Partial sums and Lebesgue integral Let $D = [-1/2,1/2]^n$ be the unit cube. Let $f:\mathbb{R}^n$ be a function such that the Riemann integral
$$\int_{\rho D} f(x)\, \text{d}x$$
exists for all $\rho > 0.$
Under what conditions can we say that
(i)
$$\lim_{\rho \to \infty} \int_{\rho D} f(x) \,\text{d}x =  \int_{\mathbb{R}^n} f(x) \,\text{d}x,$$
passing to the Lebesgue integral on the right-hand side and
(ii) $$\int_{\mathbb{R}^n} f(x)\, \text{d}x = \lim_{k\to\infty} \sum_{x \in \mathbb{Z}^n} f(x/k)k^{-n} \ ?$$
For instance, for the Riemann integral it is not hard to see that:
$$ \int_{\rho D} f(x) \,\text{d}x = \lim_{k\to\infty} \sum_{x \in \mathbb{Z}^n} f(x/k)k^{-n}\mathbb{1}_{\rho(D)}(x)  $$
 A: Part (i)
If $f$ is Lebesgue integrable over $\mathbb{R}^n$ (which requires absolute integrability) then 
$$-\infty < \int_{\mathbb{R}^n}f  = \int_{\mathbb{R}^n}f^+ -  \int_{\mathbb{R}^n}f^-  < +\infty.$$
where the positive and negative parts are $f^+(x) = \max(f(x),0)$ and $f^-(x) = -\min(f(x), 0),$ and individually
$$0 \leqslant \int_{\mathbb{R}^n}f^+  < +\infty, \\ 0 \leqslant \int_{\mathbb{R}^n}f^-  < +\infty.$$
The Riemann and Lebesgue integrals coincide on bounded rectangles.  Hence, by the monotone convergence theorem
$$\lim_{\rho \to \infty} \int_{\rho D} f^+(x) \, dx = \lim_{\rho \to \infty} \int_{\rho D} f^+  = \lim_{\rho \to \infty} \int_{\mathbb{R^n}} f^+ \chi_{\rho D} =  \int_{\mathbb{R^n}} f^+ .$$
A similar result holds for $f^-$ and, consequently
$$\lim_{\rho \to \infty} \int_{\rho D} f(x) \, dx = \int_{\mathbb{R^n}} f. $$
If $f$ is not Lebesgue integrable over $\mathbb{R}^n$, but Riemann integrable over bounded rectangles, then the limit on the left might converge to an improper Riemann  integral, despite the non-existence of a finite Lebesgue integral.  For $n=1$, the typical example is $f(x) = \sin x / x$.
Part (ii)
It seems you are asking if the limit of Riemann sums over an expanding region can be evaluated as a double limit with the order switched and converge to the integral over the infinite domain.  
For $n = 1$ and integration over $[0,\infty)$ this would translate to 
$$\lim_{k \to \infty} \lim_{m \to \infty} k^{-1} \sum_{j=1}^{mk} f(j/k) = \lim_{m \to \infty} \lim_{k \to \infty} k^{-1} \sum_{j=1}^{mk} f(j/k) = \lim_{m \to \infty} \int_0^m f(x) \, dx\\ = \int_0^\infty f(x) \, dx$$
I surmise this can be justified by some form of monotone or dominated convergence.
