Measures $\psi$ is the smallest under all other measures with this property Let $\alpha$ and $\beta$ be two finite measures, in particualar $\alpha(A\cup B)=\alpha(A)+\alpha(B)$ if $A\cap B=\varnothing$ (and also for $\beta$) and if $A_1,A_2,\ldots$ are a decreasing sequence of such that $\bigcap(A_n)=\varnothing$, then $\alpha(A_n)\rightarrow0$ (and also for $\beta$). Assume that both finite measures are defined on an $\sigma$-algebra $\Sigma$. Define now: $\psi(X)=\sup\{\alpha(A)+\beta(B):A,B\in\Sigma,A\cup B\subset X, A\cap B=\varnothing\}$ for all $X\in\Sigma$. 
I want to prove that $\mu$ is also a finite measure, thus in particular additive and if $A_1,A_2,\ldots$ are a decreasing sequence of such that $\bigcap(A_n)=\varnothing$, then $\psi(A_n)\rightarrow0$. Moreover i have to prove that $\psi$ is the smallest under all finite measures $\theta$ with the property: $\theta\geq\alpha$ and $\theta\geq\beta$.
I have already problems by proving the additivity if i write out all definitions. Intuitive this is correct but i can not write it down on a good way.
Can someone help me with this problems?
Thank you very much :)
 A: Your conditions imply that $\alpha, \beta, \psi$ are all in fact sigma additive measures - This is a (finger) exercise in typing:
Say $\{A_n: n \geq 1\}$ is a disjoint collection with $A = \bigcup_{n \geq 1} A_n$. Let $B_n = \bigcup_{i \leq n} A_i$. Then since $A \backslash B_n$ descends to empty set, your assumption implies $\alpha(A\backslash B_n) = \alpha(A) - \sum_{i < n} \alpha(A_i)$ goes to zero. Hence $\alpha(A) = \sum_{n \geq 1} A_n$. So $\alpha, \beta$ are sigma additive measures.
You can now show that $\psi$ is also a sigma additive measure. So let $\{A_n: n \geq 1\}$ be a disjoint collection and $A = \bigcup_{n \geq 1} A_n$. Suppose $X, Y$ partition $A$. Put $X_n = A_n \cap X$, $Y_n = A_n \cap Y$. Then $\alpha(X) + \beta(Y) = \sum_{n \geq 1} \alpha(X_n) + \sum_{n \geq 1} \beta(Y_n) = \sum_{n \geq 1} \alpha(X_n) + \beta(Y_n) \leq \sum_{n \geq 1} \psi(A_n)$. Taking supremum over all such pairs $X, Y$, we get $\psi(A) \leq \sum_{n \geq 1} \psi(A_n)$. The other inequality is $\epsilon$-similar.
Finally note that since $A, \phi$ is a partition of $A$, $\psi(A) \geq \alpha(A), \beta(A)$. Also if for some $A$, $\theta(A) < \psi(A)$ then, for some partition $X, Y$ of $A$, $\theta(A) = \theta(X) + \theta(Y) < \alpha(X) + \beta(Y)$ so that either $\theta(X) < \alpha(X)$ or $\theta(Y) < \beta(Y)$.
Let me add that if you drop the descending condition, you still get (via same argument) that $\psi$ is the smallest finitely additive measure above $\alpha$ and $\beta$.
