Understanding proof of "if $f$ is non-negative, then it is the supremum of simple functions."

Question

I am studying measure theory based on the book "Probability and Measure Theory" by Ash and have a question about understanding one of the theorems. Here it goes:

(Page 43) Theorem 1.5.9 (d) If $h \ge 0$ and $B \in \mathcal{F}$, then $$\int_B h \ d\mu = \sup \left\{ \int_B s \ d\mu \; | \; 0 \le s \le h, s \text{ is simple} \right\}$$

Proof:

If $0 \le s \le h$, then $\int_B s \ d\mu \le \int_B h d\mu$ (take this as given; previous theorem). Hence, $$\int_B h \ d\mu \ge \sup \left\{\int_B s \ d\mu \ | \ 0 \le s \le h \right\}.$$

If $t$ is simple such that $0 \le t \le h I_B$ ($I$ denotes indicator function), then $t = tI_B \le h$ so $$\color{red}{\int_\Omega t \ d\mu \le \sup \left\{\int_\Omega s I_B \ d\mu \ | \ 0 \le s \le h, s \text{ is simple} \right\}.}$$ Take the $\color{blue}{\text{sup over } t \text{ to obtain } \int_B h \ d\mu \le \sup \left\{\int_B s \ d\mu \ | \ 0 \le s \le h, s \text{ is simple} \right\}}$.

My questions are as follows:

1. (Red part.) We could extend this to the below, right? $$\int_\Omega t \ d\mu \le \sup \{ \cdots\} \le \int_\Omega hI_B \ d\mu\le \int_\Omega h \ d\mu$$
2. (Blue part.) Then, I am not sure how the last inequality holds. Moreover, I am not sure how taking the sup over $t$ results in that inequality.

Any help would be very much appreciated. Thank you in advance!

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Read @angryavian's comment and come back to the edits below.

EDIT 1: (What we know from above: $t$ is simple function s.t. $0 \le t \le hI_B$.)

If the extension follows (under Q1), by taking the sup over $t$, we would just have that extension. In other words:

$$\sup_t \int_\Omega t \ d\mu = \sup \left\{ \int_\Omega sI_B \ d\mu \; | \; 0 \le s \le h, s \text{ is simple} \right\} \le \int_\Omega hI_B \ d\mu\le \int_\Omega h \ d\mu$$ Equivalently, we would not have $$\int_B h \ d\mu \le \sup \left\{\int_B s \ d\mu \ | \ 0 \le s \le h, s \text{ is simple} \right\} $$

EDIT 2: Or, is it that by taking the sup over $t$, we have the following steps?

Step 1. By definition of integral of non-neg function $h$, and following from our previous proof: $$ \int_\Omega t \ d\mu \le \sup \left\{\int_\Omega s I_B \ d\mu \ | \ 0 \le s \le h, s \text{ is simple} \right\} = \int_\Omega hI_B \ d\mu = \int_B h \ d\mu $$

Step 2. Taking sup over $t$: $$ \int_\Omega t \ d\mu \le \sup \left\{\int_\Omega s I_B \ d\mu \ | \ 0 \le s \le h, s \text{ is simple} \right\} \le \sup_t \int_\Omega t \ d\mu, \text{ where } 0 \le t \le hI_B, $$

because the $\{ \cdots \}$ constrains to $I_B$, and henceforth the result follows as $\sup \{\cdots \} = \int_B h \ d\mu$.

Answer 1

For your first question, yes that is true, but I am not sure why that would be helpful in proving $\int_B h \, d\mu \le \sup\{\cdots\}$.

For your second question: This follows from the definition of the integral (see page 38). Specifically, $\int_\Omega hI_B \, d\mu$ is defined as $\sup_t \int_\Omega t \, d\mu$ for $0 \le t \le hI_B$. The purpose of proving this theorem (Theorem 1.5.9(d)) is to show that essentially the same statement holds if you replace $\Omega$ with a measurable set $B$.

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You already have $$\int_\Omega t \, d\mu \le \sup\left\{\int_\Omega sI_B \mid 0 \le s \le h, \text{$s$ simple}\right\}.$$ This inequality holds for any $t$ satisfying $0 \le t \le hI_B$. So if we take the supremum over all such $t$, the left-hand side becomes $\int_\Omega hI_B \, d\mu$, and inequality still holds with the right-hand side unchanged.