I am trying to prove the following, but I must say I am not totally convinced by my proof. Could you check it and provide me a feedback?
Problem: Consider a pre-measure $\mu$ on a finite and not empty set $\Omega$, $\mu$ = $\sum_{\omega \in \Omega}\epsilon_\omega$ on P($\Omega$). $\epsilon_\omega$ is the indicator function. Then $\mu$(A) is the number of elements in every subset A $\in$ $\Omega$. Show that every pre-measure $\mu$ on P($\Omega$) has the form $\mu$ = $\sum_{\omega \in \Omega}\alpha_\omega\epsilon_\omega$ where $\alpha_\omega$ := $\mu$({$\omega$})
My proof:
Since $\Omega$ is finite, it is also a countable set and every subset A in P($\Omega$) is also countable. There is also for every A a counting where no element is counted several times.
Since $\mu$ is a pre-measure on P($\Omega$), we can write: $\mu(A)$ = $\mu$($\cup_{\omega_i \in A}${$\omega_i$}).$\epsilon_{\omega_i}$(A) + $\mu$($\cup_{\omega_j \notin A}${$\omega_j$}).$\epsilon_{\omega_j}$(A) (the second part of the sum should 0)
Because of the properties of a pre-measure, we have
$\mu($($\cup_{\omega_i \in A}${$\omega_i$}) = $\sum_{\omega_i \in A}$$\mu$({$\omega_i$}), since the $\omega_i$ are all distinct and also
$\mu($($\cup_{\omega_j \notin A}${$\omega_j$}) = $\sum_{\omega_j \notin A}$$\mu$({$\omega_j$})
Consequently: $\mu$ = $\mu(A)$ = $\sum_{\omega_i \in A}$$\mu$({$\omega_i$}).$\epsilon_{\omega_i}$(A) = $\sum_{\omega \in \Omega}\alpha_\omega\epsilon_\omega$ where $\alpha_\omega$ = $\mu$({$\omega_i$}) and $\epsilon_\omega$ = $\epsilon_{\omega_i}$(A)
Looks okay, but is a bit messy. One mistake (or typo as I hope) is that of course we do not have $A\in\Omega$ as you write but $A\in\wp(\Omega)$.
It is enough to state that for a pre-measure on $\wp(\Omega)$ where $\Omega$ is countable the following is true:
$$\mu(A)=\mu(\bigcup_{\omega\in A}\{\omega\})=\sum_{\omega\in A}\mu(\{\omega\})=\sum_{\omega\in\Omega}\mu(\{\omega\})\epsilon_{\omega}(A)=\left(\sum_{\omega\in\Omega}\mu(\{\omega\})\epsilon_{\omega}\right)(A)$$