Integration

Mirroring Durrett’s development.

Given a measure $\mu$ defined on a $\sigma$ -algebra $\mathcal{F}$ of a set $\Omega$ , the measure of a measurable function $f : (\Omega,\mathcal{F}) \rightarrow (\mathbb{R},\mathcal{B})$ is a multistep extension procedure:

Extend the definition of $\mu$ to

(bounded) indicator functions
(bounded) simple functions
bounded functions
non-negative functions
general functions

Indicator Functions

definition: when $A$ is measurable, $\mu(\mathbb{I}_A) = \mu(A)$ .

lemma: if $\mathbb{I}_A \le \mathbb{I}_B$ then $\mu(\mathbb{I}_A) \le \mu(\mathbb{I}_B)$

This is just a restatement of the monotonicity property of measures.

lemma: the measure of a countable sum of disjoint indicator functions of measurable sets is the measure of the indicator function of the countable union of sets comprising the indicator functions of the sum. Furthermore, the definition is independent of the sum.

this is just a restatement of the countable additivity property of measure in terms of disjoint indicator functions:

$\mu(\mathbb{I}_{\cup_i A_i}) = \mu(\cup_i A_i) = \sum_i \mu(A_i) = \sum_i \mu(\mathbb{I}_{A_i})$ .

Bounded Indicator Functions

definition: a bounded indicator function is an indicator function on a measurable set of bounded measure.

lemma: the finite sum of disjoint bounded indicator functions is a bounded indicator function.

Since $\sum_{i=1..n} \mathbb{I}_{A_i} = \mathbb{I}_{\cup_{i=1..n} A_i}$ when the sets are disjoint, and the measure of a finite union of sets bounded by the finite sum of the measures of the sets comprising the union (finite subadditivity) the conclusion is immediate.

lemma: the measure of the finite sum of disjoint bounded indicator functions is the measure of the bounded indicator function of the union of the sets of the bounded indicator functions.

Almost Everywhere Properties

definition: a property $P$ . holds $\mu$ -a.e. if $\mu \mathbb{I}_{\neg P} = 0$ .

lemma: if $P,Q$ holds $\mu$ -a.e. then $P \vee Q$ holds $\mu$ -a.e.

Since $\mathbb{I}_{\neg P \vee \neg Q} = \mathbb{I}_{\neg P} + \mathbb{I}_{P \wedge \neg Q} \le \mathbb{I}_{\neg P} + \mathbb{I}_{\neg Q}$ it follows that $\mu(\mathbb{I}_{\neg P \vee \neg Q}) \le 0$ .

definition: two measurable functions $f \overset{a.e.}{=} g$ if $\mu \mathbb{I}_{f \neq g} = 0$ .

lemma: $\overset{a.e}{=}$ is an equivalence relation on measurable functions.

Clearly $\overset{a.e}{=}$ is reflexive and symmetric. If $f \overset{a.e}{=} g \overset{a.e}{=} h$ then $f \neq h$ iff $f = g \neq h$ or $f \neq g, f \neq h$ and implies $g \neq h$ or $f \neq g$ . But this means that $\mu \mathbb{I}_{f \neq h} \le \mu \mathbb{I}_{f \neq g} + \mu \mathbb{I}_{g \neq h} = 0$ and so $f\overset{a.e}{=} h$ . Hence $\overset{a.e}{=}$ is transitive.

definition: the support function of a measurable function $f$ is the indicator function $\mathbb{I}_{f \neq 0}$ . A function has bounded support if $\mu \mathbb{I}_{f \neq 0} < \infty$ .

lemma: if $f = g$ then $\mathbb{I}_{f \neq 0} = \mathbb{I}_{g \neq 0}$ .

Let $f = g$ . Then $\mathbb{I}_{f \neq 0} \le \mathbb{I}_{g \neq 0} \le \mathbb{I}_{f \neq 0}$

Simple Functions

definition: A (bounded) simple function is a measurable function with a finite range (and bounded support).

lemma: A simple function is a linear combination of disjoint bounded indicator functions:

$s = \sum_{i=1..n} a_i \mathbb{I}_{A_i}$

where $n$ is a non-negative integer, $a_1,\ldots,a_n$ is a list of real values, and $A_1,\ldots, A_n$ is a disjoint list of measurable sets of finite measure.

Clearly a simple function can be expressed as a finite sum of disjoint indicator functions from which the zero value is deleted:

$s = \sum_{a \in \text{Ra}(s)} a \mathbb{I}_{s = a} = 0 \cdot \mathbb{I}_{s = 0} + \sum_{a \in \text{Ra}(s), a\neq 0} a \mathbb{I}_{s = a} = \sum_{a \in \text{Ra}(s), a\neq 0} a \mathbb{I}_{s = a}$

Since for any $a \in \text{Ra}(s) : a \neq 0$ , $\mathbb{I}_{s = a} \le \mathbb{I}_{s \neq 0}$ , $\mu \mathbb{I}_{s = a} \le \mu \mathbb{I}_{s \neq 0} < \infty$ , $s$ the indicator functions are bounded indicator functions.

Conversely, suppose a function $s$ is a linear combination of bounded indicator functions. Let $j_1,\ldots,j_m$ be the indices of first occurrences of nonzero values in $a_1,\ldots,a_n$ . Then

$s = \sum_{k = 1..m} a_{j_k} \sum_{i : a_i = a_{j_k}} \mathbb{I}_{A_i} = \sum_{k = 1..m} a_{j_k} \mathbb{I}_{\cup_{i : a_i = a_{j_k}}A_i}$

and the range of $s$ is the set $\{ a_{j_k} : k = 1..m\}$ . Since the disjoint sum of bounded indicator functions is a bounded indicator function, this representation is that of a simple function.

lemma: simple functions form a linear space over the ring of real numbers.

If $s$ is simple, then $as$ is simple. Let $f,g$ be simple functions. Then they can be expressed as linear combinations of bounded indicator functions:

$f = \sum_{i=1..n} a_i \mathbb{I}_{A_i}$

$g = \sum_{j = 1..m} b_j \mathbb{I}_{B_j}$

define $A_0 = \Omega - \cup_{i=1..n} A_i$ and $B_0 = \Omega - \cup_{j=1..m} B_j$ . Then

$f = \sum_{i=1..n, j=0..m} a_i \mathbb{I}_{A_i \cap B_j}$

$g = \sum_{j=1..m, i=0..n} b_j \mathbb{I}_{A_i \cap B_j}$

and so

$f + g = \sum_{i=1..n,j=1..m} (a_i + b_j) \mathbb{I}_{A_i \cap B_j} + \sum_{i=1..n} a_i \mathbb{I}_{A_i \cap B_0} + \sum_{j=1..m} b_j \mathbb{I}_{B_j \cap A_0}$

which is a simple function since all sets are of finite measure and are mutually disjoint.

definition : $\mu(\sum_{i=1..n} a_i \mathbb{I}_{A_i}) = \sum_{i=1..n} a_i \mu(\mathbb{I}_{A_i})$

lemma: the measure of a simple function is independent of the representation

Let $f = \sum_{i=1..n} a_i \mathbb{I}_{A_i}$ and $g = \sum_{j = 1..m} b_j \mathbb{I}_{B_j}$ be such that $f = g$ and where no coefficient is zero. From the prior lemma,

$f = \sum_{i=1..n,j=1..m} a_i \mathbb{I}_{A_i \cap B_j} + \sum_{i=1..n} a_i \mathbb{I}_{A_i \cap B_0}$

$g = \sum_{i=1..n,j=1..m} b_j \mathbb{I}_{A_i \cap B_j} + \sum_{j=1..m} b_j \mathbb{I}_{B_j \cap A_0}$

If $x \in A_i \cap B_0$ then $f(x) = a_i = g(x) = 0$ , which is false. So $A_i \cap B_0 = \emptyset$ . Analagously, $B_j \cap A_0 = \emptyset$ .

$f = \sum_{i=1..n,j=1..m} a_i \mathbb{I}_{A_i \cap B_j}$

$g = \sum_{i=1..n,j=1..m} b_j \mathbb{I}_{A_i \cap B_j}$

If $A_i \cap B_j \neq \emptyset$ then when $x \in A_i \cap B_j$ , $f(x) = a_i = g(x) = b_j$ . So

$f = \sum_{i=1..n,j=1..m : A_i \cap B_j \neq \emptyset} a_i \mathbb{I}_{A_i \cap B_j} = g$

$\mu(f) = \sum_{i = 1..n} a_i \mu(\mathbb{I}_{A_i}) = \sum_{i=1..n,j=1..m : A_i \cap B_j \neq \emptyset} a_i \mu(\mathbb{I}_{A_i \cap B_j}) + \sum_{i=1..n,j=1..m : A_i \cap B_j = \emptyset} a_i \mu(\mathbb{I}_{A_i \cap B_j}) + \sum_{i=1..n} a_i \mu(\mathbb{I}_{A_i \cap B_0})$ and analogously for $\mu(g)$ .

Hence $\mu(f) = \mu(g)$

definition: for measurable functions $f,g$ let $f \ge g$ $\mu$ -a.e if $\mu( \mathbb{I}_{f < g} ) = 0$ .

lemma: $\mu$ is a.e. positive linear functional on simple functions.

To be shown:

when $f$ is simple, $f \ge 0$ a.e then $\mu(f) \ge 0$ .
$\mu$ is a linear functional defined on the linear space of simple functions.

Let $f = \sum_i a_i \mathbb{I}_{A_i}$ and $f \ge 0$ a.e. then if $a_i < 0$ $\mu(A_i) = 0$ , and so

$\mu(f) = \sum_i a_i \mu(A_i) = \sum_{i : a_i \ge 0} \mu(A_i) \ge 0$ .

By line in prior result,

$f + g = \sum_{i=1..n,j=1..m} (a_i + b_j) \mathbb{I}_{A_i \cap B_j} + \sum_{i=1..n} a_i \mathbb{I}_{A_i \cap B_0} + \sum_{j=1..m} b_j \mathbb{I}_{B_j \cap A_0}$

and so

$\mu(f+g) = \sum_{i=1..n,j=1..m} (a_i + b_j) \mu(\mathbb{I}_{A_i \cap B_j}) + \sum_{i=1..n} a_i \mu(\mathbb{I}_{A_i \cap B_0}) + \sum_{j=1..m} b_j \mu(\mathbb{I}_{B_j \cap A_0})$

$= \sum_{i=1..n} a_i \sum_{j=0..m} \mu(\mathbb{I}_{A_i \cap B_j}) + \sum_{j=1..m} b_j \sum_{i=0..n} \mu(\mathbb{I}_{A_i \cap B_j})$

$= \mu(f) + \mu(g)$

corollary: for simple functions $f,g$ ,

if $f \le g$ a.e then $\mu(f) \le \mu(g)$
if $f = g$ a.e. then $\mu(f) = \mu(g)$
$| \mu(f) | \le \mu(|f|)$

Since $f \le g$ a.e. iff $g - f \ge 0$ a.e. $\mu(g-f) = \mu(g) - \mu(f) \ge 0$ . Since $f = g$ a.e. iff $f \ge g \ge f a.e$ , it follows from the first result that $\mu(f) \ge \mu(g) \ge \mu(f)$ and so $\mu(f) = \mu(g)$ . Since $-f,f \le |f|$ and $|f|$ is a simple function, $\mu(-f),\mu(f) \le \mu(|f|)$ . Since $\mu(-f) = -\mu(f)$ , it follows that $| \mu(f)| \le \mu(|f|)$ .

Bounded Functions

definition: $f$ is a bounded measurable function if its support is a set of finite measure $E$ and on this set $f$ is uniformly bounded.

lemma: the set of bounded measurable functions is a linear space over the ring of real numbers.

It is clear that if $|f| \le M$ then $|af| \le |a|M$ . Since $\mu(\mathbb{I}(af \neq 0)) = \mu(\mathbb{I}(f \neq 0)) \mathbb{I}(a \neq 0)$ It follows that the scalar multiple of a bounded measurable function is a bounded measurable function.

Let $f,g$ be bounded measurable functions. If $|f| \le M, |g| \le N$ then $|f+g| \le |f| + |g| \le M + N$ . Also since $f=0, g = 0$ implies $f + g = 0$ , it follows that $\mu(\mathbb{I}(f + g \neq 0)) \le \mu(\mathbb{I}(f \neq 0)) + \mu(\mathbb{I}(g \neq 0)) < \infty$ . Hence the sum of bounded measurable functions has finite support.

definition: if $\sup\{\mu(s) : s \le f, s \; \text{simple}\} = \inf\{ \mu(s) : f \le s, s \; \text{simple}\}$ then

$\mu(f) = \sup\{\mu(s) : s \le f, s \; \text{simple}\}$

lemma: if $f$ is a bounded measurable function, then $\mu(f)$ exists.

There is an $M < \infty$ such that $|f| \le M$ on $E$ the support of $f$ . Define $f_n = n^{-1} \lfloor n f \rfloor$ and $g_n = (f_n + n^{-1})\mathbb{I}_E$ . Then $f_n,g_n$ are simple functions such that $f_n \le f \le g_n$ and $\mu(g_n - f_n) = n^{-1} \mu(\mathbb{I}_E)$ . By linearity, $\mu(g_n) -\mu(f_n) = n^{-1} \mu(\mathbb{I}_E)$ .

Define $U = \sup\{\mu(s) : s \le f, s \; \text{simple}\}$ and $L = \inf\{ \mu(s) : f \le s, s \; \text{simple}\}$

and since $L \le \mu(g_n)$ and $\mu(f_n) \le U$ ,

$L - U \le n^{-1} \mu(\mathbb{I}_E)$ . Since this inequality is valid for all $n$ , $L = U$ . Hence $\mu(f)$ is defined and $\mu(f) = U$ .

lemma: $\mu$ is a.e. positive linear functional on bounded measurable functions.

Let $f$ be a bounded measurable function. Then $af$ is a bounded measurable function.

Either $a > 0$ , $a = 0$ or $a < 0$ .

If $a > 0$ , Let $s$ be a simple function such that $s \le f$ . Then $as \le af$ and so $\mu(as) \le \mu(af)$ and $\mu(s) \le a^{-1} \mu(af)$ . Since this bound is valid for all simple functions such that $s \le f$ , $\mu(f) \le a^{-1} \mu(af)$ , so $a \mu(f) \le \mu(af)$ . Let $s$ be a simple function such that $s \le af$ . Then $a^{-1} s$ is a simple function such that $a^{-1} s \le f$ , and so $\mu(a^{-1} s) \le a \mu(f)$ . So $\mu(s) \le a \mu(f)$ . But this implies that $\mu(af) \le a \mu(f)$ and so $\mu(af) = a \mu(f)$ .

If $a = 0$ , then $af = 0$ is a simple function and $\mu(af) = 0 = 0 \mu(f)$ .

If $a < 0$ , let $s$ be a simple function such that $f \le s$ . Then $as \le af$ and so $a \mu(s) = \mu(as) \le \mu(af)$ and so $a^{-1} \mu(af) \le \mu(s)$ . Since this is true for all such $s$ , $a^{-1} \mu(af) \le \mu(f)$ , and $a \mu(f) \le \mu(af)$ . Dually, $\mu(af) \le a \mu(f)$ . Hence $\mu(af) = a \mu(f)$ .

Let $f,g$ be bounded measurable functions. Then $f + g$ is a bounded measurable function. Let $s,t$ be simple functions such that $s \le f, latex t \le g$ . Then $s+t$ is a simple function such that $s+t \le f + g$ , and so $\mu(s) + \mu(t) = \mu(s+t) \le \mu(f+g)$ and this implies that $\mu(f) + \mu(g) \le \mu(f+g)$ . Dually, let $s,t$ be simple functions such that $f \le s, g \le t$ . Then $f + g \le s + t$ and so $\mu(f+g) \le s + t$ and so $\mu(f+g) \le \mu(f) + \mu(g)$ . Hence $\mu(f+g) = \mu(f) + \mu(g)$ .

Let $f$ be a bounded measurable function such that $f \ge 0$ a.e. Then $f = f \mathbb{I}_{f \ge 0} + f \mathbb{I}_{f < 0}$ is the sum of bounded measurable functions, and so $\mu(f) = \mu(f \mathbb{I}_{f \ge 0}) + \mu(f \mathbb{I}_{f < 0})$ . But Since $|f| \le M$ for some $M$ , it follows that $-M \mathbb{I}_{f < 0} \le f \mathbb{I}_{f < 0} \le M \mathbb{I}_{f < 0}$ and so $\mu(f \mathbb{I}_{f < 0}) = 0$ . Since $f \mathbb{I}_{f \ge 0} \ge 0$ , it follows that $\mu(f \mathbb{I}_{f \ge 0} ) \ge 0$ , and so $\mu(f) \ge 0$ ..

corollary: for bounded measurable functions $f,g$ ,

if $f \le g$ a.e then $\mu(f) \le \mu(g)$
if $f = g$ a.e. then $\mu(f) = \mu(g)$
$| \mu(f) | \le \mu(|f|)$

For any bounded simple function $\phi$ such that $\phi \le f$ a.e., it follows that $\phi \le g$ a.e., $\mu(phi) \le \mu(g)$ and so $\mu(f) \le \mu(g)$ .

The second property is immediate consequence of the first property.

Since $f, -f \le |f|$ it follows for any bounded simple function $\phi \le f$ that $\phi \le |f|$ and so $\mu(\phi) \le \mu(|f|)$ and so $\mu(f) \le \mu(|f|)$ . Analogously, $-\mu(f) = \mu(-f) \le \mu(|f|)$ and so $|\mu(f)| \le \mu(|f|)$ .

Non-negative Functions

definition: a non-negative measurable function is a measurable function such that $f \ge 0$ .

lemma: the set of non-negative measurable functions is closed under addition and multiplication by non-negative scalars.

definition: $\mu(f) = \sup \{ \mu(g) : 0 \le g \le f, g \; \text{bounded measurable function}\}$ .

lemma: Let $\left< E_n \right>$ be a non-decreasing sequence of bounded measurable sets such that $\cup E_n = \Omega$ . Then $\mu( \min(f,n) \mathbb{I}_{E_n}) \uparrow \mu(f)$

Since $0 \le \min(f,n)\mathbb{I}_{E_n} \le \min(f,n+1)\mathbb{I}_{E_{n+1}})$ and $\min(f,n)\mathbb{I}_{E_n}$ is a non-negative bounded measurable function, it follows that $\mu(\min(f,n)\mathbb{I}_{E_n}) \le \mu(\min(f,n+1)\mathbb{I}_{E_{n+1}}) \le \mu(f)$ . Hence $\liminf_n \mu(\min(f,n)\mathbb{I}_{E_n}) \le \mu(f)$ .

If $h$ is a non-negative bounded measurable function with bounded support such that $h \le f$ , then then there is an $M < \infty$ such that $h \le M$ and so $h \le \min(f,n)$ for all $n \ge M$ . So $\mu(\min(f,n) \mathbb{I}_{E_n}) \ge \mu(h \mathbb{I}_{E_n})$ . But $h = h \mathbb{I}_{E_n} + h \mathbb{I}_{E_n^C}$ is the sum of bounded measurable functions and so $\mu(h) = \mu(h \mathbb{I}_{E_n}) + \mu(h \mathbb{I}_{E_n^C})$ . Hence $\mu(\min(f,n) \mathbb{I}_{E_n}) \ge \mu(h) - \mu(h \mathbb{I}_{E_n^C})$ . But $\mu(h \mathbb{I}_{E_n^C}) \le M \mu(\mathbb{I}_{E_n^C} \mathbb{I}_{h > 0})$ and $\lim_n \mu(E_n^C) = 0$ and so $\liminf_n \mu(\min(f,n) \mathbb{I}_{E_n}) \ge \mu(h)$ . Hence $\liminf_n \mu(\min(f,n) \mathbb{I}_{E_n}) \ge \mu(f)$ .

lemma: $\mu$ is a non-negative functional on non-negative measurable functions that is a linear functional on the semiring of positive real numbers.

When $0 \le h \le f$ where $h$ is a bounded non-negative measurable function with bounded support, it follows that $0 \le \mu(h)$ and $\mu(h) \le \mu(f)$ . Hence $0 \le \mu(f)$ .

Let $0 \le h \le f$ and $a > 0$ . Then $0 \le ah \le af$ and so $a\mu(h) = \mu(ah) \le \mu(af)$ . Hence $\mu(h) \le a^{-1} \mu(af)$ for all $h$ such that $0 \le h \le f$ and so $\mu(f) \le a^{-1} \mu(af)$ . and so $a \mu(f) \le \mu(af)$ . Analogously, $\mu(af) \le a \mu(f)$ , and so $a\mu(f) = \mu(af)$ . Let $f,g$ be non-negative measurable functions. Then when $0 \le h \le f, 0 \le k \le g$ , for bounded measurable functions $h,k$ it follows that $0 \le h + k \le f + g$ and so $\mu(h+k) \le \mu(f + g)$ . Hence $\mu(h) + \mu(k) \le \mu(f + g)$ for all such $h,k$ But this implies that $\mu(f) + \mu(g) \le \mu(f+g)$ . Conversely, since $\min(f+g,n) \le \min(f,n) + \min(g,n)$ , using the prior lemma, $\mu(f + g) \le \mu(f) + mu(g)$ and so $\mu(f + g) = \mu(f) + \mu(g)$ .

corollary: If $0 \le f \le g$ a.e. then $\mu(f) \le \mu(g)$ and if $f = g$ a.e. then $\mu(f) = \mu(g)$ .

Since $0 \le f \le g$ a.e. implies $0 \le f \mathbb{I}_{f \le g} \le g$ and so $0 \le g - f \mathbb{I}_{f \le g}$ and so $0 \le \mu(g - f \mathbb{I}_{f \le g}) = \mu(g) - \mu(f \mathbb{I}_{f \le g})) = \mu(g) - \mu(f)$

Integrable Functions

definition: a measurable function $f$ is integrable if $\mu(|f|) < \infty$.

definition: the positive and negative parts of $f$ are the non-negative functions $f^+ = \max(f,0)$ and $f^- = \max(-f,0)$

lemma: $f = f^+ - f^-$ and $|f| = f^+ + f^-$ .

true since $f = \max(f,0) + \min(f,0) = f^+ - \min(-f,0) = f^+ - f^-$ and since $|f| = \max(f,0) + \min(-f,0)$

lemma: the set of integrable functions is a linear space over the ring of real numbrs.

Let $a$ be a real number and let $f,g$ be integrable functions. Then $|af| = |a| |f|$ and so $\mu(|af|) = |a| \mu(|f|) < \infty$ . Also $|f + g| \le |f| + |g|$ and so $\mu(|f+ g|) \le \mu(|f|) + \mu(|g|) < \infty$ .

definition: when $f$ is integrable $\mu(f) = \mu(f^+) - \mu(f^-)$ .

The measure of an integrable function is well-defined since $f^+,f^- \le |f|$ and so the positive and negative parts of an integrable function are integrable functions.

lemma: If $f = f_1 - f_2$ where $f_1,f_2$ are non-negative and integrable functions, then $\mu(f) = \mu(f_1) - \mu(f_2)$

Since $f = f^+ - f^- = f_1 - f_2$ then $f^+ + f_2 = f_1 + f^-$ . and so $\mu(f^+) + \mu(f_2) = \mu(f_1) + \mu(f^-)$ . Since all measures are finite, subtraction is allowed and $\mu(f) = \mu(f_1) - \mu(f_2)$ .

lemma: $\mu$ is a non-negative a.e linear functional

Let $f$ be an integrable function and $a$ a real number. If $a>0$ then $(af)^+ = af^+$ and $(af)^- = af^-$ and so $\mu(af) = \mu(af^+) - \mu(af^-) = a \mu(f^+) - a \mu(f^-) = a \mu(f)$ . If $a < 0$ then $(af)^+ = (-a) f^-$ and $(af)^- = (-a) f^+$ and so $\mu(af) = \mu((-a) f^-) - \mu((-a)f^+) = (-a) (-\mu(f)) = a \mu(f)$

Let $f,g$ be integrable functions. Since $f = f^+ - f^-$ and $g = g^+ - g^-$ and so $f + g = (f+g)^+ - (f+g)^- = f^+ - f^- + g^+ - g^-$ and

$(f+g)^+ + f^- + g^- = (f+g)^- + f^+ + g^+$ are all non-negative integrable functions and so

$\mu((f+g)^) + \mu(f^-) + \mu(g^-) = \mu((f+g)^-) + \mu(f^+) + \mu(g^+)$ are all finite since $f+g,f,g$ are integrable. Hence on gathering terms, $\mu(f+g) =\mu(f) + \mu(g)$ .

COMMENTS: a few loose ends but overall okay