Stage One: Simple Functions

Let $E \in \mathscr {F}$.

Let $t, s, s_{1}, s_{2}, \ldots \in \mathbb {S}$. Why can't we say that $\mathfrak {I}_{E}^{s}(a s+ b t)=a \mathfrak {I}_{E}^{s}(s)+b \mathfrak {I}_{E}^{s}(t)$ for every $a, b \in \mathbb {R}$, as compared to saying that the result holds for every $0 \leqslant a, b<\infty$ ? Also, why can't we necessarily write $\mathcal{I}_{E}^{s}\left(\sum_{i=1}^{\infty } c_{i} s_{i}\right)=\sum_{i=1}^{\infty } c_{i} \mathcal{I}_{E}^{s}\left(s_{i}\right)$ ?

Let $E \in \mathscr {F}$ be such that $\mu (E)>0$. Then $\mathcal{I}_{E}^{s}(s)=0$ iff $s=0 \mu$-a. e. on E. In particular, $\mathcal{I}^{\mathcal{s}}(s)=0$ iff $s=0 \mu$-a. e. [on $\Omega$ ].

Let $E \in \mathcal{F}$ and $s \in \mathbb {S}$.

Suppose that $\left\{ E_{n}\right\}_{n=1}^{\infty }$ is a nonincreasing sequence of $\mathcal{F}$-sets. Also, let $s \in \mathbb {S}$.

Let $s \in \mathbb {S}$ and recall (S7), where $v_{s}$ is the measure on $\mathcal{F}$ with $\nu_{s}(B)=\mathcal{I}_{B}^{s}(s)$ for every $B \in \mathcal{F}$. Then, for any $E \in \mathcal{F}$ and $t \in \mathfrak {S}$, we have $\mathcal{I}_{E}^{s}\left(t ; \nu_{s}\right)=\mathcal{I}_{E}^{s}(t s ; \mu )$.

Let $A \in \mathcal{F}$ be nonempty, and let $s \in \mathfrak {S}$. Let $\mathcal{F}_{A}=\left\{ E \subseteq A: E \in \mathcal{F}\right\}$, and let $\mu_{A}$ denote the restriction of $\mu$ from $\mathcal{F}$ to $\mathcal{F}_{A}$. Finally, let $s_{A}$ denote the restriction of $s$ from $\Omega$ to $A$.

Let $A \in \mathcal{F} \backslash \left\{ \varnothing \right\}$, and suppose that $s: A \rightarrow [0, \infty )$ is simple and $\mathcal{F} / \mathcal{B}^{*}$-measurable. Let $E \in \mathcal{F}_{A}$, where $\mathcal{F}_{A}$ is defined in the previous exercise. Consider two programs:

Program 1 Extend $s$ from $A$ to $\Omega$ as follows: Let $s^{*} \in \mathbb {S}$ be such that $s^{*}=s$ on $A$ and $s^{*}=0$ on $A^{c}$.

Program 2 Do not extend $s$ from $A$ to $\Omega$ as in Program 1, but instead view $s$ as a function defined everywhere relative to the measure space ( $A, \mathscr {F}_{A}, \mu_{A}$ ), where the notation is as in the previous exercise.

These two programs are equivalent in the sense that $\mathcal{I}_{E}^{s}\left(s^{*} ; \mu \right)=\mathscr {I}_{E}\left(s ; \mu_{A}\right)$, where $\mathfrak {I}_{E}^{s}$ is this section's functional relative to the measure space ( $\Omega , \mathcal{F}, \mu$ ), and $\mathscr {J}_{E}$ is this section's functional relative to the measure space ( $A, \mathscr {F}_{A}, \mu_{A}$ ).

Quickly prove the following "almost everywhere" modification of (S3) and (S4). As usual, $s, t \in \mathbb {S}$ and all sets are in $\mathcal{F}$.

[(S3): if $s=t$ on $E$, then $\mathcal{I}_{F}^{s}(s)=\mathcal{I}_{F}^{s}(t)$ for every $F \subseteq E$. (S4): $s \leqslant t$ [on $\Omega$] implies $\mathcal{I}_{E}^{s}(s) \leqslant \mathcal{I}_{E}^{s}(t)$; more generally, if $s \leqslant t$ on $E$, then $\mathcal{I}_{F}^{s}(s) \leqslant \mathcal{I}_{F}^{s}(t)$ for every $F \subseteq E$.]