A lecturer made a passing reference to this... he said that finite additivity plus continuity is equivalent to countable additivity... could someone offer a proof of this or else direct me to one?

I don't understand why the continuity criterion is necessary

If $\displaystyle F_i\cap F_j = \emptyset \ \forall i\neq j \Rightarrow \mu(\bigcup_{i=1}^NF_i) = \sum_{i=1}^N\mu(F_i)$

Then don't we get countable additivity straight away just by induction?

Can someone provide a counterexample? Thanks

2. $\displaystyle \mu$ is a finitely additive measure?

What induction already gave you is that "two-set additivity" $\displaystyle A \cap B = 0 \rightarrow \mu(A \cup B) = \mu(A)+\mu(B)$ implies finite additivity, it can't give you more (e.g. sigma-additivity).

For counterexample, take an ultrafilter $\displaystyle \mathcal{U}$ containing Fréchet filter on $\displaystyle \mathbb{N}$ (we don't want $\displaystyle \mathcal{U}$ to be trivial) and for every $\displaystyle A \subseteq \mathbb{N}$ set
$\displaystyle \mu(A) = \begin{cases} 1 & \mbox{if } A \in \mathcal{U} \\ 0 & \mbox{otherwise} \end{cases}$
It is straightforward to verify that $\displaystyle \mu$ is a finitely additive measure but is not sigma additive.

I don't quite understand the remark about continuity though.