How to fill a simplex with almost disjoint cuboids? There is an algorithm that give us cuboids in $\mathbb{R}^3$, say $Q_1,Q_2,\ldots$, such that $\cup_{i=1}^{\infty} Q_i$ is the simplex with vertices $(0,0,0), (1,0,0) , (0,1,0), (0,0,1)$, and the $Q_i$'s are almost disjoint (i.e $\lambda(Q_i\cap Q_j)=0$ if $i\neq j$)? 
In $\mathbb{R}^2$ there are many easy ways to fill a triangle with almost disjoint-rectangles but I had not find the way to generalize this to higher dimensions. Do you have any ideas?
This will be very helpful, for example, to approximate the cumulative distribution function of a sum of 3 or more random variables that are not necessarily independents. 
 A: OK, since we finally have figured out what Andres is asking and since 600 characters is a bit too restrictive, I'll post this as an answer.
The following Asymptote code will draw the filling except I used the size 4 simplex instead of size 1 one here:

size(400);
import three;
import graph3;


pen[] q={red,green,magenta,blue,black};
q.cyclic(true);

int N=4;
triple O=(0,0,0),A=(4,0,0),B=(0,4,0),C=(0,0,4);
draw(O--A--B--C--A--C--O--B);
for(int n=0;N>n;++n)

{
real s=1/2^n;
for(real x=4-3*s;x>=0;x-=2s)
for(real y=4-x-3*s;y>=0;y-=2s)
{
draw(shift((x,y,4-x-y-3*s))*scale3(s)*unitcube,q[n]);
draw(shift((x-s,y,4-x-y-3*s))*scale3(s)*unitcube,q[n]);
draw(shift((x-s,y+s,4-x-y-3*s))*scale3(s)*unitcube,q[n]);
draw(shift((x-s,y,4-x-y-2*s))*scale3(s)*unitcube,q[n]);
}
}

The unitcube is just $[0,1]^3$, the rest should be self-explanatory.
A: Here's a (highly nonoriginal) method: overlay a finite hyperplane grid onto the region of interest.  For R^2 this divides the plane into rectangles, for R^3 into cuboids (a.k.a rectangular parallelipipeds), and into higher dimensional intervals for the space of your choice.  Now, mark all those bounded grid areas which lie fully inside your region; if you have some space left over not inside a marked area, add finitely many more hyperplanes to your grid, and repeat.  Doing this countably many times will get you within epsilon in measure to your region.  I leave the formalizing of the algorithm to you.
Oops, in using the word measure I gave it away.  Cf. Lesbesgue or Riemann for technical details.
Gerhard "Ask Me About System Design" Paseman, 2010.02.17
