# Condition of being a measure

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!

• Hint: For $j\le-1$ it's not $\sigma$-additive. – Berci Sep 20 '18 at 23:26
• I don't really see how $j \leq -1$ implies that it's not countably additive. – Sank Sep 21 '18 at 0:07

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.
• Is the only issue the fact that {$1,2,4,8,16,...$} will have a finite measure if $j \leq -1$, which contradicts the construction of $\mu$? Is there anything else I should consider/worry about? – Sank Sep 21 '18 at 0:32
• Yes, that's the only issue: if the measures of singletons are too low, infinite of them might sum up only to a finite number, while $\mu$ measures every infinite set as infinite. – Berci Sep 21 '18 at 7:39