Dtyjlui

Given a finitely additive measure $mu : mathcal{B} rightarrow [0,infty]$. If we want to show that it is also countably additive, finite additivity implies
$$mu(E) geq sum_{n=1}^infty mu(E_n)$$
where $cup E_n = E$ with $E_n$ disjoint automatically.

For the inequality in the other direction, clearly countably subadditive is sufficient. I was wondering if there are weaker conditions that would also be sufficient. (Just take $mathcal{B}$ to be the Borel $sigma$-algebra on $mathbb{R}$)

I know for fact that if $mu$ is a Radon measure would also be enough (maybe $sigma$-finite with some regularity properties).

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 $A_1subset A_2subsetcdots$ and $A=bigcup_nA_n$,
then $mu(A_n)to A$),
then $mu$ is countably additive: Take ${A_n}$ disjoint,
and define $B_n=A_1cupcdotscup A_n$.
Then,
$B_1subset B_2subsetcdots$ and $bigcup_nB_n=bigcup_nA_n$,
and thus,
finite additivity implies that
$$lim_{ntoinfty}sum_{i=1}^nmu(A_i)=lim_{ntoinfty}mu(B_n)=muleft(bigcup_{n}A_nright).$$

2. If $mu$ is finitely additive and continuous from above at $varnothing$ (that is,
if $A_1supset A_2supsetcdots$ and $bigcap_nA_n=varnothing$,
then $mu(A_n)to0$),
then $mu$ is countably additive:
Take ${A_n}$ disjoint,
and define $B_n=A_1cupcdotscup A_n$.
Then,
$bigcup_nB_n=bigcup_nA_n$,
and for every $m$,
we have by finite additivity that
$$muleft(bigcup_nA_nright)=muleft(B_mcupleft(bigcup_nA_nright)setminus B_mright)=muleft(B_mright)+muleft(left(bigcup_nA_nright)setminus B_mright).$$
Now,
we see that $left(bigcup_nA_nright)setminus B_1supsetleft(bigcup_nA_nright)setminus B_2supsetcdots$,
and that $bigcap_mleft(left(bigcup_nA_nright)setminus B_mright)=varnothing$.
Thus,
$$muleft(bigcup_nA_nright)=lim_{mtoinfty}left[muleft(B_mright)+muleft(left(bigcup_nA_nright)setminus B_mright)right]=lim_{mtoinfty}mu(B_m)+0.$$
Since $mu(B_m)=sum_{i=1}^mmu(A_i)$ for each $m$,
this implies countable additivity.

user78270's answer is good, but leaves out some details. The following is from Measure Theory by F. R. Halmos, first edition, section 9, page 38. Remark on notation

$
text{Let ${E_{n}}$ be a sequence, then}\
$
$$
limlimits_{nrightarrowinfty}E_{n} = bigcaplimits_{n=1}^{infty}bigcuplimits_{mgeq n}^{infty}E_{m} = bigcuplimits_{n=1}^{infty}bigcaplimits_{mgeq n}^{infty}E_{m}
$$
$text{i.e. } limlimits_{nrightarrowinfty}E_{n}=liminflimits_{nrightarrowinfty}E_{n}=limsuplimits_{nrightarrowinfty}E_{n}$

Theorem:

$quad$If $mu$ is a measure on a ring $mathcal{R}$, and if ${E_{n}}$ is an increasing sequence of sets in $mathcal{R}$ such that $limlimits_{nrightarrowinfty}E_{n} in mathcal{R}$, then $$mu(limlimits_{nrightarrowinfty}E_{n}) = limlimits_{nrightarrowinfty}mu(E_{n})$$

Theorem:

$quad$If $mu$ is a measure on a ring $mathcal{R}$, and if ${E_{n}}$ is an decreasing sequence of sets in $mathcal{R}$ of which at least one has finite measure and for which $limlimits_{nrightarrowinfty}E_{n} in mathcal{R}$, then $$mu(limlimits_{nrightarrowinfty}E_{n}) = limlimits_{nrightarrowinfty}mu(E_{n})$$