I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 5 on p.85 in Exercises 3A in this book.
Exercise 5
Verify the assertion that integration with respect to counting measure is summation (Example 3.6).
3.6 Example integration with respect to counting measure is summation
Suppose $\mu$ is counting measure on $\mathbb{Z}^+$ and $b_1,b_2,\dots$ is a sequence of nonnegative numbers. Think of $b$ as the function from $\mathbb{Z}^+$ to $[0,\infty)$ defined by $b(k)=b_k.$ Then $$\int b d\mu=\sum_{k=1}^\infty b_k,$$ as you should verify.
Definitions in this book:
3.1 Definition $\mathcal{S}$-partition
Suppose $\mathcal{S}$ is a $\sigma$-algebra on a set $X.$ An $\mathcal{S}$-partition of $X$ is a finite collection $A_1,\dots,A_m$ of disjoint sets in $\mathcal{S}$ such that $A_1\cup\dots\cup A_m=X.$
3.2 Definition lower Lebesgue sum
Suppose $(X,\mathcal{S},\mu)$ is a measure space, $f:X\to [0,\infty]$ is an $\mathcal{S}$-measurable function, and $P$ is an $\mathcal{S}$-partition $A_1,\dots,A_m$ of $X.$ The lower Lebesgue sum $\mathcal{L}(f,P)$ is defined by $$\mathcal{L}(f,P)=\sum_{j=1}^m \mu(A_j)\inf_{A_j} f.$$
3.3 Definition integral of a nonegative function
Suppose $(X,\mathcal{S},\mu)$ is a measure space and $f:X\to [0,\infty]$ is an $\mathcal{S}$-measurable function. The integral of $f$ with respect to $\mu$, denoted $\int f d\mu$, is defined by $$\int f d\mu=\sup\{\mathcal{L}(f,P):P\text{ is an }\mathcal{S}\text{-partition of }X\}.$$
My solution:
$$\int b d\mu:=\sup\{\mathcal{L}(b,P):P\text{ is a }2^{\mathbb{Z}^+}\text{-partition of }\mathbb{Z}^+\}.$$
Let $P_1$ be a $2^{\mathbb{Z}^+}$-partition such that $P_1=\{\{1\},\{2\},\dots,\{n\},\{n+1,n+2,\dots\}\}.$
$$\mathcal{L}(b,P_1)=\\\mu(\{1\})\cdot\inf_{\{1\}} b + \mu(\{2\})\cdot\inf_{\{2\}} b + \dots + \mu(\{n\})\cdot\inf_{\{n\}} b + \mu(\{n+1,n+2,\dots\})\cdot\inf_{\{n+1,n+2,\dots\}} b\\=1\cdot b_1+1\cdot b_2+\dots+1\cdot b_n+\infty\cdot\inf\{b_{n+1},b_{n+2},\dots\}\\=b_1+b_2+\dots+b_n+\infty\cdot\inf\{b_{n+1},b_{n+2},\dots\}.$$
(Case 1) We consider the case in which $\sum_{k=1}^\infty b_k =\infty.$
In this case, $b_1+b_2+\dots+b_n$ can be arbitrary large if $n$ is sufficiently large because $\sum_{k=1}^\infty b_k =\infty.$
So, $\int b d\mu:=\sup\{\mathcal{L}(b,P):P\text{ is a }2^{\mathbb{Z}^+}\text{-partition of }\mathbb{Z}^+\}=\infty=\sum_{k=1}^\infty b_k.$
(Case 2) We consider the case in which $\sum_{k=1}^\infty b_k \in\{x\in\mathbb{R}:0\leq x\}.$
In this case, we know $\lim_{k\to\infty}b_k=0$ holds by a result in Calculus.
So, $\inf\{b_{n+1},b_{n+2},\dots\}=0.$
So, $\infty\cdot\inf\{b_{n+1},b_{n+2},\dots\}=0$ by a famous convention.
So, $\mathcal{L}(b,P_1)=b_1+b_2+\dots+b_n.$
Let $P_2$ be an arbitrary $2^{\mathbb{Z}^+}$-partition such that $P_2=\{A_1,A_2,\dots,A_n\}.$
If $A_i$ is an infinite set for some $i\in\{1,\dots,n\}$, then $\mu(A_i)=\infty$ by the definition of counting measure and $\inf_{A_i} b=0$ since $\lim_{k\to\infty}b_k=0$.
So, $\mu(A_i)\cdot\inf_{A_i} b=0.$
Let $T:=\{i\in\{1,\dots,n\}:A_i\text{ is a finite set.}\}=\{i_1,\dots,i_j\}.$
Then, $$\mathcal{L}(b,P_2)=\\\mu(A_{i_1})\inf_{A_{i_1}} b + \dots +\mu(A_{i_j})\inf_{A_{i_j}} b=\\\mu(A_{i_1})\min_{A_{i_1}} b + \dots +\mu(A_{i_j})\min_{A_{i_j}} b\leq\\\sum_{i\in A_{i_1}} b_i+\dots+\sum_{i\in A_{i_j}} b_i=\\\sum_{i\in A_{i_1}\cup\dots\cup A_{i_j}} b_i\leq\\\sum_{i=1}^{\max A_{i_1}\cup\dots\cup A_{i_j}} b_i\leq\\\sum_{k=1}^\infty b_k.$$
Therefore, from the above argument, $$\int b d\mu:=\sup\{\mathcal{L}(b,P):P\text{ is a }2^{\mathbb{Z}^+}\text{-partition of }\mathbb{Z}^+\}=\sum_{k=1}^\infty b_k.$$