Deducing a weight function from a set function. I was working on this problem for which I think I have almost the solution, but if you could help me finish it, I would be so grateful.
${\bf{ Problem:}}$
So, $P$ is a nonempty collection of subsets of $\{1,\dots,n\}$ such that 
i) If two sets are in $P$, so is their intersection and union.
ii) If $S$ is a set in the collection $P$ that is not the empty set, there is another set in $P$ with exactly one element less than $S$.
Let $f$ be a GIVEN function from $P$ to the real numbers such that $f(\phi)=0$ and $$ f(S\cup S')= f(S)+f(S')-f(S\cap S')\text{ for all }S,S'\in P. $$
 Can we find real numbers $f_1 \cdots f_n$ such that for any $S \in P$, $$f(S)=\sum_{i \in S}f_i$$
$\mathbf{My Solution:}$
I think we can.
Using the union property in i), I deduced that $P$ had to have a unique maximal element $M$ that every other element was a subset of, and that $\phi \in P$.
Then, we define the following sequence by repeatedly applying property ii) and choosing the $a_i$ accordingly
$M_1 = M =\{a_1,\cdots a_k\}$
$M_2 = \{a_2 \cdots a_k \}$ and so on until
$M_k = \{ a_k\}$
$M_{k+1}=\phi$.
Then, I define $f_{a_i}$ as $f(M_{i})-f(M_{i+1})$ for $1\leq i \leq n$ and set all other $f_i$s to be zero. Then, I can verify that this implies the required property above for the sets $M_i$, but not for all the elements of $P$ in general..
Any help, guys? Thank you in advance! 
 A: Hint: First, show that the $f_i$ take on the same values no matter what descending chain of sets you decide their value is based on. (This is the hardest part - you'll have to use the functional equation for $f$ a couple times.) With this in hand you can induct on $|S|$ to show that $$f(S) = \sum_{i \in S} f_i$$
EDIT: here's the proof of well-definedness.
Let's assume we have two different inclusions $M_i \subset M_{i+1}$ and $N_i \subset N_{i+1}$ that satisfy our conditions - that $M_{i+1} \backslash M_i = N_{i+1} \backslash N_i = \{k\}$. For well-defined-ness, we want $f(N_{i+1}) - f(N_i) = f(M_{i+1}) - f(M_i)$. Now, by the definition of our function $f$, we know 
$$f(M_{i} \cup N_{i+1}) = f(M_i) + f(N_{i+1}) - f(M_i \cap N_{i+1})$$
$$f(M_{i+1} \cup N_i) = f(M_{i+1}) + f(N_i) - f(M_{i+1} \cap N_i)$$
The first thing to note is that the LHS in both equations is the same - they're both equal to $M_{i+1} \cup N_{i+1}$ (because $M_{i+1} \backslash M_i = \{k\}$, and we already 'get' the $k$ from $N_{i+1}$; similarly for the other equality), and we also have that $M_{i+1} \cap N_i = M_i \cap N_{i+1}$ for similar reasons. So, subtracing one equation from another, we ultimately get $f(M_{i+1}) - f(M_i) = f(N_{i+1}) - f(N_i)$, which is what we wanted all along.
