# Krein-Milman theorem and dividing pizza with toppings

In this question the OP mentions the following problem as an exercise on Krein-Milman theorem:

You have a great circular pizza with $n$ toppings. Show that you can divide the pizza equitably among $k$ persons, which means every person gets a piece of pizza with exactly $\frac{1}{k}$ of any of the $n$ topping on it.

I am not sure whether I interpreted it correctly. To me it seems that mathematical reformulation of the problem could be that I have functions $f_i\colon X\to [0,1]$ for $i=1,\dots,n$ corresponding to how the toppings are distributed on the surface of pizza - the space $X$. And I would like to obtain a decomposition $X=A_1,\dots,A_k$ such that for any $i=1,\dots,n$ I have $\int_{A_1} f_i = \dots = \int_{A_k} f_i$, i.e., each person has the same portion from each of the toppings.

I am not sure whether this is the correct formalization. (Certainly the fact that Krein-Milman is supposed to be used is some kind of a hint on how the problem is supposed to be interpreted.)

But even if my interpretation is correct, I am not sure where to start. To apply Krein-Milman theorem, I need some locally convex topological vector space and some compact convex subset. I am not sure what space I could choose, since based on the above formulation it seems that the elements of this spaces should be somehow related to divisions of the circle.

Another possible interpretation I can think of is that I work with the functions $\{1,2,\dots,k\} \to \mathbb R^n$ representing how much of each topping gets the $k$-the person. It is certainly possible that $k$-the person gets the whole pizza, which would correspond to $f(k)=(1,1,\dots,1)$ and $f(i)=(0,0,\dots,0)$ for $i\ne k$. By I do not see how to show that the set of all admissible possibilities is convex. And if I can show that it is convex, then I do not really need Krein-Milman theorem.

I think this refers to the range of vector valued measures: If $\mu_1,\ldots,\mu_n$ are real nonatomic measures on a measurable space $(\Omega,\mathcal B)$ then $\mu:\mathcal B \to \mathbb R^n$, $B\mapsto (\mu_1(B),\ldots,\mu_n(B))$ has a compact and convex range.

This is Theorem 5.5 in Rudin's Functional Analysis and indeed an application of the Krein-Milman theorem.

• Thanks for your answer. This certainly sounds like a reasonable interpretation of that question. After your post I was able to find that this is sometimes called Lyapunov Convexity theorem (for example, Theorem 13.33 in the book by Aliprantis and Border) and there is even a post on this site with some references for various proofs. – Martin Sleziak Jun 7 '17 at 10:03
• I am curious about Rudin's Theorem 5.5, surely each $\mu_k$ must also be finite? Clearly if we take each $\mu_k$ to be the Lebesgue measure, then the range of the above function is the (non compact) line $t \mapsto (t,\cdots,t)$? I presume this is a misprint in the book, but am unable to find an erratum. – copper.hat Dec 2 '18 at 20:24
• A * real* measure has real values. – Jochen Dec 2 '18 at 20:52
• @MartinSleziak Could you please explain why this theorem fits the analogy? if $(\Omega, \mathcal{B})$ is the pizza, and $\mu_1,\dots ,\mu_n$ measure the individual toppings, what does convexity say(and compactness)? – pitariver Jan 30 at 18:49
• @pitariver From $\emptyset,\Omega\in\mathcal B$ we get that both $(0,0,\dots,0)$ and $(1,1,\dots,1)$ are in the range. From convexity we get that $(\frac1k,\frac1k,\dots,\frac1k)$ belongs in the range, i.e., there is a set $B\in\mathcal B$ such that $\mu_1(B)=\mu_2(B)=\dots=\mu_n(B)=\frac1k$, so if we split the $B$ for one person, they get the right amount from each topping. We can then continue with dividing $\Omega\setminus B$. – Martin Sleziak Jan 30 at 19:13