Extrema of a $\sigma$-additive function are attained. I've written down a proof for the following theorem which appears in Loeve's book of probability (page 86). I believe I'm missing something because Loeve's proof is much more complicated but I can't see where is my mistake:

If $\phi$ is $\sigma$-aditive function on a $\sigma$-field, then there
  exist sets C and D of the $\sigma$-field such that $\phi(C) = sup \ \phi(B)$ and $\phi(D) = inf \ \phi(B)$.

So, here is my proof: (we can assume that $\phi$ is finite)
Let's consider a sequence $A_n$ such that $\phi(A_n) > sup \ \phi(B) - 1/n$ for each $n \in \mathbb{N}$ which exist by the definition of supremum.
Let's take $A_n'=A_n\setminus\cup_{k=1}^{n-1}A_k'$ clearly $A_n'$ are disjoint and $\sum_{k=1}^{N}A_n'=A_N$.
It follows using $\sigma$-aditivity that:
$\phi(\cup_{k=1}^\infty A_k)=\phi(\sum_{k=1}^\infty A_k') = \sum_{k=1}^\infty \phi(A_n')= \lim_{N} \sum_{k=1}^N \phi(A_k')= \lim \phi(\sum_{k=1}^N A_k') = \lim \phi(A_N) \ge sup \ \phi(B)$
where the last equality follows taking limits in the first inequality of the proof.
Can you see any mistakes?

That $A_N =\sum_{k=1}^N A_k '$ is only true if the $A_n$ form a non-decreasing sequence. In general, you will only have $\sum_{k=1}^N A_k ' = \bigcup_{k=1}^N A_k$, where the set on the right might be very different from $A_N$.
Also, there are typos in your definition of $A_k '$.
