The meaning of $\int\limits_\mathbb{R}$ Which of 
$\lim_{a\to\infty} \int_{-a}^{a}$    and   $\lim_{r\to-\infty} \lim_{s \to \infty} \int_{r}^{s}$
does $\int_{\mathbb{R}}$
 mean? 
 A: Rather the second one:
$$\int\limits_\mathbb{R}{f(x)\,dx}=\int\limits_{-\infty}^{\infty}{f(x)\,dx}=\lim_{r\to-\infty} \lim_{s \to \infty} \int_{r}^{s}{f(x)\,dx}$$
A: It really depends on the context.
It is possible that it is just the Lebesgue integral, which is a "proper integral" without needing any limits.
As lhf said in comments, see the wikipedia article for improper integrals. Specifically, check out the section titled Types of Integrals.
If the function is non-negative and "nice" in some way, the various improper integrals and the Lebesgue integral will all agree.
The first form is the least likely. It has the obvious disadvantage that it potentially gives different results when  $g(x)=f(x+c)$ for some constant $c$. It would also compute an integral of $0$ for any odd function. Would you want:
$$\int_{\mathbb R} xdx = 0?$$
That seems like an unfortunate result, somehow.
Possibly better than your double limit is to pick a fixed $a\in \mathbb R$ and define it as:
$$\lim_{s\to -\infty} \int_s^{a} + \lim_{r\to +\infty} \int_a^r$$
This has the advantage that it is symmetric in $r,s$ while the $\lim_r\lim_s$ definition is not (obviously) so. (The two limits are the same thing, it's just a little clearer that the $r,s$ limits are independent if you break at some $a$ as I have.)
