Integral

We consider a measure space $(X, \Omega, \mu)$, where $X$ is a set of elements, $\Omega$ is a collection of measurable sets and $\mu : \Omega \rightarrow R$ a measure defined on $\mu$. Note that by definition $\Omega$ is a $\sigma$ algebra and $\mu$ satisfies

1. (Zero element) $\mu(\emptyset) = 0$,
2. (Non negativity) $\mu(E) \ge 0$ for all $E \in \Omega$,
3. ($\sigma$-additivity) $\mu( \cup_{i \ge 1} E_i) = \sum_{i = 1}^\infty \mu(E_i)$ for a countable sequence of mutually disjoint sets $\{E_i\}$.

Simple Function

A characteristic function $[ \cdot ] \rightarrow \{0, 1\}$ is a function that takes value 1 if its argument is true and zero otherwise. A characteristic function $\chi_E : X \rightarrow \{0, 1\}$ for a set $E \subset X$ is denoted as $\chi_E \doteq [x \in E]$.

A simple function $f : X \rightarrow R$ is summation of finite terms:

where $\{E_1, E_2,... E_n \}$ are measurable sets and constitute a partition of $X$, i.e., $\cup_{i \in [n] } E_i = X$ and $E_i \cap E_j = \emptyset$ for $i \neq j$.

Simple Integrable Function

A simple function $f=\sum_{ i \in [n] } a_i \chi_{E_i }$ is said to be integrable if $\mu(E_i ) < \infty$ for all the indices $i$ for which $a_i \neq 0$, i.e., $a_i \neq 0 \rightarrow \mu(E_i) < \infty$.
The integral of $f$ is the real number $\sum_{ i \in [n] } a_i \mu(E_i)$, where we agree to take $a_i \mu(E_i ) = 0$ for all the indices $i$ for which $a_i=0, \mu(E_i )= \infty$. We denote this sum by

Theorem The integration is well defined and independent of the representation of $f$.

Proof. Let $f = \sum_{i \in [n] } a_i \chi_{E_i}$ and $g = \sum_{j = 1}^m b_j \chi_{F_j}$. Suppose that $f \equiv g$, the goal is to show that $\int f = \int g$. Define $S_{ij} = E_i \cap F_j$. If $S_{ij} \neq \emptyset$, we must have $a_i = b_j$. To see this, let $s \in S_{i,j}$. Since $\{E_1, E_2,... E_n \}$ is a partition of $X$,only one of the value $\{ \chi_{E_i} (s) : i \in [n] \}$ takes value 1 and $f(s) = \sum_{i \in [n] } a_i \chi_{E_i}(x) = a_i \chi_{E_i}(x) = a_i$. Similarly, $g(s) = b_j$. We have $a_i = b_j$ as $f \equiv g$.

Now, let $c_{i,j} = a_i = b_j$ if $S_{i,j} \neq \emptyset$ and $c_{i,j}= 0$ otherwise. Using the fact that $\{ F_1, ..., F_m \}$ also constitutes a partition of $X$, we have $E_i = \cup_{j = 1}^m E_i \cap F_j$ for $i \in [1, n]$. By finite additivity of measure,

By symmetry, we can show $\int g = \sum_{ i \in [n] } \sum_{j=1}^m c_{i,j} \mu( S_{i,j} )$, which implies that $\int f = \int g$, as desired.

NOTE: in the proof we see that $\sum_{ i \in [n] } \sum_{j=1}^m c_{i,j} \mu( S_{i,j} ) = \int f < \infty$. It follows that $\mu(S_{i, j}) = \infty \rightarrow c_{i,j} = 0$

$\blacksquare$

Theorem. (Integration on a measurable set). If $E$ is any measurable set and $f$ is an integrable simple function, then $\chi_E f \doteq \chi_{E} \sum_{i \in [n] } a_i \chi_{E_i}$ is also an integrable simple function.

Proof.

By monotonicity of measure, if $a_i \neq 0$, then $\mu(E_i \cap E) \le \mu(E_i) < \infty$. Therefore, $\chi_E f$ is integrable.

$\blacksquare$

Definition The integration of $\chi_E f$ is abbreviated as

$$\int_E f = \int \chi_E f.$$

Now we show some basic properties of simple integration. These properties really inherit from the non-negativity and complete additivity of measure $\mu$.

Basic Properties

Let $f=\sum_{ i \in [n] } a_i \chi_{E_i }$ and $g=\sum_{j=1}^m b_j \chi_{F_j }$ be two integrable simple functions, and $\alpha, \beta \in R$ two real values.

1. Linearity. Integrable simple functions forms a vector space, i.e., $\int \alpha f + \beta g =a \int f + \beta \int g$
   Proof. Define $S_{ij} = E_i \cap F_j$. Now $\{ S_{i,j} \}$ is a partition of $X$. Let $c_{i,j} = a_i = b_j$ if $S_{i,j} \neq \emptyset$ and $c_{i,j}= 0$ otherwise. Moreover, we have $c_{i,j} \neq 0 \implies \mu(S_{i,j} ) = 0$.

   $$\begin{aligned} \alpha f + \beta g &= \alpha \sum_{i \in [n] } a_i \chi_{E_i} + \beta \sum_{j = 1}^m b_j \chi_{F_j} = \sum_{i,j} (\alpha + \beta) c_{i,j} \chi_{S_{i,j} } \,, \\ \int \alpha f + \beta g &= \sum_{i,j} (\alpha + β) c_{i,j} \mu(S_{i,j}) \\ &= \alpha \sum_{i,j} c_{i,j} \mu(S_{i,j} ) + β \sum_{i,j} c_{i,j} \mu( S_{i,j} ) \\ &= \alpha \int f + \beta \int g \qquad \qquad \qquad \qquad \qquad \blacksquare \end{aligned}$$

2. If $f \ge 0 \ a.e.$, then $\int f \ge 0$.
   Proof. Let $f = \sum_{i \in [n] } a_i \chi_{E_i}$. We claim that $a_i \mu(E_i) \ge 0$ for all $i \ge 1$.

   • If $\mu(E_i) = \infty$, then $a_i = 0$, since $f$ is measurable.

   • If $\mu(E_i) > 0$, then $a_i \ge 0$. Otherwise, we have $f < 0$ on a set $E_i$ with positive measure, contradicting $f \ge 0 \ a.e.$.

   • If $\mu(E_i) = 0$, the claim is trivially true.
     Therefore, $\int f = \sum_i a_i \mu(E_i ) \ge0$.

   $\blacksquare$

3. If $f \ge g \ a.e$, then $\int f \ge \int g$.
   Proof. Observe that

   $$\begin{aligned} \int f - \int g &= \sum_{i \in [n] } a_i \mu(E_i ) -\sum_{j \in [m] } b_j \mu(F_j ) \\ &= \sum_{i \in [n], j \in [m] } (a_i - b_j ) \mu(E_i \cap F_j) \end{aligned}$$

   and that

   $$a_i-b_j < 0 \implies \mu(E_i \cap F_j ) = 0$$

   NOTE: Another way to prove this is by defining $h \doteq f - g$ which if non-negative almost everywhere and use conclusion of (1).
   $\blacksquare$

4. $|f|$ is an integrable simple function, and $|\int f|\le \int |f|$.
   Proof. Let $|f|=\sum_i |a_i | \chi_{E_i }$. Therefore, $|a_i| ≠ 0 \implies \mu(E_i )=0$. $|f|$ is simple and integrable.

   $$\int|f| =\sum_i |a_i |\mu(E_i ) \ge \left| \sum_i a_i \mu(E_i ) \right| = \left| \int f \right|. \qquad \blacksquare$$

5. $\int|f+g | \le \int |f| + \int |g|$.
   Proof.

   $$\int|f+g| \le \int(|f|+|g|) =\int|f| +\int|g|. \qquad \blacksquare$$

6. If $m \le f \le M \ a.e.$ on a measurable set $E$ with $\mu(E) < \infty$, then $m \mu(E) \le \int_E f \le M \mu(E)$.
   Proof. The condition $\mu(E) < \infty$ ensures that $m \cdot \mu(E) < \infty$. Note that

   $$\int_E f=\sum_i a_i \mu(E_i \cap E)$$

   and that

   $$a_i < m \implies \mu(E_i \cap E) = 0 \implies \sum_i (a_i-m) \mu(E_i \cap E) \ge 0.$$

   In a similar manner we can prove the other inequality.

   NOTE: another way is to utilize that $\mu \chi_E$ is simple, integrable $\mu \chi_E \le f$ almost everywhere.
   $\blacksquare$

7. If $f \ge 0,\ a.e.$, and if $E$ and $F$ are measurable sets with $E⊂F$, then $\int_E f \le \int_F f$.
   Proof.

   $$\int_E f = \sum_i a_i \mu(E_i \cap E) \le \sum_i a_i \mu(E_i \cap F) =\int_F f.$$

   Then inequality holds since it is always true that

   $$a_i \mu(E_i \cap E) \le a_i \mu(F_i \cap E).$$
   • If $\mu(E_i \cap E) \le \mu(E_i \cap F) = \infty$, then $a_i = 0$.
   • If $0 < \mu(E_i \cap E) \le \mu(E_i \cap E) < \infty$, then $a_i \ge 0$.
   • If $0 = \mu(E_i \cap E) = \mu(E_i \cap E)$, then it is trivially true.

   NOTE: another way is to utilize that $\chi_E f \le \chi_F f$ almost everywhere.
   $\blacksquare$

8. Monotone convergence. If $F = +_{m \ge 1} F_m$ is a disjoint union of measurable sets, then

   $$\int_F f = \int f \cdot \Big( \sum_{m \ge 1} \chi_{ F_m } \Big) = \sum_{m \ge 1} \int f \cdot \chi_{ F_m } = \sum_{m \ge 1} \int_{F_m} f$$

   Proof. This is a consequence of $\sigma$-additivity of measure. Let $f = \sum_{i \in [n] } a_i \chi_{E_i}$,

   $$\begin{aligned} \int_F f &= \sum_{ i \in [n] } a_i \mu(E_i \cap F) \\ &=\sum_{ i \in [n] } a_i \mu(E_i \cap ( +_{m \ge 1} F_m )) \\ &=\sum_{ i \in [n] } a_i \mu( +_{m \ge 1} E_i \cap F_m ) \\ &=\sum_{ i \in [n] } a_i \sum_{m \ge 1} \mu(E_i \cap F_m) & \text{ by $\sigma$-additivity of measure }\\ &=\sum_{ i \in [n] } a_i \lim_{k \rightarrow \infty} ⁡\sum_{m=1}^k \mu(E_i \cap F_m) \\ &= \lim_{k \rightarrow \infty} ⁡\sum_{ i \in [n] } a_i \sum_{m=1}^k \mu(E_i \cap F_m) & \text{ exchange order of limit and finite summation} \\ &= \lim_{k \rightarrow \infty} ⁡ \sum_{m=1}^k \sum_{ i \in [n] } a_i \mu(E_i \cap F_m) \\ &= \lim_{k \rightarrow \infty} ⁡\sum_{i=1}^k \int_{F_m} f \qquad \blacksquare \end{aligned}$$

9. Monotone convergence. If $F_1 \subset F_2 \subset F_3 \subset ...$ is a sequence of increasing measurable sets and let $F = \cup_{n} F_n$, then $\lim_n \int_{F_n} f = \int_F f$.
   Proof. Let $F_n' = F_n - F_{n - 1}$. The result follows immediately from the previous property.
   $\blacksquare$

10. If $F_1 \supset F_2 \supset F_3 \supset ...$ is a sequence of decreasing measurable sets and let $F = \cap_{n} F_n$, then $\lim_n \int_{F_n} f = \int_F f$.

    Proof. Let $f = \sum_{i = 1}^N a_i \chi_{E_i}$. It follow that

    $$a_i \neq 0 \implies \mu(E_i \cap F_1) < \mu(E_i) < \infty.$$

    Using continuity of measure $\mu$,

    $$\lim_n \mu(E_i \cap F_n) = \mu(E_i \cap \lim_n F_n) = \mu (E_i \cap F),$$

    which implies that $\lim_n \int_{F_n} f = \int_F f$.

    Remark. Since $f$ is integrable, it has nonzero values on a finite support. We don't even require that $\mu(F_1) < \infty$.
    $\blacksquare$

11. Let $f = \sum_{i \in [n] } a_i \chi_{E_i}$ and define the support $N(f)$ of simple integrable function $f$ to be the set of elements with non-zero values $N(f) = \{ x \in X : f(x) \neq 0 \} = \cup_{i, a_i \neq 0} E_i$, then $\mu( N(f) ) \le \infty$.

    Proof. Since $f$ is integrable, $a_i \neq 0$ implies $\mu(E_i) < \infty$. Now

    $$\mu( N(F) ) = \sum_{1 \le i \le n, a_i \neq 0} \mu(E_i) < \infty \qquad \qquad \blacksquare$$

With simple integrable functions as building blocks, we are ready to define integration for functions in general.

Fist we introduce a notation of convergence we need in the following discussion.

Cauchy In Mean

Definition. A sequence $\{ f_n \}$ of simple functions is called Cauchy in mean if

$$\lim_{n, m} \int |f_n - f_m| = 0$$

Cauchy in mean does not implies Cauchy almost everywhere. Consider the sequence of functions

1. $\chi_{[0, 1/2]}, \chi_{[1 / 2, 1]}$,
2. $\chi_{ [0, 1 / 4 ] }, \chi_{ [1 / 4, 1 / 2] }, \chi_{ [ 1 / 2, 3 / 4 ] }, \chi_{ [3 / 4, 1] }$,
3. ...
4. $... \chi_{ [ i / 2^n, (i + 1) / 2^n ] } ...,(i \in \{0, 1, 2, ..., 2^{n} - 1 \} )$

They are Cauchy in mean but don's Cauchy almost everywhere. But it is Cauchy in measure.

Theorem If a sequence $\{ f_n \}$ of simple functions is Cauchy in mean, it is Cauchy in measure.

Proof. Let $h_{n, m} = |f_n - f_m|$. As $f_n$ and $f_m$ are simple and integrable, $h_{n, m}$ is simple and integrable. Moreover, it is non-negative. We write it as $h_{n,m} = \sum_{j \in [k] } b_j \chi_{F_j}$, where $b_j \ge 0$. For all $\delta > 0$, $b_j \ge \delta > 0$ implies that $\mu(F_j) < \infty$. Now

Hence, $\mu( \{ x \in X: h_{n,m}(x) \ge \delta\} ) \le \frac{ \int |f_n - f_m| }{ \delta }$. For any fix $\delta$, this can be arbitrary small as $n, m \rightarrow \infty$.

$\blacksquare$

NOTE: the converse is not true.

Example Consider this sequence of functions:

$$\{ n \cdot \chi_{[0, 1 / n] } : n \in \mathbf{N} \}$$

It converges in measure to $f \equiv 0$. But $\int n \cdot \chi_{[0, 1 / n] } = 1$ for all $n$. $\lim_n \int n \cdot \chi_{[0, 1 / n] } \neq \int f$.

The implication here is that we need additional condition more than converge in mean to define the integral for a general function $f$. We need to connect the sequence $\{ f_n \}$ with $f$, the function we want to define integral. Can we say something like $\{ f_n \}$ converges to $f$ in mean, i.e., $\lim_n \int |f_n - f| = 0$? But we have not define integral of non-simple function yet. Here is how we establish the connection.

General Integrable Functions

Definition. An extended real-valued, measurable function $f$, on a measure space $(X, \Omega, \mu)$ is said to be integrable if there exists a sequence $\{ f_n \}$ of integrable simple functions having the following properties:

1. $\{ f_n \}$ is a Cauchy sequence in the mean.
2. $\lim_{n} f_n (x) = f(x) \quad a.e.$.

The integral $\int f$ of $f$ is defined as

$$\int f = \lim_{n} \int f_n.$$

There are two issues we need to answer for the definition to be well-defined. The first one is that the limit of $\int f_n$ indeed exists. Second, we need to show that for any sequence satisfying condition (1) and (2), their integral converges to the same limit.

Theorem (Existence of Limit). $\lim_n \int f_n$ exists.

Proof.

Therefore, $\{ \int f_n : n \in \mathbf{N} \}$ (note that each element in this set is a real number) is a Cauchy sequence and its limit exists.

$\blacksquare$

Theorem (Uniqueness of Limit). Let $\{ f_n \}$ and $\{ g_n \}$ be two Cauchy sequences in the mean of integrable simple functions, such that

$$\lim_n f_n = \lim_n g_n = f \qquad a.e.$$

Then

$$\lim_n \int f_n = \lim_n \int g_n.$$

Proof. Define $h_n \doteq f_n - g_n$ for $n \in \mathbf{N}$. As $f_n$ and $g_n$ are simple and integrable, $h_n$ is simple and integrable.

Step 1.

We claim that

1. $\{ h_n \}$ is Cauchy in mean.
2. $\lim_n h_n = 0 \quad a.e.$.

By monotonicity and linearity of integration of simple functions,

The integral can be arbitrary small as $n, m \rightarrow \infty$.

The second claim results from the $\lim_n f_n = f$ and $\lim_n g_n = f$ $a.e.$.

We are going to prove $\lim_n \int h_n = 0$, which then implies

Step 2.

Our goal is to prove that $\int h_n \rightarrow 0$. Specifically,

• $\forall \epsilon > 0$, $\exists N > 0$, s.t., $\forall n > N, | \int h_n | < O(\epsilon)$.

The trick here is to restrict the discussion of $\int h$ on a $\sigma$-finite measurable set. Define the support $\mathbb{Supp} ( f_n )$ of $f_n$ to be the set with non-zero values:

As $f_n$ is a simple integrable function, $\mu( \mathbb{Supp} ( f_n )) < \infty$. Let $E_n = \cup_{ k \in [n] } \Big( \mathbb{Supp} (f_k) \cup \mathbb{Supp} (g_k) \Big)$ and $E = \cup_n E_n$. We have $\mu(E_n) < \infty$, $E_n \uparrow E$ and $E$ is $\sigma$-finite. By monotone convergence property of simple integrable functions, we have, $\forall n$,

First, as $\{ h_n \}$ is Cauchy in mean, for $\epsilon > 0$, $\exists N_1 > 0$, s.t, for $n, k > N_1$, we have $\int |h_n - h_k| \le \epsilon$ . We choose an integer $k > N_1$. As $h_k$ is a simple integrable function, and as $E_m \uparrow E$, it holds that

For fix $k$, we choose an integer $m$, s..t, $\left| \int h_k - \int_{E_m} h_k \right| \le \epsilon$.

For a fix $k > N_1$, we claim that for any $n > N_1$, we have $| \int h_n - \int_{E_m} h_n | < 3\epsilon$,

We have successfully restrict our discussion on a set with finite measure. The key here is that $\int_{E_m} h_n$ converges uniformly to $\int h_n$, as $m$ increases, independent of the choice of $n$, as long as $\{ h_n \}$ is Cauchy in mean.

Indeed, what we have shown here is that

Second, we prove that

So far we have not consider the condition $\lim_n h_n = 0 \ a.e.$ It is time to turn to it. Define

Observe that $\lim_n \cup_{i \ge n} S_n \downarrow$. As $\mu( \cup_{i \ge 1} S_n) \le \mu (E_m) < \infty$, by continuity of $\mu$,

As $h_k$ is a simple function, it is bounded by some integer $K \ a.e.$. As $\mu(E_m) < \infty$, $\exists N_2 > N_1$, s.t., for $n > N_2$, we have $\mu(S_n) \le \mu(\cup_{i \ge n} S_n ) \le \epsilon / K$. Now,

Finally, we argue that $\left| \int_{S_n} h_n \right|$ is small enough.

Indeed, the technique demonstrated here shows that $\{ \int_E h_n \}$ is uniformly absolutely continuous with respect to $\mu(E)$.

$\blacksquare$

Corollary 1 Let $\{ h_n \}$ be a sequence of simple integrable function that are Cauchy in mean and $E_1 \subset E_2 \subset E_3 ...$ a sequence of increasing measurable sets that converge to $E = \cup_{i \ge 1} E_i$. Then $\int_{E_m} h_n \rightarrow \int_{E} h_n$ uniformly, i.e., $\forall \epsilon > 0, \exists M > 0$, s.t., for $m > M$, we have

$$\left| \int_{E_m} h_n - \int_E h_n \right| \le \epsilon$$

Corollary 2 Let $\{ h_n \}$ be a sequence of simple integrable function that are Cauchy in mean. $\forall \epsilon > 0, \exists \delta > 0$, s.t, for any measurable set $E$ with $\mu(E) < \delta$, we have $|\int_E h_n| < \epsilon$ independent of $n$.

Definition

Let $f$ be an integrable function and $E$ be a measurable set. Then $\chi_E f$ is integrable and its integral is denoted as

Remark:
If $f$ is integrable, then there is a sequence simple, integrable function $\{f_n \}$ that are Cauchy in mean and converges to $f$ almost every. It is easy to see that $\{ \chi_E f_n \}$ is simple, integrable, Cauchy in mean and converges to $\chi_E f$ almost everywhere, therefore $\chi_E f$ is integrable. An immediately consequence is the following lemma

Lemma 1

$\int_E f_n$ converges to $\int_E f$ uniformly irrespective of $E$.
Proof. It suffices to show the sequence $\{ \int_E f_n : n \in \mathbf{N} \}$ is a uniformly Cauchy irrespectively of $E$,

Therefore, for $\forall E \subset \Omega$, $\exists N > 0$, for all $n，m > N$, we have

Taking limit with respect to $m$, we get $\lim_m \int_E f_m - \epsilon \le \int_E f_n \le \lim_m \int_E f_m + \epsilon$, i.e.,

as desired.

Combining Corollary 2 and Lemma 1, we have the following theorem.

Theorem. Integration as Finite Signed Measure

If $f$ is integrable, then

is absolute continuous.

Proof. Let $\{ f_n \}$ as defined before. For $\forall \epsilon > 0$, $\exists N_1 > 0$, for all $n > N_1$, we have $| \int_E f - \int_E f_n | \le \epsilon$. On the other hand, For $\forall \epsilon > 0$, $\exists \delta > 0$, for all $E$ with $\mu(E) < \delta$, we have $|\int_E f_n| < \epsilon$. In combination,

$\blacksquare$

Next we discussion the convergence behavior of a sequence of integrable functions.

Lemma

If $\{ f_k \}$ is a sequence of simple, integrable function that are

1. Cauchy in the mean,
2. Converge to $f$ almost everywhere.

Then

Proof.
Note that $|f_k - f |$ may not be a simple function. To investigate the value of its integral, we need to characterize it according to the definition of integral for general functions. Here is how we do it. For a fix $k$, the sequence $\{ |f_k - f_n| : n \in \mathbf{N} \}$ is simple, integrable, and converges to $|f_k - f |$ almost everywhere. Moreover, as $\{ f_k \}$ itself is Cauchy in mean, for $\forall \epsilon > 0$, $\exists N > 0$, s.t., $\forall k, n, m > N$,

As a consequence $\{|f_k - f_n| : n \in \mathbf{N} \}$ is Cauchy in the mean. $|f_k - f|$ is integrable and $\int |f_k - f| = \lim_n \int |f_k - f_n|$. But for $k, n, m > N$,

As $\epsilon$ could be arbitrary small, we have $\lim_k \int |f_k - f| = 0$.

$\blacksquare$

Theorem

If ${g_k}$ is a sequence of integrable functions satisfying the conditions

1. ${g_k}$ is a Cauchy sequence in the mean.
2. ${g_k}$ converges in measure to $f$.

then $f$ is integrable and $\int f = \lim⁡_k \int g_k$.

Proof. For each $g_k$, there is a sequence $\{ g_{km} : m ∈ \mathbf{N} \}$ simple integrable functions,

1. $\{ g_{km} \}$ is a Cauchy sequence in the mean.

2. $\{ g_{km} \}$ converges in measure to $g_k$.

3. $\int g_k = \lim_m ⁡\int g_{km}$

For fix $k$, we choose such an integer $m$, such that $\int |g_k - \int g_{k m} |\le 1 / k$ and $\mu(\{ x:|g_k (x) - g_{km} (x) | \ge 1 / k \} ) \le 1 / k$. For convenience, denote $g_{k m}$ as $f_k$. Now we claim the sequence $\{ f_{k} \}$ of simple integrable functions satisfies