I want to find some condition to construct a continuous finitely additive measure on the natural numbers, i.e. $f:P(\mathbb{N})\rightarrow [0,1]$ such that $f(\{n\})=0$, and $f$ is an additive measure.


I know in ZFC we can use an ultrafilter $U$ and define $f$ by $f(A)=1\Leftrightarrow A\in U$, but this is too trival.


How about ZF? or some other condition? like large cardinal.



How to construct a continuous finite additive measure on the natural numbers
In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$
(giving all singletons measure 0 and $\mathbb N$ measure 1).
The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property,
and such a set cannot be constructed without some help of the axiom of choice.
In other words, there is no explicit "construction" of a non-trivial finitely additive measure.


Since you don't like the ultrafilter measure, here is one that is less trivial:


Instead of $\mathbb N$, we use the integers $\mathbb Z$ as the underlying set.
For $n\in\mathbb N$ let $d\_n(A)=\frac{|A\cap\{-n,\dots,n\}|}{2n+1}$.
The sequence $(d\_n(A))\_{n\in\mathbb N}$ typically does not converge.
We fix this using an ultrafilter $U$ on $\mathbb N$ as follows:


For each $A\subseteq\mathbb Z$ let $\mu(A)$ be the unique element of the set
$\bigcap\_{S\in U}\mbox{cl}(\{d\_n(A):n\in S\})$.
Here the closure is taken in $\mathbb R$.
Now $\mu$ is a finitely additive measure on $\mathcal P(\mathbb Z)$ that is even translation invariant and hence very different from the ultrafilter measure that you find too trivial. 


Notice that I have just proved the well known fact that the integers are an amenable group.


