Exercise 1.4.39 (Summmation is discrete integration)

We have to demonstrate that

Since $X$ is countable, we will enumerate its elements by ${(x_i)}_{i\in \mathbb{N}}$ in what follows.

• Let $h$ be an arbitrary simple function minorising $f$ and takes on distinct values $c_1,\dots ,c_k\in [0,+\infty ]$. We then have:

  $$\mathrm{\text{Simp}}{\int \; <!--\; nolimits\; -->}_X{h}_{n}d\#=\sum _{i=1}^k{c}_i\times \#h^{-1}(c_i)=\sum _{i=1}^k\sum _{x\in h^{-1}(c_i)}{c}_i\le \sum _{i=1}^k\sum _{x\in h^{-1}(c_i)}f(x)=\sum _{x\in \text{⋃}_{i=1}^k{h}^{-1}(c_i)}f(x)=\sum _{x\in X}f(x)$$

  where we have used several properties of double summation. Taking the supremums preserves the inequality.

• We demonstrate that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{x\in X}f(x)-𝜖$ is contained in the left-hand side set for every $𝜖>0$; thus, ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{x\in X}f(x)$ must be at most as big as the supremum on the left-hand side. To do so, enumerate the elements of $X$ by ${(x_i)}_{i\in \mathbb{N}}$ and define for each $n\in \mathbb{N}$:

  $$h_n:X\to [0,+\infty ],x\mapsto \{\begin{array}{cc}f(x)\hfill & \text{if }x\in \{x_1,\dots ,x_n\}\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$$

  Then $h_n$ are simple functions minorising $f$, and we have for $c_i\in \{h_n(x_j):j\le n\}$:

  $$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{\text{Simp}}{\int \; <!--\; nolimits\; -->}_X{h}_{n}d\#=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sum _{i=1}^n{c}_i\cdot \#(h_n^{-1}(c_i))=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sum _{i=1}^n\sum _{x\in h^{-1}(c_i)}f(x)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sum _{x\in \{x_1,\dots ,x_n\}}f(x)=\sum _{i=1}^{\infty }f(x_i)=\sum _{x\in X}f(x)$$

  Since $\mathrm{\text{Simp}}{\int }_X{h}_{n}d\#$ converges to ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{x\in X}$ from below, taking supremums yields the inequality.