There is the definition which we need for $4.15$
There is the theorem :
There is $4.15$:
We want to drop the finiteness condition that appears in Theorem $4.14$ without even restricting ourselves to sets that are necessarily countable. For this reason it seems advisable to clarify the meaning of the symbol $\sum_{\alpha \in A} \varphi(\alpha)$ when $\alpha$ ranges over an arbitrary set $A$.
Assume $0$ $\leq$ $\varphi(\alpha)$ $\in$ $\infty$ for each $\alpha$ $\in$ $A$.
Then $\sum_{\alpha \in A} \varphi(\alpha)$
denotes the supremum of the set of all finite sums $\varphi(\alpha_1)$ $+$ ... $+$ $\varphi(\alpha_n)$, where $\alpha_1$,...$\alpha_n$ are distinct members of $A$.
A moment's consideration will show that the sum $\sum_{\alpha \in A} \varphi(\alpha)$ is thus precisely the Lebesgue integral of $\varphi$ relative to the counting measure $\mu$ on $A$.
I don't understand why is the sum $\sum_{\alpha \in A} \varphi(\alpha)$ the Lebesgue integral of $\varphi$ relative to the counting measure $\mu$ on $A$.
Any help would be appreciated.
By definition, $\int \varphi \ d\mu = \sup \int s \ d\mu$, where the supremum is taken over all positive simple functions $s \leq \varphi$. That's how we define integrals over positive functions.
Since $\mu$ is the counting measure, any function $A \to [-\infty, \infty]$ is measurable. The positive simple functions are precisely the functions $A \to \mathbb R$ that take finitely many distinct values. A positive simple function therefore takes the form $$ s = \sum_{i=1}^n c_i \mathbf 1_{E_i},$$ where the $E_i$'s are disjoint subsets of $A$ and the $c_i$'s are strictly positive real numbers.
The integral of this simple function is $$ \int s \ d\mu = \sum_{i=1}^n c_i \ \mu(E_i) = \sum_{i=1}^n c_i \ |E_i|,$$ where $|E_i|$ is the cardinality of $E_i$. (The fact that $\mu(E_i) = |E_i|$ follows from the fact that $\mu$ is the counting measure.)
Now let's return to the evaluation of $\int \varphi \ d\mu$.
There are two cases.
Case 1: there exists some positive simple function $s = \sum_{i=1}^n c_i \mathbf 1_{E_i} $ bounded by $\varphi$ where at least one of the $E_i$'s is an infinite set. In this situation, $\int s \ d\mu = \infty$, and therefore $\int \varphi \ d\mu = \infty$.
Without loss of generality, let's say that $E_1$ is infinite. Let $\alpha_1, \alpha_2, \alpha_3, \dots$ be a sequence of distinct elements of $E_1$. Then for any $K \in \mathbb N$, we have $$ \sum_{k = 1}^K \varphi(\alpha_k) \geq K \times c_1 $$ so $$ \sup_{k \in \mathbb N} \sum_{k = 1}^K \varphi(\alpha_k) = \infty.$$
Thus $$ \sum_{\alpha \in A} \varphi(\alpha) = \infty = \int \varphi \ d\mu$$ in this case.
Case 2: For all positive simple functions $s = \sum_{i=1}^n c_i \mathbf 1_{E_i} $ bounded by $\varphi$, all of the $E_i$'s are finite sets.
In this case, each positive simple function bounded by $\varphi$ can be written as $\sum_{j = 1}^J c_j \mathbf 1_{\{\alpha_j\}}$, where each $\{\alpha_j\}$ is a singleton set, and where $0 < c_j \leq \varphi(\alpha_j)$. The sum over $j$ is finite, and the $\alpha_j$'s are distinct. (The $\alpha_j$'s in this sum are precisely the members of the $E_i$'s in the previous sum. The fact that all of the $E_i$'s are finite sets means that the sum over $j$ is finite.)
The integral of this simple function is $ \sum_{j=1}^J c_j $.
All positive simple functions $s \leq \varphi$ can be written in the form $s = \sum_{j = 1}^J c_j \mathbf 1_{\{\alpha_j\}}$ where each $0 < c_j \leq \varphi(\alpha_j)$ and where the $\alpha_j$'s are distinct. Conversely, any function of the form $s = \sum_{j = 1}^J c_j \mathbf 1_{\{\alpha_j\}}$ where $0 < c_j \leq \varphi(\alpha_j)$ and the $\alpha_j$'s are distinct is a positive simple function $\leq \varphi$. Taking the supremum over the integrals of all such positive simple functions, we see that $$ \int \varphi \ d\mu = \sum_{\alpha \in A} \varphi(\alpha)$$ in this case too.