# Counting measure on the power set sigma-algebra for the natural numbers

My textbook does not provide much about counting measures and integration. So I decided to set up integration on space $$(N , P(N) , \mu_c ,R)$$ myself, imitating the construction of the Lebesgue integral. (All functions are measurable since we are considering the power set.) I started with simple functions, which are basically sequences with finite range set.

$$\int f\,d\mu \;=\; \sum_{i=1}^{n} a_i \mu (A_i) \;=\; \sum_{k\in N} f(k).$$

Then I defined integration for $$[0,\infty]$$-valued sequences:

$$\int f\,d\mu \;=\; \sup\left\{\int g\,d\mu: \text{g \leq f and g is simple and [0,\infty]-valued }\right\}.$$

The above expression is simply summation of all terms since $$f$$ can be approached by a sequence of simple non-decreasing functions $$f_n=fX_{(-n,n)}$$.

Then for $$[-\infty,\infty]$$-valued sequences we define

$$\int f\,d\mu \;=\; \int f^+-\int f^- \;=\; \sum_{k\in N}f(k).$$

So I deduced that the integral of any function on $$N$$ is just summation of the terms of the sequence of its image. $$f$$ is said to be integrable if $$\int f^+$$ and $$\int f^-$$ are finite. Hence $$\sum_{k\in N} f(k)$$ must converge.

Now here I got confused. We know that every measurable function $$f$$ is integrable if and only if $$|f|$$ is integrable. So from the above construction, does it imply that every convergent sequence is absolutely convergent? (We know this ain't true)

Where did I make a fault in my above imitation?

• What you have is that A $f:\mathbb{N}\rightarrow\mathbb{R}$ in the counting measure iff $\sum_n|f(n)|$ is integrable. There are functions $f:\mathbb{N}\rightarrow\mathbb{R}$ for which $\lim_n\sum^n_{j=1}f(j)$ exists but $\sum^\infty_{j=0}|f(n)|=\infty$ (the so called conditionally convergent series) Those functions are not integrable. This is similar to the well known example in $L_1(\mathbb{R})$ given by $f(x)=\frac{\sin x}{x}\mathbb{1}_{[0,\infty)}(x)$. It is known that $\lim_{R\rightarrow\infty}\int^R_0f$ exists but $\int|f|=\infty$. Commented Jul 21, 2021 at 19:07

Below I will sometimes use the word "sequence" to denote the corresponding function.

Notice that $$\int f\,d\mu \;:=\; \int f^+-\int f^- \;$$. Thus when summing a sequence as an integral with respect to your measure (it's called the counting measure) you sum the positive and the negative parts of the sequence separately.
Then the (possibly infinite) integral is defined if and only if at least one of the numbers $$\int f^+, \int f^-$$ is finite. "Integrable" means that both of those numbers are finite. This corresponds to an absolutely convergent sequence.
So the key is that you can only integrate sequences with at least one of their positive/negative parts finite. And the sequence is "integrable" if and only if both of those parts are finite, which is equivalent to saying that it's absolutely convergent.

The intuition here is that by definition the integral doesn't know much about order of a sequence, thus you can't define it properly for conditionally convergent sequences. See f.i.: https://en.wikipedia.org/wiki/Riemann_series_theorem .

Remember this: not every convergent series is absolutely convergent. For example, if you take an alternating harmonic series it converges conditionally but not absolutely. Mathematically, it's expressed through the fact that if the conditional convergence is present, then $$\sum_{k\in\mathbb{N}}f(k)$$ converges. For absolute convergence, it's $$\sum_{k\in\mathbb{N}}|f(k)|$$ that must converge. Therefore, for a function $$f$$ to be integrable, $$\sum_{k\in\mathbb{N}}|f(k)|$$ must be finite, thus we have $$\sum_{k\in\mathbb{N}}|f(k)|<\infty$$.

I hope I clarified your doubts, let me know for anything! Have a good studying session :))))