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Let $F$ be the power set of $\Bbb{N}$ and consider the measurable space $(\Bbb{N}, F)$. Then what does it mean to take the integral with respect to the measure $\mu(A) = \sum_{a \in A} \frac{1}{a}$. What would $\int f \ d\mu$ represent, where $f$ is some function $f: \Bbb{N} \to \Bbb{R}$?

My attempt. Take the simplest integrand which is usually $1$ and integrate to get $\int 1 \ d\mu = 1 \mu(\Bbb{N}) = \infty$. This means what?

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    $\begingroup$ It is just a fancy way of saying weighted sums (in the case, weighted by $\frac{1}{a}$). IMHO, the most useful thing of this sort of setup is the huge machineary on Lebesgue-integral (e.g DCT, MCT ) now works for weighted sums... $\endgroup$ – achille hui May 23 '15 at 1:15
  • $\begingroup$ What do you mean by “what does it mean”? $\endgroup$ – Carsten S May 23 '15 at 8:46
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$\int f d\mu = \sum_a f(a)\mu(\{a\}) = \sum_a \frac{f(a)}{a}$

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  • $\begingroup$ Although note that, as a Lebesgue integral, this is only well-defined if the sum is absolutely convergent. $\endgroup$ – Hurkyl May 23 '15 at 13:13
  • $\begingroup$ Well, it's a sum over a set, not a sequence, so it can't be conditionally convergent. $\endgroup$ – Ben Millwood May 23 '15 at 16:06
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Consider $f: A \to \Bbb N$. Now write: $$ \int_A f(n) \,{\rm d}\mu(n) = \sum_{n \in A}\int_{\{n\}}f(n)\,{\rm d}\mu(n) = \sum_{n \in A}f(n)\mu(\{n\}) = \sum_{n \in A} \frac{f(n)}{n}. $$

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  • $\begingroup$ What does it mean though, if you don't mind that type of question? $\endgroup$ – Shine On You Crazy Diamond May 23 '15 at 0:51
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    $\begingroup$ This one I would see just like a "weighted sum". A more realistic interpretation is if you use the counting measure instead of that one - this way, integrating over $\Bbb N$ gives you the sum of a series. $\endgroup$ – Ivo Terek May 23 '15 at 0:58
  • $\begingroup$ I don't know if much more can be said regarding "what it means"; however, we might note that since the harmonic series lies "at the boundary" between convergence and divergence, this measure has the interesting property that $\int_A f(n) \,{\rm d}\mu(n) <\infty $ iff $\lim_{n\to \infty} f(n)=0$. $\endgroup$ – ApproximatelyTrue May 23 '15 at 7:32
  • $\begingroup$ @ApproximatelyTrue Not really, consider $f(n)=1/\log n$. $\endgroup$ – Chris Culter May 23 '15 at 10:25
  • $\begingroup$ @ChrisCulter Ah yes, you are right. I was imagining stuff that goes to 0 as $n^{-\epsilon}$, but it seems the claim is not true in general. In that case, I really don't know what more can be said about the measure that might be interesting. $\endgroup$ – ApproximatelyTrue May 23 '15 at 20:04

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