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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!

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

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  • $\begingroup$ 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? $\endgroup$ – Sank Sep 21 '18 at 0:32
  • $\begingroup$ 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. $\endgroup$ – Berci Sep 21 '18 at 7:39

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