Does countably monotone imply finitely monotone and vice versa? If a set function $\mu$ is finitely monotone and has the property that $\mu(\phi)=0$, does it imply it is countably monotone?
Royden claims that if a countably monotone function $\mu$ has the property that $\mu(\phi)=0$, then it is finitely monotone. Why is this the case?
In general, is finitely monotone a stronger requirement or countably monotone?
 A: 
Lemma Any set function $\mu$ which is countable monotone and which satisfies $\mu(\emptyset)=0$ is finitely monotone.

Proof: Let $B_1,\ldots,B_n$ be measurable sets. Define a sequence $(A_k)_{k \in \mathbb{N}}$ by $$A_k := \begin{cases} B_k, & k \leq n, \\ \emptyset, & k >n. \end{cases}$$ Using the countably monotonicity we find $$\mu \left( \bigcup_{k=1}^n B_k \right) = \mu \left( \bigcup_{k=1}^{\infty} A_k \right) \leq \sum_{k=1}^{\infty} \mu(A_k).$$ By assumption, we have $\mu(A_k) = \mu(\emptyset)=0$ for all $k >n$ and therefore we get $$\mu \left( \bigcup_{k=1}^n B_k \right) \leq \mu(B_1)+\ldots+\mu(B_n).$$

A set function which is finitely monotone is, in general, not countably monotone. Consider, for example,
$$\mu(A) := \begin{cases} 0, & \text{$A$ is a finite set}, \\ \infty & \text{otherwise} \end{cases} $$
for $A \subseteq \mathbb{N}$. It is not difficult to see that $\mu$ is finitely monotone; indeed: If $B_1,\ldots,B_n \subseteq \mathbb{N}$, then either


*

*all $B_j$ are finite, then also $\bigcup_{j=1}^n B_j$ is a finite set and so $$\mu \left( \bigcup_{j=1}^n B_j \right) = 0 = \sum_{j=1}^n \mu(B_j).$$


or


*

*at least one of the $B_j$ is not a finite set, then also $\bigcup_{j=1}^n B_j$ is not a finite set and so $$\mu \left( \bigcup_{j=1}^n B_j \right) = \infty = \sum_{j=1}^n \mu(B_j).$$


However, $\mu$ is not countably additive (just consider $B_j := \{j\}$, then $\sum_{j=1}^{\infty} \mu(B_j)=0$, but $\mu \left( \bigcup_{j \geq 1} B_j \right) = \infty$.)
