Does the following function
$f:\mathcal{P}(\mathbb{N})\rightarrow\{0,1\}$ exist :  

$f(\mathbb{N})=1$,  

$f(A\cup B)=f(A)+f(B)$ for $A\cap B=\emptyset$,  

$f(A)=0$ for finite $A$ 



Existence of a special density
Since the suggestion to close the question has not been
successful, allow me to offer an answer as a way of
bringing the question to a conclusion.


Many mathematicians find the situation of the question,
where one has a finitely additive measure on the natural
numbers, to be both fascinating and disturbing.


Nevertheless, one may indeed construct such a function $f$
as you describe. If $\mu$ is any [nonprincipal
ultrafilter](http://en.wikipedia.org/wiki/Ultrafilter) on
$\mathbb{N}$, we may define that $f(A)=1$, if $A\in\mu$,
and otherwise $f(A)=0$. Ultrafilters are often described as
provided a sense of "largeness", and this conforms with the
measure-theoretic sense of your question, since the sets in
$\mu$ get measure $1$, and the other sets get measure $0$.
The reason it works out for additivity is that disjoint
sets are never both large, and so the requested $f(A)+f(B)$ always has the form $0+0$, $0+1$ or $1+0$. Specifically, the fact that
$\mu$ is nonprincipal implies that finite sets get measure
$0$; since $\mathbb{N}\in\mu$, we see that
$f(\mathbb{N})=1$; and the additivity property follows from
the fact any two sets in $\mu$ have a nonempty
intersection, so if $A\cap B=\varnothing$, then at most one of
them is in $\mu$.


But in fact, the existence of a function $f$ as you
describe is exactly equivalent to a nonprincipal
ultrafilter on $\mathbb{N}$, since for any such function
$f$ the set of $A\subset\mathbb{N}$ with $f(A)=1$ must
contain every set or its complement by the additivity
property and contains no finite set by your other
hypothesis. Ultimately, it is a nonprincipal ultrafilter on
$\mathbb{N}$.


There is nothing special about $\mathbb{N}$ in the argument
above, and we might consider nonprincipal ultrafilters on
any set $X$, which are equivalent to the existence of a
finitely-additive measure $f$ on $X$ as in your question,
giving measure $1$ to $X$ and measure $0$ to any finite
subset of $X$.


There are a variety of ways to prove that there is a
nonprincipal ultrafilter on a set $X$.


* One may first observe that the collection of co-finite
subsets of $X$ forms a filter, often called the Frechet filter, and then by Zorn's lemma
we may extend this to a maximal filter, which one then proves
contains every set or its complement.
* Instead, one may enumerate all subsets of $X$ in a well
ordered sequence $X\_\alpha$, and then decide at stage
$\alpha$ of a transfinite recursion to add $X\_\alpha$ to the filter being constructed,
if this has infinite intersection with all the sets added
so far, and otherwise adding the complement of $X\_\alpha$.
The result is a nonprincipal ultrafilter on $X$.
* Another way to do it is by appealing to the compactness
theorem of first-order logic. Write down the theory axiomatizing the desired properties of $f$
in the language consisting of a function symbol for $f$,
plus constants $A$ for each $A\subset\mathbb{N}$ asserting
that $f(A)$ is either $0$ or $1$, that additivity holds,
that finite sets get measure $0$ and that $\mathbb{N}$ gets
measure $1$. This theory is finitely consistent, and so by the compactness theorem it
has a consistent completion, which provides an actual
measure $f$, as desired.
* In the case of $\mathbb{N}$, one could consider a
nonstandard model of arithmetic $\mathbb{N}^\ast$, and fix
a nonstandard infinite integer $N\in\mathbb{N}^\ast$. Now,
define that $f(A)=1$, if $N\in A^\ast$, where $A^\ast$ is
the nonstandard analogue of $A$, by the transfer principle, and otherwise $f(A)=0$.
Thus, the measure $f$ is the trace on the standard model
of the nonstandard measure concentrating on the point $N$.
This is finitely additive, gives full measure to
$\mathbb{N}$ and measure $0$ to finite sets, as desired.
* In fact, every $f$ as you requested is exactly like the previous example,
since if $\mu$ is the ultrafilter of $f$-measure one sets,
then the nonstandard number $N=[id]\_\mu$ in the ultrapower
of $\mathbb{N}$ by $\mu$ has exactly the property that
$f(A)=1$ when $N\in A^\ast$.


Finally, it should be mentioned that the proofs I have
given above that there is a nonprincipal ultrafilter make
essential use of the axiom of choice, and it is consistent
with the ZF axioms without choice that there are no
nonprincipal ultrafilters and hence no such functions $f$
as you requested. A more subtle fact is that the existence
of ultrafilters extending any given filter is a strictly
weaker principle than the axiom of choice, and is
equivalent to the [Boolean prime ideal
theorem](http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem).
Thus, one should not expect completely explicit
constructions of examples of such functions $f$ as you have
requested, although in the constructible universe $L$ one
can find them within fairly low levels of the projective
hierarchy of definability.


Lastly, let me mention that the existence of a [measurable
cardinal](http://en.wikipedia.org/wiki/Measurable_cardinal),
one of the prominent large cardinal hypotheses, is exactly
equivalent to asking whether there is a function $f$
defined on subsets of a set $X$ as you request, which
moreover is countably additive, instead of merely finitely
additive.


