Is the following property true:
If $\mu$ is a finitely additive function on an algebra (note: not $\sigma-$algebra) $\mathcal{F}$. And $\mu$ has the following property: If $A_i$ is a decreasing sequence of sets in $\mathcal{F}$ with $\inf_n \mu(A_n) >0$ then $\cap_{n \geq 1}A_n \neq \phi$. Then $\mu$ is countably additive on $\mathcal{F}$.
This question comes to me because I am unable to understand the following part in the proof of Kolmogorv extension theorem given in this source:
Since $\mu^{\prime}$ is a finitely-additive measure, to prove countable additivity it will suffice to show that if $A_1, A_2, \ldots$ are a decreasing sequence of sets in $\mathcal{F}$ with $\inf_n \mu^{\prime}(A_n)>0$ then $\cap_{n \geq 1} A_n \neq \phi$.
(Here $\phi$ denotes the empty set.)
Countable additivity of a measure $\mu$ is equivalent to the following:
For all decreasing sequences of sets $A_1,A_2,...$ in $\mathcal{F}$ with $\bigcap_{n=1}^\infty A_n=\phi$, we have $\mu(A_n)\to 0$ as $n\to\infty$.
Hence, if $\inf_n \mu'(A_n)>0$ for a decreasing sequence $(A_n)_{n=1}^\infty$, then we cannot have $\bigcap_{n=1}^\infty A_n=\phi$ by the above equivalence.
This is sometimes stated as "countable additivity of finite measures equivalent to continuity from above". A similar equivalence holds for continuity from below, where we instead consider increasing sequences of sets.