Integration of Positive Functions

In this section, $\mathcal{M}$ will be a $\sigma$ -algebra in a set $X$ and $\mu$ will be a positive measure on $\mathcal{M}$ .

1.23 Definition If $s: X \to [0, \infty)$ is a measurable simple function, of the form

where $\alpha_1, \dots, \alpha_n$ are the distinct values of $s$ (compare Definition 1.16), and if $E \in \mathcal{M}$ , we define

The convention $0 \cdot \infty = 0$ is used here; it may happen that $\alpha_i = 0$ for some $i$ and that $\mu(A_i \cap E) = \infty$ .

If $f: X \to [0, \infty]$ is measurable, and $E \in \mathcal{M}$ , we define

the supremum being taken over all simple measurable functions $s$ such that $0 \le s \le f$ .

The left member of (3) is called the Lebesgue integral of $f$ over $E$ , with respect to the measure $\mu$ . It is a number in $[0, \infty]$ .

Observe that we apparently have two definitions for $\int_E f \, d\mu$ if $f$ is simple, namely, (2) and (3). However, these assign the same value to the integral, since $f$ is, in this case, the largest of the functions $s$ which occur on the right of (3).

1.24 The following propositions are immediate consequences of the definitions. The functions and sets occurring in them are assumed to be measurable:

(a) If $0 \le f \le g$ , then $\int_E f \, d\mu \le \int_E g \, d\mu$ . (b) If $A \subset B$ and $f \ge 0$ , then $\int_A f \, d\mu \le \int_B f \, d\mu$ .

(c) If $f \ge 0$ and $c$ is a constant, $0 \le c < \infty$ , then

(d) If $f(x) = 0$ for all $x \in E$ , then $\int_E f \, d\mu = 0$ , even if $\mu(E) = \infty$ .

(e) If $\mu(E) = 0$ , then $\int_E f \, d\mu = 0$ , even if $f(x) = \infty$ for every $x \in E$ .

(f) If $f \ge 0$ , then $\int_E f \, d\mu = \int_X \chi_E f \, d\mu$ .

This last result shows that we could have restricted our definition of integration to integrals over all of $X$ , without losing any generality. If we wanted to integrate over subsets, we could then use (f) as the definition. It is purely a matter of taste which definition is preferred.

One may also remark here that every measurable subset $E$ of a measure space $X$ is again a measure space, in a perfectly natural way: The new measurable sets are simply those measurable subsets of $X$ which lie in $E$ , and the measure is unchanged, except that its domain is restricted. This shows again that as soon as we have integration defined over every measure space, we automatically have it defined over every measurable subset of every measure space.

1.25 Proposition Let $s$ and $t$ be nonnegative measurable simple functions on $X$ . For $E \in \mathcal{M}$ , define

Then $\varphi$ is a measure on $\mathcal{M}$ . Also

(This proposition contains provisional forms of Theorems 1.27 and 1.29.)

PROOF If $s$ is as in Definition 1.23, and if $E_1, E_2, \dots$ are disjoint members of $\mathcal{M}$ whose union is $E$ , the countable additivity of $\mu$ shows that

Also, $\varphi(\emptyset) = 0$ , so that $\varphi$ is not identically $\infty$ .

Next, let $s$ be as before, let $\beta_1, \dots, \beta_m$ be the distinct values of $t$ , and let $B_j = \{x: t(x) = \beta_j\}$ . If $E_{ij} = A_i \cap B_j$ , then

and

Thus (2) holds with $E_{ij}$ in place of $X$ . Since $X$ is the disjoint union of the sets $E_{ij}$ ( $1 \le i \le n$ , $1 \le j \le m$ ), the first half of our proposition implies that (2) holds. ////

We now come to the interesting part of the theory. One of its most remarkable features is the ease with which it handles limit operations.

1.26 Lebesgue’s Monotone Convergence Theorem Let $\{f_n\}$ be a sequence of measurable functions on $X$ , and suppose that

(a) $0 \le f_1(x) \le f_2(x) \le \dots \le \infty$ for every $x \in X$ , (b) $f_n(x) \to f(x)$ as $n \to \infty$ , for every $x \in X$ .

Then $f$ is measurable, and

PROOF Since $\int f_n \le \int f_{n+1}$ , there exists an $\alpha \in [0, \infty]$ such that

By Theorem 1.14, $f$ is measurable. Since $f_n \le f$ , we have $\int f_n \le \int f$ for every $n$ , so (1) implies

Let $s$ be any simple measurable function such that $0 \le s \le f$ , let $c$ be a constant, $0 < c < 1$ , and define

Each $E_n$ is measurable, $E_1 \subset E_2 \subset E_3 \subset \dots$ , and $X = \bigcup E_n$ . To see this equality, consider some $x \in X$ . If $f(x) = 0$ , then $x \in E_1$ ; if $f(x) > 0$ , then $cs(x) < f(x)$ , since $c < 1$ ; hence $x \in E_n$ for some $n$ . Also

Let $n \to \infty$ , applying Proposition 1.25 and Theorem 1.19(d) to the last integral in (4). The result is

Since (5) holds for every $c < 1$ , we have

for every simple measurable $s$ satisfying $0 \le s \le f$ , so that

The theorem follows from (1), (2), and (7).

1.27 Theorem If $f_n: X \to [0, \infty]$ is measurable, for $n = 1, 2, 3, \dots$ , and

then

PROOF First, there are sequences $\{s_i'\}$ , $\{s_i''\}$ of simple measurable functions such that $s_i' \to f_1$ and $s_i'' \to f_2$ , as in Theorem 1.17. If $s_i = s_i' + s_i''$ , then $s_i \to f_1 + f_2$ , and the monotone convergence theorem, combined with Proposition 1.25, shows that

Next, put $g_N = f_1 + \dots + f_N$ . The sequence $\{g_N\}$ converges monotonically to $f$ , and if we apply induction to (3) we see that

Applying the monotone convergence theorem once more, we obtain (2), and the proof is complete.

If we let $\mu$ be the counting measure on a countable set, Theorem 1.27 is a statement about double series of nonnegative real numbers (which can of course be proved by more elementary means):

Corollary If $a_{ij} \ge 0$ for $i$ and $j = 1, 2, 3, \dots$ , then

1.28 Fatou’s Lemma If $f_n: X \to [0, \infty]$ is measurable, for each positive integer $n$ , then

Strict inequality can occur in (1); see Exercise 8.

PROOF Put

Then $g_k \le f_k$ , so that

Also, $0 \le g_1 \le g_2 \le \dots$ , each $g_k$ is measurable, by Theorem 1.14, and $g_k(x) \to \liminf f_n(x)$ as $k \to \infty$ , by Definition 1.13. The monotone convergence theorem shows therefore that the left side of (3) tends to the left side of (1), as $k \to \infty$ . Hence (1) follows from (3). ////

1.29 Theorem Suppose $f: X \to [0, \infty]$ is measurable, and

Then $\varphi$ is a measure on $\mathfrak{M}$ , and

for every measurable $g$ on $X$ with range in $[0, \infty]$ .

PROOF Let $E_1, E_2, E_3, \dots$ be disjoint members of $\mathfrak{M}$ whose union is $E$ . Observe that

and that

It now follows from Theorem 1.27 that

Since $\varphi(\emptyset) = 0$ , (5) proves that $\varphi$ is a measure.

Next, (1) shows that (2) holds whenever $g = \chi_E$ for some $E \in \mathcal{M}$ . Hence (2) holds for every simple measurable function $g$ , and the general case follows from the monotone convergence theorem. ////

Remark The second assertion of Theorem 1.29 is sometimes written in the form

We assign no independent meaning to the symbols $d\varphi$ and $d\mu$ ; (6) merely means that (2) holds for every measurable $g \ge 0$ .

Theorem 1.29 has a very important converse, the Radon-Nikodym theorem, which will be proved in Chap. 6.