Given a measure space $X$ with positive measure $m$, and a decreasing sequence of sets, i.e., $A_{i} \supset A_{i+1}$, it might not be true that we have continuity of measure from above: $m(\cap A_{i}) = \lim m(A_{i})$. For instance, if each $A_i$ has infinite $m$-measure, there are simple examples where $m(\cap A_{i}) \ne \lim m(A_{i})$. (Convergence is always understood to be in the extended real line.)
Suppose now $X$ is a set with a $\sigma$-algebra $\mathcal A$ and a set function $m \colon \mathcal A \to [0,\infty]$ such that $m(\emptyset)=0$ and $m$ is finitely additive. For $(X,\mathcal A,m)$, suppose the continuity theorem holds for every decreasing sequence of sets. Then can we say that $(X,\mathcal A,m)$ is a measure space?
I have been able to find an example where the measure is not finite and the condition holds. However, I believe that there must be an example of a finitely additive set function $m$ on a $\sigma$-algebra $\mathcal A$ on a set $X$, such that $m$ is continuous from above, but $m$ is not a measure.
Any help is appreciated.
This question is inspired by an exercise in Royden.
I am not sure I understand your question. If it is ("...then can we say that...") then the answer is positive, i.e. if $\cal A$ is a $\sigma$-algebra and $m:{\cal A}\to[0,+\infty]$ is finitely additive and $m(\varnothing)=0$ and for any decreasing sequence $(C_n)$ of elements of $\cal A$, $m(\cap C_n)=\lim m(C_n)$, then $m$ is $\sigma$-additive.
Proof : Let $(A_n)$ be a sequence of pairwise disjoint elements of $\cal A$.
Let $B_n=\cup_{k\le n}A_k$ and $C_n=\cup_{k>n}A_k$.
Hence (by finite additivity again) $\sum m(A_k)=\lim m(B_n\cup C_n)$, i.e. $\sum m(A_k)=m(\cup A_k)$.