Moving from Finitely Additive to Countably Additive I have a question on basic measure theory here:

Let $(X, \mathscr A)$ be measurable space and $\mu$ is a non-negative set function that is finitely additive and such that $\mu (\emptyset) = 0$. Suppose that whenever $A_i$ is an increasing sequence of sets in $\mathscr A$, then $\mu(\cup_i A_i) = \lim_{i \to \infty} \mu(A_i)$. Show that $\mu$ is a measure.

Here are the questions I would like you to help:
(1) First, background question: In this problem, "finitely additive" means
$$\mu(\bigcup_{j=1}^n A_j) = \sum_{j=1}^n \mu(A_j), \quad \text{where }A_j \text{'s are pairwisely disjoint.}$$
Correct me if I am wrong here.
(2) Another background: The phrase "Suppose that whenever $A_i$ is an increasing sequence of sets..." means 
$$\text{if } A_i \uparrow \text{ then } \mu(\cup_i A_i) = \lim_{i \to \infty} \mu(A_i), \quad \text{where } \ldots A_{i-1}\subset A_i \subset A_{i+1} \ldots$$
Again, correct me if I am wrong here.
(3) Done with the background. To solve this problem, my understanding is to "move" from finitely additive as given in (1) to countably additive as given in a measure's definition, but I do not how to proceed. Do please let me know the friendliest way of solving this problem.
Thank you for your time and effort.
 A: The idea is to turn a countable disjoint union into an increasing union. To do that, you'll need to use different sets. So suppose $\{ A_n \}_{n=1}^\infty$ is a disjoint collection of measurable sets. Consider $\{ B_m \}_{m=1}^\infty$ defined by $B_m = \bigcup_{n=1}^m A_n$. This is an increasing union, so it has the limit property which was assumed. Use this to conclude the limit property you want for the measure of the union of the $A_n$.
A: (1) Yes.
(2) "Suppose that whenever $A_i$ is an increasing sequence of sets..." means $A_i \nearrow$, i.e., $A_1 \subseteq A_2 \subseteq \cdots$.
(3) To show $\mu$ is a measure, you need to show two things:


*

*$\mu(\emptyset) = 0$

*If $A_i \in \mathcal A$, $i = 1, 2, \ldots$ are pairwise disjoing, then $$\mu\bigg( \bigcup_{i = 1}^\infty A_i \bigg) = \sum_{i = 1}^\infty \mu(A_i).$$


We know it holds for the finite case by hypothesis, i.e., replace $\infty$ by $n \in \mathbb N$. You now need to show it holds for $\infty$.

Proof: 
Observe that $\mu(\emptyset) = 0$ by hypothesis. 
Let $\{A_k\}$ be a collection of disjoint measurable sets. 
Define $B_n = \bigcup_{k = 1}^n A_k$. 
Notice that $B_n \nearrow \bigcup_{k = 1}^\infty A_k$ is an increasing sequence of sets in $\mathcal A$. 
Observe that 
\begin{align*}
\mu \bigg( \bigcup_{n = 1}^\infty B_n \bigg) &= \lim_{n \to \infty} \mu(B_n) \\
&= \lim_{n \to \infty} \mu \bigg( \sum_{k = 1}^n A_k \bigg) \\
&= \lim_{n \to \infty} \sum_{k = 1}^n \mu(A_k) \\
&= \sum_{k = 1}^\infty \mu(A_k)
\end{align*}
Conclude by definition that $\mu$ is a measure.
