Integral with respect to counting measure on infinite set $\mathbb{N}$ I want to prove that the integral with respect to counting measure is the summation:

Suppose $\mu$ is counting measure on $\mathbb{N}$ and the function $b(k) = b_k$ from $\mathbb{N} \to [0, \infty)$ with $b_k$ is a non-negative number.
Then $\int b \,d\mu = \sum \limits_{k = 0}^{\infty} b_k$

I want to use the definition of Lower Lebesgue sum and the definition of integral of a non-negative function:

Definition of lower Lebesgue sum



Definition of integral of non-negative integral


My strategy is to prove that $\int b \,d\mu \ge \sum \limits_{k = 0}^{\infty} b_k$ (by using the supremum definition) and after that, prove $\sum \limits_{k = 0}^{\infty} b_k$ is $\ge$ than any Lower Lebesgue sum of function $b(k)$ on $\mathbb{N}$
However, I am stuck with the fact that $\mathbb{N}$ is an infinite set whereas the Lower Lebesgue sum is only defined for a finite partition of the set $\mathbb{N}$ (so that I can't have the infinite sum $\sum \limits_{k = 0}^{\infty} b_k$ by using the definition of Lower Lebesgue sum).
Could you please suggest me a way to prove that $\int b \,d\mu \ge \sum \limits_{k = 0}^{\infty} b_k$ ?
Thank you very much for the help!
 A: Since we can only take finite partitions, a natural choice would be to consider the partition $P_m$ of $\mathbb{N}$ that consists of $ \{0\}, \{1\}, \{2\}, \{3\}, \dots, \{m\}$ and $\{m+1,m+2,\dots\}$.
Then the singleton sets have counting measure $1$ while the last set has counting measure $\infty$, so that
$$\mathcal{L}(f, P_m) = \sum_{k=0}^m 1\cdot b_k + \infty\cdot \inf_{k>m} b_k.$$
(with the usual convention that $\infty \cdot 0 = 0$).
There are now two cases:

*

*if $\inf_{k>m} b_k =c>0$ for some $m$, then the Lebesgue sum is $\infty$, but we also know that $\sum_{k=0}^\infty b_k$ diverges because $b_k \geq c$ for infinitely many $k$. Hence, $$\int b \,d\mu \geq \mathcal{L}(f, P_m) = \infty = \sum_{k=0}^\infty b_k.$$

*if $\inf_{k>m} b_k =0$ for all $m$, then $\mathcal{L}(f, P_m) = \sum_{i=0}^m b_k$ are simply the partial sums and by definition we have
$$\mathcal{L}(f, P_m) = \sum_{k=0}^m b_k \to \sum_{k=0}^\infty b_k \quad \text{as }  m\to\infty.$$
In particular it follows that
$$\int b \,d\mu \geq \lim_{m\to\infty} \mathcal{L}(f, P_m) = \sum_{k=0}^\infty b_k.$$
