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. 



How to fill a simplex with almost disjoint cuboids?
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.


