Difficulty in understanding converse part of proof of a propostion in Andrew Browder's Mathematical Analysis 
Proposition: Let $\mu$ be finitely additive set function, defined on the algebra $\mathscr A$. Then $\mu$ is countably additive if and only if its has following property: if $A_n \in \mathscr A$ and $A_n \subset A_{n+1}$ for each positive integer $n$, and if $\cup_{n=1}^{\infty} A_n \in \mathscr A$, then $\mu(\cup_{n=1}^{\infty} A_n)= \lim_{n\to \infty} \mu(A_n)$.

To prove converse, that is to prove that $\mu$ is countably additive the proof given is as below:

Suppose that $\mu$ is finitely additive, and has the given property(called as "continuity from below" property). If $A_n$ is a sequence of pairwise disjoint sets in $\mathscr A$, and we put $B_n=\cup_{k=1}^{n}A_k$, then $B_n \subset B_{n+1}$ for every $n$, so $\mu(B_n) \to \mu(B)$, where $B=\cup_{n=1}^{\infty} B_n=\cup_{n=1}^{\infty}A_n$. But since $\mu$ is finitely additive, $\mu(B_n)=\sum_{k=1}^{n} \mu(A_k)$, so $\mu(B_n) \to \sum_{k=1}^{\infty} \mu(A_k)$. Thus, $\mu$ is countably additive.

My difficulties:


*

*Where was the "Continuity from below" property used?

*Since we are working on an algebra i.e. $\mathscr A$, how come $\cup_{n=1}^{\infty} B_n \in \mathscr A$?(This difficulty came to me because $\mu(B)$ is mentioned in the proof but for that to happen, $B$ must belong to $\mathscr A$. Also $\cup_{n=1}^{\infty}A_n \in \mathscr A$ is one of the statements in the hypothesis of the "continuity from below" property so $B\in \mathscr A$ should happen in order to get the result.)


I am seeking for a different approach to this converse part too. 
 A: It's probably good to clarify some definitions. If $\mathscr{A}$ is an algebra of sets, and $\mu \colon \mathscr{A} \to [0,+\infty]$ is a finitely additive set function, then we say that $\mu$ is $\sigma$-additive, if whenever $(A_n)_{n\in \mathbb{N}}$ is a sequence of pairwise disjoint sets in $\mathscr{A}$ such that $\bigcup\limits_{n\in\mathbb{N}} A_n \in \mathscr{A}$, then
$$\mu \biggl(\bigcup_{n\in \mathbb{N}} A_n\biggr) = \sum_{n\in \mathbb{N}} \mu(A_n).$$
Since $\mu$ is only defined on $\mathscr{A}$, the condition only makes sense if the union is again a member of $\mathscr{A}$. In general, the union of a countable family of members of $\mathscr{A}$ need not belong to $\mathscr{A}$, and for every algebra that is not a $\sigma$-algebra at least one countable family whose union doesn't belong to the algebra exists.
Similarly, for the "continuity from below" property, we can only consider ascending sequences of members of $\mathscr{A}$ whose union is again a member of $\mathscr{A}$.
Then, to prove the equivalence of $\sigma$-additivity and continuity from below, note that we can write every union of an ascending sequence in $\mathscr{A}$ as the union of a sequence of pairwise disjoint members of $\mathscr{A}$: If $A_n \subset A_{n+1}$ for all $n\in \mathbb{N}$, then since $\mathscr{A}$ is an algebra the sets $B_n = A_n \setminus A_{n-1}$ (where we set $A_{-1} = \varnothing$) belong to $\mathscr{A}$, and by construction they are pairwise disjoint, and
$$\bigcup_{n = 0}^\infty A_n = \bigcup_{n = 0}^\infty B_n.$$
And conversely, we can write every infinite union of members of $\mathscr{A}$ as the union of an ascending sequence in $\mathscr{A}$, if $A_n \in \mathscr{A}$ for all $n$, then $B_n = \bigcup\limits_{k = 0}^n A_k \in \mathscr{A}$ for all $n$, $B_n \subset B_{n+1}$, and
$$\bigcup_{n = 0}^\infty A_n = \bigcup_{n = 0}^\infty B_n.$$
So if $\mu$ is $\sigma$-additive, and $A_n$ is an ascending sequence in $\mathscr{A}$ whose union again belongs to $\mathscr{A}$, then by $\sigma$-additivity we have
\begin{align}
\mu\biggl(\bigcup_{n = 0}^\infty A_n\biggr) &= \mu\biggl(\bigcup_{n = 0}^\infty (A_n \setminus A_{n-1})\biggr) = \sum_{n = 0}^\infty \mu(A_n \setminus A_{n-1})\\
&= \lim_{N\to \infty} \sum_{n = 0}^N \mu(A_n \setminus A_{n-1}) = \lim_{N\to \infty} \mu\biggl(\bigcup_{n = 0}^N (A_n \setminus A_{n-1})\biggr)\\
&= \lim_{N\to \infty} \mu\biggl(\bigcup_{n = 0}^N A_n\biggr) = \lim_{N\to \infty} \mu(A_N),
\end{align}
which is the continuity from below.
Conversely, if $(A_n)$ is a sequence of pairwise disjoint members of $\mathscr{A}$ whose union belongs to $\mathscr{A}$, then, with $B_n = \bigcup\limits_{k = 0}^n A_k$ we have
\begin{align}
\mu\biggl(\bigcup_{n = 0}^\infty A_n\biggr) &= \mu\biggl(\bigcup_{n = 0}^\infty B_n\biggr)\\
&= \lim_{n \to \infty} \mu(B_n) \tag{continuity from below}\\
&= \lim_{n \to \infty} \mu\biggl(\bigcup_{k = 0}^n A_k\biggr)\\
&= \lim_{n \to \infty} \sum_{k = 0}^n \mu(A_k)\tag{finite additivity}\\
&= \sum_{k = 0}^\infty \mu(A_k),
\end{align}
which is the $\sigma$-additivity.
