Drew Olbrich (http://www.traipse.com/) has this hack

that figures out what a nonagon should look like.

I decided to implement this on a whim.

It 'finds' the tetrahedron, cube, dodecahedron, but finds

a d10-like shape for n=8, a triangular prism for n=5,

a pentagonal prism for n=7. But it finds a non-d10-like shape

for n=10 (square, two rows of 4 pentagons, square) finds

*something* for n=11, gets the same result for n=9 as

Olbrich, but gets n=20 horribly wrong. Maybe I need to

tweak parameters to get it to converge better.

Generally nifty, though.

Hm, there's an easy proof that in any reasonable (i.e. T1)

space, if there exists a finitely additive integral,

then there exists an additive one. For suppose we have C

and c \in C. Then IC = I(C u x) = IC u Ic hence

Ix \subseteq IC. So UIc \subseteq IC, consq.

DUIx \subseteq DIC = C. But obviously

DUIx \supseteq UDIc = Uc = C.

So DUIc = C, and we can define I*C = UIc,

and I* is an additive integral.

Another property that might be worth investigating:

an integral I is *decomposable* if, for every closed

C and c \in C, there exists a subset S \subseteq IC

such that DS = {c}. An additive integral is clearly

decomposable, but I'm not sure of the relationship

between finitely additivity and decomposibility.