I have read, that it is not possible to define an equal measure on the natural numbers. While taking this literaly, this is true, but when trying to capture what is meant, I came to the following conclusion, which is inspired by the answer to this question, and I will repeat the definition:
Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function. For each natural number $n$ consider a finite probability space $\Sigma_n$ with probability density funciton $P_n(v)$ such that for each $n$:
$$\lim_{n\rightarrow \infty}P_n(v) = 0$$
and
$$\sum_{v=1}^n P_n(v) = 1$$
We define the Riemann integral of $f$ with respect to $P$ as ($a,b \in \mathbb{R}$):
$$\int_a^b f(x) d_P(x):=(b-a) \lim_{n\rightarrow \infty} \sum_{v=1}^n f(v) P_n(v) = (b-a) \lim_{n \rightarrow \infty} E_n(f(x))$$
where $E_n(f(x)) = \sum_{v=1}^n f(v)P_n(v)$ is the expected value in each finite probability space.
Then using the definition of this integral and properties of finite expected values, we get for example:
$$\int_a^b \alpha f(x)+g(x)d_P(x) = \alpha \int_a^b f(x) d_p(x) + \int_a^b g(x) d_p(x)$$
$$\int_a^c f(x)d_P(x) = \int_a^b f(x)d_P(x)+\int_b^c f(x)d_P(x), \text{ if } a \le b \le c$$
$$\int_b^a f(x)d_P(x) = -\int_a^b f(x)d_P(x), \text{ if } a \le b$$
$$\int_a^b \alpha d_P(x) = \alpha (b-a)$$
If $f(x) \le g(x) \le h(x)$ for all $x \in \mathbb{N}$, the by monotony of expectation:
$$\int_a^b f(x) d_P(x) \le \int_a^b g(x) d_P(x) \le \int_a^b h(x) d_P(x)$$
Examples:
Taking $P_n(v)$ as in the question above and $f(v) = (-1)^{\Omega(v)}$ then in the question it is proved that:
$$\int_a^b (-1)^{\Omega(v)} d_P(v) = 0$$
Taking $P_n(v) = 1/v \cdot 1/H_n, H_n:=\sum_{v=1}^n 1/v$, $s>1$ then we find that:
$$\int_0^x \frac{1}{t^{s}} d_P(t) = (x-0) \lim_{n\rightarrow \infty} \frac{1}{H_n}\sum_{v=1}^n \frac{1}{v^{s}} \frac{1}{v} $$
which seems to converge to $x \zeta(s+1)$ for $\operatorname{Real}(s)>1$ but I am not sure about that.
Images: I can not post, due to low reputation.
Question: Is it possible to evaluate $\lim_{n \rightarrow \infty} \frac{1}{H_n} \sum_{v=1}^n 1/v^{m+1}$?
The value of the limit is $0$ if $m >0$, $1$ if $m=0$ and $\infty$ if $m < 0$