I need help with this problem:
Let $\mathcal{M}$ be a sigma algebra of subsets of $X$ and the set function $\mu:\mathcal{M}\to [0,\infty)$ be finitely additive. Prove that $\mu$ is a measure if and only when $\{A_k\}_{k=1}^\infty$ is a descending sequence of sets in $\mathcal{M}$, then $ \mu\left(\bigcap_{k=1}^\infty A_k\right) = \lim_{k\to \infty} \mu(A_k).$
Below is my attempt:
Clearly one direction follows from continuity of measure.
Now suppose that the display equation above is true. Let $E=\cup_{k=1}^\infty E_k$, where the $E_k$ are disjoint. Set $A_k=E\setminus \cup_{k=1}^n E_k$. Then $A_k \supseteq A_{k+1}$. Also, $\cap_{k=1}^\infty A_k = \emptyset$. $\mu$ is also finitely additive, so $ \mu(A_k)= \mu(E)- \sum_{n=1}^k \mu(E_n).$ So $\begin{align*} \mu(E) & = \lim_{k\to \infty} \mu(A_k)+\lim_{k\to \infty} \sum_{n=1}^k \mu(E_n)\\ & = \mu \left(\cap_{k=1}^\infty A_k\right) + \sum_{k=1}^\infty \mu(E_k)\\ & = \sum_{k=1}^\infty \mu(E_k) \end{align*} $
Please, does this look right?