Finiteness requirement regarding a statement on measures I've seen this as a problem in a few different places :

Let $(X,\mathcal A)$ be a measurable space and $\mu:\mathcal A\to \mathbb R^+$ be a function that is finitely additive and such that $\mu(\emptyset)=0$ and $\mu(X)<\infty$. Suppose that whenever $A_i$ is a sequence of sets in $\mathcal A$ that decreases to $\emptyset$, then $\lim_n \mu(A_n)=0$. Prove that $\mu$ is a measure.

Here's my proof: it suffices to prove that for disjoint $A_i$ in $\mathcal A$, $\mu(\cup_i A_i) =\sum_{i=1}^\infty \mu(A_i)$.
Let $n\in \mathbb N$. Since $\mu$ is finitely additive, $$\mu(\cup_i A_i) = \mu(\cup_{i=1}^nA_i) +\mu(\cup_{i=n+1}^{\infty}A_i)= \sum_{i=1}^n \mu(A_i)+\mu(\cup_{i=n+1}^{\infty}A_i)$$ 
$\left(\cup_{i=n+1}^{\infty}A_i\right)$ is a sequence of sets indexed by $n$  that decreases to  $\emptyset$ (essentially because the $A_i$ are disjoint). 
Hence $\displaystyle \lim_{n\to \infty} \mu(\cup_{i=n+1}^{\infty}A_i) =0$. Therefore $\mu(\cup_i A_i)  = \sum_{i=1}^{\infty} \mu(A_i)$.

What looks suspicious here is that I didn't use the hypothesis $\mu(X)<\infty$. Is my proof flawed ?

 A: I found no flaw in the proof, but the result can also be found in Proposition 1.2.6. in Measure theory 2ed. by Donald L. Cohn. 
The reason you might have the $\mu(X)<\infty$ condition is that "continuity at zero" is equivalent to countable additivity for $\mathbb{R_+}$-valued measures. For that statement the condition $\mu(X)<\infty$ is needed to show the converse of your implication, since it is not in general true for measures with values in $[0,\infty]$. E.g. let $\mu$ be the countably additive $[0,\infty]$-valued counting measure on $(\mathbb{N},\mathcal{P}(\mathbb{N}))$ and let
$$
A_n = \bigcap_{k=1}^n \{m\in \mathbb{N}:m\geq k\} = \{m\in\mathbb{N}:m\geq n\}, \quad \quad \text{ for }n\geq 1.
$$
We note that $A_n\supset A_{n+1}$ for all $n\in\mathbb{N}$ and that
$
A_n \to_n \bigcap_{k=1}^\infty \{m\in \mathbb{N}:m\geq k\} =\emptyset
$
but
$
\mu(A_n)=\infty, \, \forall n\geq 1  \implies \lim_{n\to\infty}\mu(A_n)\not= 0
$,
in which case countable additivity of $\mu$ does not imply "continuity at zero" of $\mu$.
