# Integrating over the naturals. What does it mean?

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?

• 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... – achille hui May 23 '15 at 1:15
• What do you mean by “what does it mean”? – Carsten S May 23 '15 at 8:46

## 2 Answers

$\int f d\mu = \sum_a f(a)\mu(\{a\}) = \sum_a \frac{f(a)}{a}$

• Although note that, as a Lebesgue integral, this is only well-defined if the sum is absolutely convergent. – Hurkyl May 23 '15 at 13:13
• Well, it's a sum over a set, not a sequence, so it can't be conditionally convergent. – Ben Millwood May 23 '15 at 16:06

Consider $f: A \to \Bbb N$. Now write: $$\int_A f(n) \,{\rm d}\mu(n) = \sum_{n \in A}\int_{\{n\}}f(n)\,{\rm d}\mu(n) = \sum_{n \in A}f(n)\mu(\{n\}) = \sum_{n \in A} \frac{f(n)}{n}.$$

• What does it mean though, if you don't mind that type of question? – Shine On You Crazy Diamond May 23 '15 at 0:51
• 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. – Ivo Terek May 23 '15 at 0:58
• 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 $\lim_{n\to \infty} f(n)=0$. – ApproximatelyTrue May 23 '15 at 7:32
• @ApproximatelyTrue Not really, consider $f(n)=1/\log n$. – Chris Culter May 23 '15 at 10:25
• @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. – ApproximatelyTrue May 23 '15 at 20:04