The Lebesgue Integral

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 $F \subseteq E$.]

Prove

It was claimed in (N5) that $\mathcal{I}_{E}^{n}(c f)=c \mathcal{I}_{E}^{n}(f)$ for every $c \in [0, \infty )$. This result in fact holds in $c=\infty$ as well: $\mathfrak {I}_{E}^{n}(\infty f)=\infty \mathfrak {I}_{E}^{n}(f)$. Therefore, (N5) holds for all $c \in [0, \infty ]$. Similarly, we may allow the numbers $c_{1}, \ldots , c_{n}$ to be in $\left[0, \infty \right]$ in the statement (N8).

[(N8)(a): for functions $g, f, f_{1}, f_{2}, \ldots \in \mathfrak {N}$, $\mathcal{I}_{E}^{n}\left(\sum_{i=1}^{n} c_{i} f_{i}\right)=\sum_{i=1}^{n} c_{i} \mathcal{I}_{E}^{n}\left(f_{i}\right)$ for every $n \in \mathbb {N}$ and choice of $c_{1}, \ldots , c_{n} \in [0, +\infty )$, so that $\mathcal{I}_{E}^{n}$ is linear in this sense; this problem extends that to $c_{1}, \ldots , c_{n} \in [0, +\infty ]$.]

This exercise concerns Fatou's Lemma.

Let $\Omega$ denote an arbitrary nonempty set, and fix attention upon a particular $\omega_{0} \in \Omega$. Let $\mathcal{F}=2^{\Omega }$, and define $\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}}$ by writing $\mu (A)=1$ if $\omega_{0} \in A$ and $\mu (A)=0$ if $\omega_{0} \notin A$.

Let $\Omega$ denote an uncountable set, and let \mathcal{F}=\left\{ A \subseteq \Omega : A is amc or $A^{c}$ is amc. Define $\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}}$ for all $A \in \mathcal{F}$ by writing $\mu (A)=0$ if $A$ is amc and $\mu (A)=1$ if $A^{c}$ is amc.

Let $\Omega$ denote an arbitrary nonempty set, let $A \subseteq \Omega$, and let $f: \Omega \rightarrow [0, \infty ]$ be given. Write

Let $\Omega$ denote a nonempty set, let $\mathcal{F}=2^{\Omega }$, and let $\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}}$ be such that $\mu (A)=$ the number of points in $A$ when $A$ is finite, and $\mu (A)=\infty$ otherwise .

The previous exercise regarding counting measure has some interesting applications to statements regarding infinite series.

Consider the measure space $(\Omega , \mathcal{F}, \mu )$, and let $E \in \mathcal{F}$. For each $f \in \mathfrak {N}$, define

where the supremum is taken over all finite disjoint collections $\left\{ A_{1}, \cdots , A_{n}\right\}$ of $\mathcal{F}$-sets with $\bigcup_{i=1}^{n} A_{i}=\Omega$. Then $\bar{\mathcal{J}}_{E}(f)=\bar{\mathcal{I}}_{E}(f)$, so that this is an equivalent formulation of what we have done in this section.

[Recall the arithmetical conventions $+\infty \times 0=0 \times +\infty =0$ and $\inf \varnothing =+\infty$ for the case when some $A_{i}=\varnothing$.]

Consider the measure space $(\Omega , \mathcal{F}, \mu )$, and let $E \in \mathcal{F}$ and $f \in \mathfrak {N}$. Define $\bar{\mathcal{J}}_{E}(f)=\lim_{n} \mathcal{I}_{E}\left(s_{n}\right)$, where $\left\{ s_{n}\right\}_{n=1}^{\infty }$ is a nondecreasing sequence of functions in $\mathfrak {S}_{f}$ such that $s_{n} \rightarrow f$. We will show that $\bar{\mathcal{J}}_{E}(f)=\bar{\mathcal{I}}_{E}(f)$ without using MCT, so that this definition could have instead served as our definition.

Let $(\Omega , \mathcal{F}, \mu )$ denote a measure space with $E \in \mathcal{F}$ such that $\mu (E)<+\infty$, and let $f \in \mathfrak {N}$. Lebesgue's original version of our operator is

Let $\Omega$ denote a nonempty set, and let $\mathcal{F}$ denote a $\sigma$-field on $\Omega$. For each $n \in \mathbb {N}$, let $\mu_{n}$ denote a measure with domain $\mathcal{F}$. For each $n \in \mathbb {N}$, let $\sum_{i=1}^{n} \mu_{i}$ denote the measure that assigns the value $\sum_{i=1}^{n} \mu_{i}(A)$ to each $A \in \mathcal{F}$.

Let $A \in \mathcal{F}$ be nonempty, and suppose that $f: A \rightarrow [0,+\infty ]$ is $\mathcal{F}/\mathcal{B}^{*}$-measurable. Let $E \in \mathcal{F}_{A}$, where $\mathcal{F}_{A}$ is the set of $\mathcal{F}$-subsets of $A$. Consider two programs:

Program 1: Extend $f$ from $A$ to $\Omega$ as follows: let $f^{*} \in \mathfrak {N}$ be such that $f^{*}=f$ on $A$ and $f^{*}=0$ on $A^{c}$.

Program 2: Do not extend $f$ from $A$ to $\Omega$ as in Program 1, but instead view $f$ as a function defined everywhere relative to the measure space $\left(A, \mathcal{F}_{A}, \mu_{A}\right)$, where $\mu_{A}$ is the restriction of $\mu$ from $\mathcal{F}$ to $\mathcal{F}_{A}$.

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

Keep ( $\Omega , \mathcal{F}, \mu$ ) general. Suppose that $\mathcal{F}_{0} \subseteq \mathcal{F}$ is a $\sigma$-field on $\Omega$, and let $\mu_{0}$ denote the restriction of $\mu$ to $\mu_{0}$. If a nonnegative $\mathcal{F} / \mathcal{B}^{*}$ measurable function $f$ defined on $\Omega$ also happens to be $\mathcal{F}_{0} / \mathcal{B}^{*}$-measurable in addition, then $\mathcal{I}_{E}^{n}(f ; \mu )=\mathcal{I}_{E}^{n}\left(f ; \mu_{0}\right)$ for every $E \in \mathcal{F}_{0}$.

(N16): Suppose that $(\Omega , \mathcal{F}, \mu )$ is a $\sigma$-finite measure space, and further suppose that $\bar{\mathcal{I}}_{E}(f)=\bar{\mathcal{I}}_{E}(g)$ for every $E \in \mathcal{F}$. Then $f=g$ $\mu$-a.e. on $\Omega$.

We have the following claims related to (N16):

Let $f$ denote a nonnegative $\mathcal{F}^{\prime } / \mathcal{B}^{*}$-measurable function. We have the two equalities

and

Let $E \in \mathcal{F}$ be such that $\mu (E)<\infty$, and let $f \in \mathfrak {M}$ be such that $f=0$ on $E^{c}$ and $m \leqslant f \leqslant M$ on $E$, where $m, M \in \mathbb {R}$. Then $\mathcal{I}^{g}(f)$ exists and is finite. Furthermore, we have $m \mu (E) \leqslant \mathcal{I}^{g}(f) \leqslant M \mu (E)$.

Let $E \in \mathscr {F}, \varepsilon >0$ and let $f \in \mathfrak {M}$ be such that $\mathcal{I}_{E}^{g}(f)$ exists and is finite. There exists a subset $F \subseteq E$ such that $\mu (F)<\infty$ and $\left|\mathcal{I}_{E}^{g}(f)-\mathcal{I}_{F}^{g}(f)\right|<\varepsilon$.

$f$ is such that $\mathcal{I}_{E}^{g}(f)$ exists and is finite iff for any $\varepsilon >0$ there are functions $g$ and $h$ in $\mathfrak {M}$ such that $h \leqslant f \leqslant g$ on $E$ and $\mathcal{I}_{E}^{g}(g-h)<\varepsilon$.

Let $f, f_{1}, f_{2}, \ldots$ denote a sequence of nonnegative functions in $\mathfrak {M}$. For each $n \in \mathbb {N}$ and $E \in \mathcal{F}$, define $v_{n}(E)=\mathcal{I}_{E}^{g}\left(f_{n} ; \mu \right)$ and $v(E)=\mathcal{I}_{E}^{g}(f ; \mu )$. Furthermore, assume that $v(\Omega ), v_{1}(\Omega ), v_{2}(\Omega ), \ldots$ are finite and $f_{n} \rightarrow f \mu$-a.e. on $\Omega$. Then

Let $(\Omega , \mathcal{F}, \mu )=\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right)$ with $E \in \mathcal{F}$, and let $f \in \mathfrak {M}$ be such that $\mathcal{I}^{g}(f)$ exists and is finite.

Suppose that $\mu (\Omega )<\infty ,\left\{ f_{n}\right\}$ is a sequence of functions in $\mathfrak {M}$ such that there exists $M \in \mathbb {R}$ with $\left|f_{n}\right| \leq M$ for each $n \in \mathbb {N}$, and let $f \in \mathfrak {M}$ be such that $f_{n} \rightarrow f$ uniformly on $E$. Then $\lim_{n} \mathcal{I}_{E}^{g}\left(f_{n}\right)=\mathcal{I}_{E}^{g}(f)$.

Let $\left\{ f_{n}\right\}_{n=1}^{\infty }$ denote a nondecreasing sequence of functions in $\mathfrak {M}$ such that $\mathcal{I}_{E}^{g}\left(f_{n}\right)$ exists and is finite for each $n \in \mathbb {N}$ and $\sup_{n \in \mathbb {N}} \mathcal{I}_{E}^{g}\left(f_{n}\right)<\infty$. Then $\mathcal{I}_{E}^{g}\left(\lim_{n} f_{n}\right)$ exists, is finite, and equals $\lim_{n} \mathcal{I}_{E}^{g}\left(f_{n}\right)$. This is one form of Beppo Levi's Theorem.

Let $\left\{ f_{n}\right\} ,\left\{ g_{n}\right\}$, and $\left\{ h_{n}\right\}$ denote sequences of functions in $\mathfrak {M}$ such that $f_{n} \xrightarrow {\text{ a.e. }} f, g_{n} \xrightarrow {\text{ a.e. }} g$, and $h_{n} \xrightarrow {\text{ a.e. }} h$ for some functions $f, g, h \in \mathfrak {M}$.

Suppose for any $p \in \left\{ g, g_{1}, g_{2}, \ldots , h, h_{1}, h_{2}, \ldots \right\}$ that $\mathcal{I}_{E}^{g}(p)$ exists and is finite. Furthermore, suppose that $\lim_{n} \mathcal{I}_{E}^{g}\left(g_{n}\right)=\mathcal{I}_{E}^{g}(g)$ and $\lim_{n} \mathcal{I}_{E}^{g}\left(h_{n}\right)=\mathcal{I}_{E}^{g}(h)$. Also, assume that $g_{n} \leqslant f_{n} \leqslant h_{n}$ for every $n \in \mathbb {N}$.

In properties (G6) and (G9), we gave a proof for the "everywhere" case. Prove the "almost everywhere" case for these two properties. [No problems should arise.]

(G6): if $f \leqslant g$ $\mu$-a.e. on $E$ and $f, g \in \mathfrak {M}(E)$, then $\overline{\overline{\mathcal{I}}}_{E}(f) \leqslant \overline{\overline{\mathcal{I}}}_{E}(g)$.

(G9): if $\overline{\overline{\mathcal{I}}}_{E}(f)$ eaif and $g$ is such that $\left|g\right| \leqslant f$ $\mu$-a.e. on $E$, then $\overline{\overline{\mathcal{I}}}_{E}(g)$ eaif. As a consequence, any $f \in \mathfrak {M}$ that is bounded $\mu$-a.e. on a set $E$ of finite measure and zero $\mu$-a.e. on $E^{c}$ is such that $\overline{\overline{\mathcal{I}}}_{F}(f)$ eaif for every $F \in \mathcal{F}$.

Suppose that for each $t \in \mathbb {R}$ we have $f_{t} \in \mathfrak {M}$, and suppose that $\lim_{t \rightarrow +\infty } f_{t}(\omega )$ exists for each $\omega \in \Omega$. Next, assume that there exists nonnegative $g \in \mathfrak {M}$ such that $\bar{\mathcal{I}}_{E}(g)$ exists and is finite; further assume that $\left|f_{t}\right| \leqslant g$ for each $t \in \mathbb {R}$.

Suppose that $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is a sequence of functions in $\mathfrak {M}$ such that $\lim_{n} f_{n}$ exists on $\Omega$. Next, suppose that $\left\{ g_{n}\right\}_{n=1}^{\infty }$ is a sequence of nonnegative functions in $\mathfrak {M}$ such that for all $n \in \mathbb {N}$ we have (i) $\bar{\mathcal{I}}_{E}\left(g_{n}\right)$ exists and is finite and (ii) $\left|f_{n}\right| \leqslant g_{n}$. Finally, suppose that (iii) $\lim_{n} g_{n}$ exists on $\Omega$, (iv) $\bar{\mathcal{I}}_{E}\left(\lim_{n} g_{n}\right)$ exists and is finite, and (v) $\bar{\mathcal{I}}_{E}\left(\lim_{n} g_{n}\right)=\lim_{n} \bar{\mathcal{I}}_{E}\left(g_{n}\right)$. Then $\bar{\mathcal{I}}_{E}\left(\lim_{n} f_{n}\right)=\lim_{n} \bar{\mathcal{I}}_{E}\left(f_{n}\right)$.