Definition of Riemann integral over the whole $\mathbb R^n$ In single variable calculus, I know that the integral over the entire real line is defined to be:
$$\int_R f(x)dx = \int_{-\infty}^a f(x)dx + \int_{a}^{+\infty} f(x)dx.$$
Then, if I consider $\mathbb R^n$, $n \geq 2$, what is the definition of $\int _{\mathbb R^n} f(x) dx?$
My initial guess was
$$\int _{\mathbb R^n} f(x) dx = \lim_ {r\to \infty} \int _{B(0,r)} f(x)dx ,$$ where $B(0,r)$ denotes a ball centred at $0$ with radius $r$.
But, I am not comfortable with this idea because by taking a ball I am choosing my domain of integration to be symmetric, and I know that even in 1D, we do not use
$$\int_\mathbb R f(x)dx = \lim _{a\to\infty} \int_{-a}^a f(x)dx.$$
 A: A "domain independent" definition may be given as follows: 
First, define what it means to integrate $\int_I f$, where $I$ is a compact interval. Now let $S$ be an arbitrary subset of $\mathbb{R}$. Define, 
$$ \int_S f = \sup_{I \subseteq S } \int_I f $$
For this definition to generalize integration we have to assume that $f\geq 0$. 
Now you can generalize to $\mathbb{R}^n$ by replacing $I$, the compact interval, by a "compact subset". 
You may ask, what if $f$ is non-negative? In this case you can split $f$ as $f=f^+ - f^-$ where $f^+ = \max(0,f)$ and $f^- = \min(0,f)$. Then apply this definition to each summand.   
A: If your function is positive, then you may define the integral:
$(1)\int_ {\mathbb{R}^n}f:=\lim_{n \rightarrow \infty} \int_{Dom_n}f$
with $dom_n$ being any sequence of domains that "converge monotonically" to the whole $\mathbb{R}^n$. This follows from the monotone convergence theorem:
$\lim_{n \rightarrow \infty} \int_{dom_n}f=\lim_{n \rightarrow \infty} \int_{\mathbb{R}^n} \chi_{dom_n} f=\int_{\mathbb{R^n}}f$
We use Lebesgue just to assert that the value is well-defined ("independent of how you go to infinity"). But the definition $(1)$ doesn't depend on Lebesgue Integration.
Aside from this, I guess you can't go deeper without getting unnecessarily messy. Riemann integral is not the proper setting for this: note that even in $\mathbb{R}$ you only really define the integral on closed and bounded intervals, and of bounded functions.
A: Well,
$$
\lim_ {r\to \infty} \int _{B(0,r)} f(x)dx ,
\tag1$$
would be a "principal value" integral.  This is fine if $f \ge 0$ or if $\int_{\mathbb R^n} |f| < \infty$.
Recall that the Riemann integral is defined for functions with domain
an "interval" of the form $I = [a_1,b_1\times[a_2,b_2]\times\cdots\times[a_n,b_n]$.  For the general improper integral in $\mathbb R^n$ we would probably require that
$$
\lim_{k\to\infty} \int_{I_k} f
\tag2$$
has the same limit for all increasing sequences $I_k$ of intervals with $\bigcup_k I_k = \mathbb R^n$.

But (let's face it) you would probably be better off with the Lebesgue integral anyway.  For example, in definition $(1)$ the integral is not invariant under translations:  it could happen that
$$
\int_{\mathbb R^n} f(\mathbf{x})\;d^n\mathbf{x}
\ne \int_{\mathbb R^n} f(\mathbf{x}+\mathbf{a})\;d^n\mathbf{x}
$$
And what about $(2)$ under rotations?
