From finitely additive to countably additive

There are several easy criterions other than those you have mentioned that, together with finite additivity, imply countable additivity (but they all end up implying countable subadditivity in the end). Here are two examples:

1. If $\mu$ is finitely additive and continuous from below (that is, if $A_1\subset A_2\subset\cdots$ and $A=\bigcup_nA_n$, then $\mu(A_n)\to A$), then $\mu$ is countably additive: Take $\{A_n\}$ disjoint, and define $B_n=A_1\cup\cdots\cup A_n$. Then, $B_1\subset B_2\subset\cdots$ and $\bigcup_nB_n=\bigcup_nA_n$, and thus, finite additivity implies that $$\lim_{n\to\infty}\sum_{i=1}^n\mu(A_i)=\lim_{n\to\infty}\mu(B_n)=\mu\left(\bigcup_{n}A_n\right).$$

2. If $\mu$ is finitely additive and continuous from above at $\varnothing$ (that is, if $A_1\supset A_2\supset\cdots$ and $\bigcap_nA_n=\varnothing$, then $\mu(A_n)\to0$), then $\mu$ is countably additive: Take $\{A_n\}$ disjoint, and define $B_n=A_1\cup\cdots\cup A_n$. Then, $\bigcup_nB_n=\bigcup_nA_n$, and for every $m$, we have by finite additivity that $$\mu\left(\bigcup_nA_n\right)=\mu\left(B_m\cup\left(\bigcup_nA_n\right)\setminus B_m\right)=\mu\left(B_m\right)+\mu\left(\left(\bigcup_nA_n\right)\setminus B_m\right).$$ Now, we see that $\left(\bigcup_nA_n\right)\setminus B_1\supset\left(\bigcup_nA_n\right)\setminus B_2\supset\cdots$, and that $\bigcap_m\left(\left(\bigcup_nA_n\right)\setminus B_m\right)=\varnothing$. Thus, $$\mu\left(\bigcup_nA_n\right)=\lim_{m\to\infty}\left[\mu\left(B_m\right)+\mu\left(\left(\bigcup_nA_n\right)\setminus B_m\right)\right]=\lim_{m\to\infty}\mu(B_m)+0.$$ Since $\mu(B_m)=\sum_{i=1}^m\mu(A_i)$ for each $m$, this implies countable additivity.