Finding possible p's for signed measure Im new here at StackExchange and hope some of you can help me with a problem I'm dealing with. 
We are considering a signed measure $\nu$ on $(\mathbb{N},2^{\mathbb{N}})$, which is given by $\nu(\{k\}):= k^p (-1)^k$ with $k \in \mathbb{N}$.
Now I have to find out, which values of $p \in \mathbb{R}$ are possible?
I know, that signed measures satisfy $\nu( \emptyset ) = 0$. Due to there is not given, what happens with empty sets, I would associate this with inserting zero for k (assuming, that $0 \notin \mathbb{N}$ ). So this doesn't give me additional information to determine possible $p$'s
Furthermore for countable $I$ and a set of pairwise disjoint $A_i$'s holds $\nu(\bigcup_{i\in I}A_i) = \sum_{i \in I} \nu(A_i)$.
I guess this is one information, that might be used to determine possible $p$'s. But my problem is, that I don't really know, how to use this, due to there is no real definition, what happens, if I put something different than singletons in $\nu$. Is there anything, that I have overseen, or is there something missing in the indication? Or is there a reason, why we can claim, that $\nu(\{k,m\}):= (k+m)^p (-1)^{k+m} = \nu(\{k+m\})$ ?
Thank you very much for your help!
 A: [Note: My definition of a signed measure requires that the measure of every measurable set is finite. This is to avoid things like $\infty - \infty$.]
There is a slick way to write out a solution to this problem if you are prepared to quote the Radon-Nikodym theorem. Let $\nu$ be a signed measure on $\mathbb N$. Applying Radon-Nikodym with $\mu$ being the standard counting measure on $\mathbb N$ (which is certainly $\sigma$-finite), we learn that there must exists a function $h : \mathbb N \to \mathbb R$ in $L^1(\mu)$ such that $ \nu (A) = \int_A  h \ d \mu $
for all $A \subset \mathbb N$. (There is no singular contribution, because the only measure that is singular with respect to $\mu$ is the zero measure.) Conversely, for any real $h \in L^1(\mu)$, the expression $\nu (A) = \int_A  h \ d \mu$ defines a signed measure $\nu$ on $\mathbb N$.
In your problem, you are given that $\nu (\{ n \}) = n^p (-1)^n $
for all $n \in \mathbb N$.  This immediately tells you that $h(n) = n^p (-1)^n.$ The $h \in L^1 (\mu)$ condition says that
 $\sum_{n=1}^\infty n^p < \infty,$ and this holds iff $p < -1.$

If you're not allowed to use Radon-Nikodym theorem, it should still be possible to solve this problem in the same spirit, following these steps:


*

*Use the axioms for a signed measure, together with what you have been told about the $\nu$ measure of singleton sets, to show that the only possible definition of $\nu$ is $\nu (A) = \sum_{n \in A} n^p (-1)^n$.

*For this definition to make sense, it must be the case that $\sum_{n \in A } n^p (-1)^n$ is convergent, finite and independent of summation ordering for any $A \subset \mathbb N$. Apply this to $A = \{1, 3, 5, \dots \}$ and $A = \{2, 4, 6, \dots \}$ to argue that $\sum_{n = 1}^\infty n^p < \infty $ is a necessary condition on $p$.

*Let $\mu$ denote the standard counting measure on $\mathbb N$. Then the condition $\sum_{n = 1}^\infty n^p < \infty $ ensures that the function $h : \mathbb N \to \mathbb R$ given by $h(n) = n^p (-1)^n$ is $\mu$-integrable. Use dominated convergence to show that if $\sum_{n = 1}^\infty n^p < \infty $, then for any $A \subset \mathbb N$, the series $\sum_{n \in A} n^p (-1)^n$ is convergent, finite, independent of ordering, and equal to $\int_A h \ d\mu$. So the condition $\sum_{n = 1}^\infty n^p < \infty $ implies that my definition for $\nu(A)$ makes sense.

*Finally, show that the condition $ \sum_{n \in A} n^p < \infty $ is sufficient to ensure that $\nu (\cup_{j \in J} A_j) = \sum_{j \in J} \nu (A_j)$ for any arbitrary countable collection of disjoint subsets $\{ A_j \}_{j \in J}$. This can again be proved by dominated convergence, using the fact that $\nu (A) = \int h \mathbb 1_A \ d\mu$ for all $A \subset \mathbb N$.
