Axiom of Finite Additivity and Axiom of Continuity imply Axiom of Countable Additivity. This exercice in particular comes from Statistical Inference by Casella and Berger, I have been following its solution manual. The exercice is 1.12 and can be found here http://exampleproblems.com/Solutions-Casella-Berger.pdf
Let $S$ be a sample space with associated $\sigma$-algebra $\mathcal{A}$. 
Finite Additivity: Let $A,B \in \mathcal{A}$, $P(A \sqcup B) = P(A) + P(B)$ 
Continuity: Let $\{A_i\}_i$ be a sequence of nested sets s.t. $A_k \downarrow \emptyset$, then $P(A_k) \rightarrow 0$.
$$P\left(\bigsqcup_{i=1}^\infty A_i\right) = P\left(\bigsqcup_{i=1}^k A\right) + P\left(\bigsqcup_{i=k}^\infty\left) = \sum_{i=1}^k P(A_i) + P\right(\bigsqcup_{i=k}^\infty A_i\right)$$
Define $A^*_k = \bigsqcup_{i=k}^\infty A_i$, then $\{A^*_k\}_k$ is a sequence of nested sets. My problem is how to see that $A^*_k \downarrow \emptyset$. The argument is that $A_k \downarrow \emptyset$ as $k \rightarrow \infty$ otherwise the sum of probabilities would be infinite.
But for example, say $S = [0,1]$, $\mathcal{A} = \mathcal{B}([0,1])$ and for every $A \in \mathcal{A}$ we define $P(A)$ as its lebesgue measure, then $(S, \mathcal{A},P)$ is a probability space. In particular we can construct a nested sequence of sets $\{A_i\}$ st $A_i = \{x_0\}$ for all $i>k$, $\sum_i P(A_i) \leq 1$ and $A_i \not\downarrow \emptyset$.
What am I missing?
 A: By continuity, $A_k \searrow A$ implies that $P(A_k) \to P(A)$ (because $A_k \setminus A \searrow \emptyset$).
Now $B_k = \bigcup_{n\geq k} A_n$ is a decreasing sequence, and $B_k \searrow A$ for some $A$.
If $P(A) > 0$, then we would have $P(\bigcup_{n \geq 1} A_k) = \infty = \sum_{n \geq 1} P(A_k)$ which doesn't make sense since $P$ is a probability.
Hence $P(A) = 0$, and we're done.
A: What happens here is not that $\displaystyle \bigsqcup_{i=k}^\infty A_i \downarrow \varnothing$ as $k\to\infty,$ but rather that $\displaystyle P\left( \bigcap_{k=1}^\infty \bigsqcup_{i=k}^\infty A_i \right) = 0.$
There are nonempty sets whose probability is $0.$ This intersection is in some cases one of those. Where Casella and Berger wrote $B_k \to \phi$ as $k\to\infty,$ they were wrong unless their textbook established a convention according to which that notation means the intersection of the descending sequence is a set whose probability is $0.$
A: This is because the $A_i$ are disjoint. Indeed, for any $\omega$, if $\omega\in A_i$ for some $i$, then $\omega\notin A_j$ for all $j>i$ and hence $\omega\notin\bigcup_{k\ge j}A_k$ for all $j>i$. This shows the intersection is empty (as any $\omega$ in the intersection would obviously have to be in at least one $A_i$) and thus proves countable additivity.
