Lebesgue Integral, Revisited.

Let $(\Omega, \mathcal{F}, \mu)$ be measure space. Further, suppose that $\mu$ is $\sigma$-finite, i.e., there is a countable sequence $\{ S_k \}$ in $\mathcal{F}$, such that

1. $S_k \uparrow \Omega$.
2. $\mu (S_k) < \infty$.

We study the integral, $\int f \, d\, \mu$, of a measurable function $f$. In what follows, we omit $d\, \mu$ and abbreviate $\int f \, d\, \mu$ as $\int f$.

Definition

Simple Function

Definition. For each $A \in \mathcal{F}$, define the indicator function $\mathbf{1}_A : \Omega \rightarrow \mathbb{R}$ as

Definition. $\varphi : \Omega \rightarrow \mathbb{R}$ is said to be a simple function if $\varphi = \sum_{i \in [n] } a_i \mathbf{1}_{ A_i }$, where $A_i \in \mathcal{F}$ are mutually disjoint.

We can assume that the $\{ A_i \}$ is a partition of $\Omega$. Otherwise, let $A_{n + 1} = \Omega - A_1 - \ldots - A_n$ and $a_{n + 1} = 0$. Then $A_1, \ldots, A_{n + 1}$ is a partition of $\Omega$, and

The definition of integral of measurable functions consists of two steps.

Nonnegative Function

Bending the notation slightly, we say that $f$ is nonnegative, denoted as $f \ge 0$, if for all $\omega \in \Omega$, $f(\omega) \ge 0$.

Definition $\diamondsuit$. Let $f \ge 0$ be measurable. Define

$$\int f \doteq \sup \left\{ \sum_{i \in [n] } a_i \mu(A_i): \begin{array}{l} \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \text{ is simple}, \\ 0 \le \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \le f \end{array} \right\}.$$

It is possible that $\int f = \infty$. We say that $f$ is integrable if $\int f < \infty$.

For each nonnegative simple function $\varphi = \sum_{i \in [ n ] } a_i \mathbf{1}_{A_i}$, define $\varphi_k \doteq \mathbf{1}_{S_k} \cdot \varphi = \sum_{i \in [n] } a_i \mathbf{1}_{A_i \cap S_k}$. Note that $\varphi_k$ has finite support and $\varphi_k \rightarrow \varphi$. By continuity of $\mu$,

Hence, the integral of $f$ has an equivalent definition.

Definition $\heartsuit$. Let $f \ge 0$ be measurable. Define

$$\int f \doteq \sup \left\{ \sum_{i \in [n] } a_i \mu(A_i): \begin{array}{ll} \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \text{ is simple}, \\ \mathbb{Supp} \Big( \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \Big) < \infty, \\ 0 \le \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \le f \end{array} \right\}.$$

General Function

A general function $f : \Omega \rightarrow \mathbb{R}$ is said to be integrable if $\int |f| < \infty$. Let $f_+ = \max \{ f, 0\}$ and $f_- = \max \{ - f, 0 \}$. Then $f = f_+ - f_-$.

Definition $\clubsuit$. Let $f$ be measurable. Define

$$\int f \doteq \int f_+ - \int f_-.$$

Integration over a subset

Let $f: \Omega \rightarrow \mathbb{R}$ be measurable and $E \in \mathcal{F}$. Then

Remark: note that if $f$ is measurable and $\mathbf{1}_{E}$ is measurable, then $f \cdot \mathbf{1}_{E}$ is measurable.

Basic Properties

1. Monotonicity. If $f \ge g$ then $\int f \le \int g$.

2. Linearity. The integral is linear on the vector space of integral functions; in particular, $\forall \alpha, \beta \in \mathbb{R}$, $\int \alpha \cdot f + \beta \cdot g = \alpha \cdot \int f + \beta \cdot \int g$.

Monotonicity follows directly from the definition. Establishing linearity is more fiddly and we leave it to the end of the article. We proceed by assuming these two properties of integration.

Properties for Limits

Suppose that $f_n \rightarrow f$ a.e., when can we argue that

i.e., when the order of the limiting and the integration operations can be exchanged?

Intuitively, there are two kinds of convergence: the monotone ones and the bounded ones.

Monotone Convergence

Theorem (Monotone Convergence). If $f_n \ge 0$ and $f_n \uparrow f$, then

$$\int f = \lim_{n \ge 1} \int f_n.$$

Corollary 1. The integral is $\sigma$-additive for nonnegative functions.

Corollary 2. If $f_n \ge g$, where $g$ is integrable, and $f_n \uparrow f$, then

$$\int f = \lim_{n \ge 1} \int f_n.$$

Proof. As $f \ge f_n$, by monotonicity of integration, $\int f \ge \int f_n$. It follows that

It is left to prove the reverse direction of the inequality. Let $\varphi \doteq \sum_{i \in [k] } a_i \mathbf{1}_{ A_i }$ be a simple function such that $\varphi \le f$. Let $c \in (0, 1)$, and define

Exercise. Prove that $E_n \in \mathcal{F}$.

Then

1. $c \cdot \varphi \cdot 1_{E_n} \le f_n$ is also a simple function.

2. $\lim_n E_n = \Omega$, since $f_n \uparrow f$.

By the definition of $\int f_n$, we have

Taking limit of both side, and by continuity of $\mu$, we get

Taking supremum over $\varphi \le f$ and letting $c \rightarrow 1$, we get

$\square$

Fatou's Lemma

Theorem (Fatou's Lemma). If $f_n \ge 0$, then

$$\int \liminf_n f_n \le \liminf_n \int f_n.$$

Corollary. If $f_n \le 0$ and $g$ is integrable, then

$$\limsup_n \int f_n \le \int \limsup_n f_n .$$

Corollary. If $f_n \ge g$ and $g$ is integrable, then

$$\int \liminf_n f_n \le \liminf_n \int f_n.$$

Corollary. If $f_n \le g$ and $g$ is integrable, then

$$\limsup_n \int f_n \le \int \limsup_n f_n .$$

Proof. For all $n \ge 1$, define $h_n = \inf_{k \ge n} f_k$. Since $h_n \le f_n$, by monotonicity of integration, $\int h_n \le \int f_n$. Taking $\liminf$ of both side, we get

Then $h_n \uparrow \liminf_n f_n$, by monotonicity of integration, $\int h_n$ is an increasing sequence. It holds that $\liminf_n \int h_n = \lim_n \int h_n$. It is left to prove

This follows exactly from the Monotone Convergence theorem, or can be proven directly by using the same technique.

$\square$

Bounded Convergence

Theorem (Bounded Convergence). Let $E \in \mathcal{F}$ with $\mu(E) < \infty$. Let $f_n$ be a sequence of functions supported on $E$ with $|f_n| \le M$. Suppose that $f_n \rightarrow f$ point-wise. Then $\int f_n \rightarrow \int f$.

Proof. The proof relies on the lemma.

Lemma. If $f_n \rightarrow f$ point-wise, then $f_n \rightarrow f$ in measure.

Suppose $f_n \rightarrow f$ in measure. Let $\epsilon > 0$. Define $A_n \doteq \{ \omega \in \Omega : |f_n - f| > \epsilon \} \subset E$. Then by linearity and monotonicity of integration,

Since $\mu(A_n) \rightarrow 0$ and $\epsilon$ is arbitrary, this shows $\int f_n \rightarrow \int f$.

$\square$

Modulus of Integrability

Theorem (Modulus of integrability). Let $f \ge 0$ be integrable. Then for any $\epsilon > 0$ there is a $\delta > 0$ such that $m(E) < \delta \implies \int_E f < \epsilon$.

Proof. Define $f_n = \min( f, n) \cdot \mathbb{1}_{S_n}$. Then $f_n \uparrow f$ and by monotone convergence theorem, when $n$ is large enough,

On the other hand, by monotonicity of integration, when $\delta < \epsilon / n$,

Applying linearity and monotonicity of integration again,

$\square$

Dominated Convergence

Theorem (Dominated Convergence). If $f_n \rightarrow f \, a.e.$, $|f_n| \le g$ for all $n$, and $g$ is integrable, then $\int f_n \rightarrow \int f$.

Proof 1.

Step 1. $\exists E \in \mathcal{F}$ with $\mu(E) < \infty$, such that $\int_{\Omega - E} g \le \epsilon$.

Define $g_k = g \cdot \mathbb{1}_{S_k}$. Then $g_k \uparrow g$. Using monotone convergence theorem and linearity of integration, there exists $K \ge 0$,

Then we let $E \leftarrow S_k$.

Step 2.
Since $f_n \rightarrow f$ a.e., $f_n \rightarrow f$ in measure. Define $A_n \doteq \{ \omega \in E : |f_n - f| > \epsilon \} \subset E$. Then $\mu(A_n) \rightarrow 0$.

Finally, by linearity and monotonicity of integration,

As $\mu(A_n) \rightarrow 0$, $\int_{A_n} g \rightarrow 0$. Since $\epsilon$ is arbitrary, this shows $\int f_n \rightarrow \int f$.

$\square$

Proof 2. There is a concise proof by Fatous' lemma.

Let $h_n = 2g - |f - f_n| \ge 0$. Using Fatous' lemma,

As $g$ is integrable, we can subtract $\int g$ from both side and get

Therefore,

$\square$

Proof 3. Since $|f_n| \le g$, we have $- g \le f_n \le g$. As $f_n - (-g) = f_n + g \ge 0$, via Fatous' lemma, we get

$\square$

Linearity

It is difficult to prove linearity by the definitions of $\diamondsuit$ or $\heartsuit$. Hence, we introduce some equivalent definitions for integration, starting from simple functions. These definitions also provide deeper intuitions.

Simple Functions

First, we prove that definition $\diamondsuit$ has an equivalent one for nonnegative simple functions.

Theorem. Let $\varphi \ge 0$ be a simple function and $\sum_{j \in [m] } b_j \mathbf{1}_{B_j}$ be one of its possible representations. Then it holds that

$$\small \int \varphi \doteq \sup \left\{ \sum_{i \in [n] } a_i \mu(A_i): \begin{array}{l} \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \text{ is simple}, \\ 0 \le \sum_{i \in [n] } a_i \mathbf{1}_{A_i} \le \varphi \end{array} \right\} = \sum_{j \in [m] } b_j \mu (B_j).$$

Proof. By the definition of $\varphi$, clearly it holds that

To prove the reverse inequality, let $\psi \doteq \sum_{i \in [n] } a_i \mathbf{1}_{A_i}$ be a simple function, where $A_i \in \mathcal{F}$ are mutually disjoint. Note that the $\{ B_j \}$ are also mutually disjoint. If $\psi \le \varphi$, it holds that for all $i, j$, either $a_i \le b_j$ or $A_i \cap B_j = \emptyset$. Using that $\mathcal{F}$ is a $\sigma$-field and that $\mu$ is a measure on $\mathcal{F}$,

Taking the supremum over all possible $\psi \le \varphi$ finishes the proof.

$\square$

Combined with definition $\clubsuit$, we get

Corollary. Let $\varphi$ be a simple function and $\sum_{j \in [m] } b_j \mathbf{1}_{B_j}$ be one of its possible representations.

$$\small \int \varphi = \sum_{j \in [m] } b_j \mu (B_j).$$

We are ready to prove the linearity of integration for simple functions.

Linearity. The integral is linear on the vector space of simple functions.

Proof. If $\varphi$ is a simple function, clearly $\int c \cdot \varphi = c \cdot \int \varphi$ for all $c \in \mathbb{R}$. Let $\varphi = \sum_{i \in [n] } a_i \mathbf{1}_{ A_i }$ and $\psi = \sum_{j \in [m] } b_j \mathbf{1}_{B_j}$. It is left to verify that $\int \varphi + \psi = \int \varphi + \int \psi$. As both the $\{ A_i \in \mathcal{F} : i \in [n] \}$ and $\{ B_j \in \mathcal{F} : j \in [m] \}$ are mutually disjoint, so are $\{ A_i \cap B_j \in \mathcal{F} : i \in [n], j \in [m] \}$. Then

By rearranging the term on the right hand side, and by finite additivity of $\mu$, we get the desired result.

$\square$

Bounded Functions

Let $E \in \mathcal{F}$ be bounded with $\mu(E) \le \infty$. Let $f$ be a function whose support $\mathbb{Supp} (f) \subset E$ and suppose that $f$ is bounded on $E$, i.e., $\exists M > 0$, s.t., $\forall \omega \in E, | f (\omega) | \le M$. It holds that

where $\varphi$ ranges over all simple functions.

Lemma. Suppose $\mathbb{Supp}(f) \subset E$ and $f$ is bounded on $E$. Then $f$ is measurable if and only if

$$\sup_{\varphi \le f } \int \varphi = \inf_{ \psi \ge f } \int \psi$$

where $\varphi$ and $\psi$ ranges over all simple functions.

Proof. If $\sup_{\varphi \le f } \int_E \varphi = \inf_{ \psi \ge f } \int_E \psi$, then we can find a sequence of $\varphi_n, \psi_n$, such that $\int_E (\varphi_n - \psi_n) \rightarrow 0$. Let $\varphi = \sup_n \varphi_n$ and $\psi = \inf_n \psi_n$. It must be the case that $\varphi = \psi \, a.e.$. Otherwise, $\exists \epsilon > 0$ and a set $A \subset E$, such that $\varphi - \psi > \epsilon$ on $A$. As the integration for simple functions is monotone,

If $f$ is measurable, we need to prove that there are simple functions $\varphi \le f \le \psi$ whose integrals are arbitrary close. For all $\epsilon > 0$, we can find a finite collection of disjoint intervals $\{ (a_i, b_i] \}$ of length less than $\epsilon$, such that it covers $[-M, M]$. Define $E_i = f^{-1} \big( (a_i, b_i] \big)$ and let $\varphi \doteq \sum_i a_i \mathbf{1}_{ E_i }$ and $\psi \doteq \sum_i b_i \mathbf{1}_{ E_j }$. Then $\int_E \psi - \int_E \le \epsilon \cdot \mu(E)$. It concludes that $\sup_{\varphi \le f } \int_E \varphi \le \inf_{ \psi \ge f } \int_E \psi$. The other direction of the inequality is trivial.

$\square$

Linearity. The integral on $E \in \mathcal{F}$ with $\mu(E) \le \infty$ is linear on the vector space of bounded and measurable functions.

Proof. Let $f, g: E \rightarrow \mathbb{R}$ is measurable and bounded. Clearly $\int c \cdot f = c \cdot \int f$ for all $c \in \mathbb{R}$. It is left to verify that $\int f + g = \int f + \int g$. First,

where the second equality follows from that integral for simple functions are linear. Similarly,

$\square$

Remark: In the proof we use the fact that for all subset $A, B \subset \mathbb{R}$, it holds that $\sup_{a \in A, b \in B} (a + b) = \sup_{a \in A} a + \sup_{b \in B} b$.

Nonnegative Functions

Let $f : \Omega \rightarrow \mathbb{R}_{\ge 0}$ be a nonnegative function. Then $\int f = \sup_{0 \le h \le f} \int h$, where $h$ ranges over all nonnegative bounded functions with finite support. Formally,

It is easy to see that this definition is consistent with definition $\heartsuit$.

Linearity. The integral is linear on the vector space of positive functions.

Proof 1. Let $f, g: \Omega \rightarrow \mathbb{R}_{\ge 0}$ be nonnegative functions. Clearly $\int c \cdot f = c \cdot \int f$ for all $c \in \mathbb{R}$. It is left to verify that $\int f + g = \int f + \int g$. First,

where the second equality follows from that integral for bounded functions with finite support are linear. To prove the reverse direction, for each bounded function $0 \le h \le f + g$, let $h_1 = \min \{ f, h\}$ and $h_2 = h - h_1$. Clearly $0 \le h_1 \le f$. On the other hand, $0 \le h_2 = h - h_1 \le h - \min \{ f, h \} = \max \{ h - f, 0 \} \le g$. Hence,

Taking supremum over $h$ finishes the proof.

$\square$

Proof 2. There is another proof. If we examine the proof of monotone convergence theorem，we can see that it does not rely on linearity of integral. It turns out that we can rely on it to prove linearity of nonnegative functions. Let $\varphi_n \uparrow f$ and $\psi_n \uparrow g$ be two sequences of nonnegative simple functions. Using monotone convergence theorem, we get

Then using only linearity of integration for simple function, we get

$\square$

General Functions

Linearity. The integral is linear on the vector space of integrable general functions.

Proof. Let $f, g: \Omega \rightarrow \mathbb{R}$ be integrable functions.

For each $c \in \mathbb{R}_+$, via the definition of integration,

For each $c \in \mathbb{R}_-$, via the definition of integration,

It is left to verify that $\int f + g = \int f + \int g$. First,

It follows that $f + g$ is integrable. Observe

Suppose that $|f| < \infty$, and $|g| < \infty$, then clearly it holds that

Suppose that $f + g = \infty$. Then $0 \le (f + g)_- \le \infty$. And since $f + g$ is well defined, we must have $f(\omega) \neq -\infty$ and $g(\omega) \neq -\infty$. Therefore, $0 \le f_- \, , \, g_- \le \infty$. We can obtain

Suppose that $f + g = -\infty$. Then $0 \le (f + g)_+ \le \infty$. And since $f + g$ is well defined, we must have $f(\omega) \neq \infty$ and $g(\omega) \neq \infty$. Therefore, $0 \le f_+ \, , \, g_+ \le \infty$. We can obtain

In all cases, we have

Integrating for both side,

Rearranging terms finishes the proof.

$\square$

Reference

[1]. McMullen, C. "Real Analysis."

[2]. R. Durrett, “Probability : theory and examples,” p. 432.

[3]. 实变函数论，周民强.