A question on the construction of the Lebesgue integral in Rudin

Question

In Rudin's Real and Complex Analysis (page 19, definition 1.23) we take a measurable function $f:X\rightarrow [0,\infty]$ in which $(X,\mathcal M,\mu)$ is a measure space with $\sigma-$algebra $\mathcal M$ and measure $\mu$. We define the integral of this function over $E\in \mathcal M$ to be: $$\int_E f\text{ }d\mu=\sup\int_E s\text{ }d\mu \tag{1},$$ in which $s$ is a simple function, $s:X\rightarrow [0,\infty)$, defined as: $$s(x)=\sum_{i=1}^n \alpha_i\chi_{A_i}, \tag{2}$$ in which $\chi_{A_i}(x)$ is the characteristic function on $X$. Rudin states that the supremum in $(1)$ is being taken over all simple measurable functions such that $0 \leq s\leq f$, I do not know what this means. Is this a pointwise comparison of the functions between $0$ and $f$?

Standard mandatory disclaimer: I am a physicist not a mathematician.

Answer 1

Break it into pieces:

$1). S$ is the collection of simple functions such that $s(x)\le f(x)$ for all $x\in X$.

$2).\ I:= \{\int_X s:s\in S\}$ is a set of numbers, with each $\int_X s$ defined as in your question.

$3).\ $ Since $I$ is a set of numbers, we can take $\sup I=\sup\{\int_Xs:s\in S\}.$ The result is a number in the extended reals, and this is the integral of $f$ by definition.