $1/4$ going from $-4$ to $4$.
Then: equally spaced points with difference $1/8$ going from $-8$ to $8$.
And so on.  
A: Another couple of recipes to build such sequences uniformly distributed on $\mathbb{R}$ (a kind of long comment). 
1. Choose a sequence of integers $(k_n)_{n\in\mathbb{N}}$  which is uniformly distributed in $\mathbb{Z}$ (with the clear meaning), and define correspondingly $$\mu(n):=\#\{ 0\le j < n\,:\, k_j=k_n \}$$ 
(that is, for any $n\in\mathbb{N}$ the integer $k_n$ has already shown $\mu(n)$ times before).
Let $(u_n)_{n\in\mathbb{N}}$ be any uniformly distibuted sequence in $[0,1]$. Then define $$x_n:= k_n + u_{\mu(n)}\, .$$ Therefore for any integer $k$, the terms of $(x_n)$ that belong to the interval $[k,k+1]$ are, in order of appearance, $k+u_0,k+u_1, k+u_2,k+u_3\dots$. As a consequence, the sequence $(x_n)$ verifies the wanted identities at least when $a$ and $b$ are a pair of consecutive integers. But this easily implies the identities for all $a < c < d < b $.
2. The following seems easy to show: if $(v_n)$ is uniformly distributed in $[-1,1]$, for some diverging sequence $(a_n)$ the sequence $a_n v_n$ is uniformly distributed in $\mathbb{R}$.
