I'm working on a measure theory problem and I'm completely stumped.
I'm trying to find out for which integers $j$, $\mu$ will be a measure on $(\mathbb Z_+, \mathcal P(\mathbb Z_+))$ where $\mu$ is defined to be: $$ \mu(E)= \begin{cases} \sum_{n\in E}n^j&\text{if}\, c(E)< \infty\\ \infty&\text{if}\, c(E) = \infty\\ \end{cases} $$ *$c$ is the counting measure here.
I think I can simply consider finite sets or countable union of disjoint sets {${A_n}$}$_{n=1}^\infty$ where for some $N$, $A_n = \varnothing$ where $n\geq N$. Satisfying $\mu(\varnothing)$=$0$ is trivial, and since we are on $\mathbb Z_+$, $\mu(E) \geq 0$ for every $E \in \mathcal P(\mathbb Z_+)$. I think countable additivity would impose some condition on $j$, but not really sure of the point of attack here.
Any help is appreciated!
Since $\mu$ is defined as a sum over the elements, it is clearly finitely additive.
If $j\ge0$, then $n^j\ge 1$ for each positive integer $n$, so when we have infinitely many such numbers, they will sum up to infinity, matching the definition of $\mu$.
To see $\mu$ is not $\sigma$-additive for negative $j$'s, it's enough to consider only singletons.
Specifically, the sum of the $\mu$ of singletons in $\{1,2,4,8,16,\dots\}$ is finite then.