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?
Best Answer
First, we assume that $\mu$ is a measure. Let $\{A_k\}\subset \mathcal M$ a decreasing sequence of measurable subsets of $X$. Put $B_k:=X\setminus A_k$ for $k\geq 0$ and $C_k:=B_k-B_{k-1}$ for $k\geq 1$, $C_0:=B_0$. Then the sets $C_k$ are pairwise disjoint, $\bigcup_{k\geq 0}C_k=\bigcup_{k\geq 0}B_k=\bigcup_{k\geq 0}A_k^c=X\setminus \bigcap_{k\geq 0}A_k$ and since $\mu(B_k)$ is finite we have $\mu(C_k)=\mu(B_k)-\mu(B_{k-1})$. Hence $$\mu\left(\bigcap_{k\geq 0}A_k\right)=\mu(X)-\mu\left(X\setminus \bigcap_{k\geq 0}A_k\right)=\mu(X)-\mu\left(\bigcup_{k\geq 0}C_k\right)\\\ =\mu(X)-\sum_{k\geq 1}(\mu(B_k)-\mu(B_{k-1}))-\mu(B_0)=\mu(X)-\lim_{k\to \infty}\mu(B_k)=\lim_{k\to \infty}\mu(A_k).$$
For the converse what you did is fine,I would just add that $\mu(\emptyset)=\mu(\emptyset\cup\emptyset)=\mu(\emptyset)+\mu(\emptyset)$ and since $ \mu(\emptyset)<\infty$ we have $\mu(\emptyset)=0$.
The fact that the measure was finite was necessary here, for example if we take $\mu(A)=+\infty$ for all $A\in\mathcal M$ we won't get a measure since we need $\mu(\emptyset)=0$.