The Lebesgue integral, vanishing lemmas and the Riemann integral

The Lebesgue integral is defined on a measure space and compared, on intervals of ${\mathbb{R},}$ with the Riemann integral.

Lebesgue integral#

The aim is a new definition of the integral. Its idea is drawn in Figure 1: first the integral of a simple function over a set ${E,}$ then simple functions placed below the graph of a function ${f.}$

Figure 1. On the left, a simple function, constant on each of the sets ${D_1,}$ $D_2$ and ${D_3,}$ with the region under its graph hatched, and a set $E$ over which that region is shaded. On the right, a function $f$ and, below its graph, the region under a simple function.

Nonnegative functions#

Definition 1 (Lebesgue integral of a simple function). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${s \colon X \to [0, +\infty]}$ be a simple function, written as in Definition 16 of the previous post,

$$s = \sum_{i=1}^{k} a_i \chi_{D_i},$$

and let ${E \in \mathcal{M}.}$ We define the Lebesgue integral of $s$ over $E$ as

$$\int_E s \, d\mu := \sum_{i=1}^{k} a_i \, \mu(E \cap D_i),$$

where ${+\infty \cdot 0 = 0,}$ as in Definition 5 of the first post.

Remark 2. Let ${(X, \mathcal{M}, \mu),}$ $E$ and $s$ be as in Definition 1. Then $\chi_E$ is simple, ${s \chi_E}$ is simple, and

$$\int_E s \, d\mu = \int_X s \chi_E \, d\mu.$$

Definition 3 (Lebesgue integral of a nonnegative function). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${f \colon X \to [0, +\infty]}$ be measurable and let ${E \in \mathcal{M}.}$ We define the Lebesgue integral of $f$ over $E$ as

$$\int_E f \, d\mu := \sup \Big\{ \int_E s \, d\mu : s \text{ simple}, \ 0 \le s \le f \Big\}.$$

Remark 4 (Basic properties). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${f_1, f_2 \colon X \to [0, +\infty]}$ be measurable, and let ${N, E, F \in \mathcal{M}.}$
(1) If ${\mu(N) = 0,}$ then ${\int_N f_1 \, d\mu = 0.}$
(2) If ${f_1 \le f_2,}$ then ${\int_E f_1 \, d\mu \le \int_E f_2 \, d\mu.}$
(3) If ${E \subset F,}$ then ${\int_E f_1 \, d\mu \le \int_F f_1 \, d\mu.}$
(4) If ${0 \le c \le +\infty,}$ then ${\int_E c f_1 \, d\mu = c \int_E f_1 \, d\mu.}$
(5) ${\int_E (f_1 + f_2) \, d\mu = \int_E f_1 \, d\mu + \int_E f_2 \, d\mu.}$
(6) If ${E \cap F = \emptyset,}$ then ${\int_E f_1 \, d\mu + \int_F f_1 \, d\mu = \int_{E \cup F} f_1 \, d\mu.}$
Properties (2) and (3) are called monotonicity, and properties (4) and (5) linearity.

Proposition 5 (Measure induced by simple functions). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${s \colon X \to [0, +\infty]}$ be simple, and let ${\nu \colon \mathcal{M} \to [0, +\infty]}$ be defined by

$$\nu(E) := \int_E s \, d\mu.$$

Then $\nu$ is a measure.

Proof (sketch). Only the $\sigma$-additivity of $\nu$ is proved. Let ${(E_m) \subset \mathcal{M}}$ be pairwise disjoint, let ${E = \bigcup_m E_m,}$ and write $s$ as in Definition 1, so that ${D_i \cap E = \bigcup_m (D_i \cap E_m)}$ for every ${i,}$ as in Figure 2. The fourth equality below is the $\sigma$⁠-additivity of ${\mu{:}}$

$$\begin{aligned} \nu(E) &= \int_E s \, d\mu = \sum_{i=1}^{k} a_i \, \mu(D_i \cap E) \\ &= \sum_{i=1}^{k} a_i \, \mu\Big( \bigcup_m (D_i \cap E_m) \Big) \\ &= \sum_{i=1}^{k} a_i \sum_{m=0}^{\infty} \mu(D_i \cap E_m) \\ &= \sum_{i=1}^{k} a_i \lim_{N \to \infty} \sum_{m=0}^{N} \mu(D_i \cap E_m) \\ &= \lim_{N \to \infty} \sum_{m=0}^{N} \sum_{i=1}^{k} a_i \, \mu(D_i \cap E_m) \\ &= \sum_{m=0}^{\infty} \sum_{i=1}^{k} a_i \, \mu(D_i \cap E_m) = \sum_{m=0}^{\infty} \int_{E_m} s \, d\mu, \end{aligned}$$

and ${\int_{E_m} s \, d\mu = \nu(E_m)}$ for every ${m.}$ $\blacksquare$

Figure 2. The drawing of the proof: a set $D_i$ and three of the pairwise disjoint sets ${E_m,}$ here ${E_1,}$ $E_2$ and ${E_3,}$ with the pieces ${D_i \cap E_m}$ shaded.

Remark 6. Let ${X = \mathbb{N},}$ ${\mathcal{M} = \mathcal{P}(\mathbb{N})}$ and ${\mu = \mu_\#,}$ the counting measure. In this case a function ${f \colon \mathbb{N} \to [0, +\infty]}$ is a sequence ${(f_m),}$ with ${f_m = f(m) \ge 0,}$ as in Figure 3, and

$$\int_{\mathbb{N}} f \, d\mu_\# = \sum_{m=0}^{\infty} f_m.$$

Figure 3. The drawing of Remark 6: the values ${f_0,}$ ${f_1,}$ $f_2$ and $f_3$ of a sequence, as bars over the points ${0,}$ ${1,}$ $2$ and $3$ of ${\mathbb{N}.}$

Remark 7. Let ${(X, \mathcal{M}, \mu)}$ be a complete measure space and let ${f, g \colon X \to [0, +\infty]}$ be measurable. If ${f = g}$ almost everywhere, then

$$\int_X f \, d\mu = \int_X g \, d\mu.$$

Proof (complete). Let ${N = \{ x \in X : f(x) \neq g(x) \},}$ so that ${\mu(N) = 0.}$ Then

$$\begin{aligned} \int_X f \, d\mu &= \int_{X \setminus N} f \, d\mu + \int_N f \, d\mu \\ &= \int_{X \setminus N} g \, d\mu \\ &= \int_{X \setminus N} g \, d\mu + \int_N g \, d\mu = \int_X g \, d\mu. \end{aligned}$$

The integrals over $N$ are $0$ by (1) of Remark 4. $\blacksquare$

Example 8. On $\mathbb{R}$ with the Lebesgue measure ${\lambda,}$ the function $\chi_{\mathbb{Q}}$ satisfies ${\chi_{\mathbb{Q}} = 0}$ almost everywhere. Hence ${\int_{\mathbb{R}} \chi_{\mathbb{Q}} \, d\lambda = 0}$ by Remark 7. The Riemann integral of this function does not exist.

Proposition 9 (Chebyshev inequality, or Markov inequality). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${f \colon X \to [0, +\infty]}$ be measurable and let ${c \in (0, +\infty).}$ Then

$$c \, \mu(\{ f > c \}) \le \int_{\{ f > c \}} f \, d\mu \le \int_X f \, d\mu,$$

where ${\{ f > c \} = \{ x \in X : f(x) > c \}.}$

Proof (complete). By monotonicity with respect to the domain of integration, (3) of Remark 4, and by monotonicity with respect to the function, (2) of Remark 4,

$$\int_X f \, d\mu \ge \int_{\{ f > c \}} f \, d\mu \ge \int_{\{ f > c \}} c \, d\mu = c \, \mu(\{ f > c \}). \tag*{$\blacksquare$}$$

From now on, the measure space ${(X, \mathcal{M}, \mu)}$ is assumed complete.

Corollary 10. Let ${(X, \mathcal{M}, \mu)}$ be a complete measure space, let ${f \colon X \to [0, +\infty]}$ be measurable and let ${E \in \mathcal{M}.}$
(1) ${\int_E f \, d\mu = 0}$ if and only if ${f = 0}$ almost everywhere on ${E.}$ This is the vanishing lemma for nonnegative functions.
(2) If ${\int_X f \, d\mu < +\infty,}$ then ${\mu(\{ f = +\infty \}) = 0.}$

Proof (sketch). Only (1) is proved. If ${f = 0}$ almost everywhere on ${E,}$ then ${\int_E f \, d\mu = 0}$ by Remark 7. Conversely, suppose that ${\int_E f \, d\mu = 0,}$ and for every ${m \in \mathbb{N}}$ with ${m \ge 1}$ let

$$E_m = \Big\{ x \in E : f(x) > \frac{1}{m} \Big\}.$$

Then

$$\bigcup_m E_m = \{ x \in E : f(x) > 0 \},$$

and the goal is to prove that this set has measure zero. By Proposition 9 with ${c = 1/m,}$ applied on ${E,}$

$$\mu(E_m) \le m \int_E f \, d\mu = 0,$$

so ${\mu(E_m) = 0}$ for every ${m.}$ Hence

$$\mu\Big( \bigcup_m E_m \Big) \le \sum_{m=1}^{\infty} \mu(E_m) = 0. \tag*{$\blacksquare$}$$

Real valued functions#

Definition 11 (Integrable function, Lebesgue integral). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f \colon X \to \mathbb{R}}$ be measurable.
(i) The function $f$ is said to be integrable if ${\int_X |f| \, d\mu < +\infty.}$

(ii) If $f$ is integrable, we define the Lebesgue integral of $f$ as the real number

$$\int_X f \, d\mu := \int_X f^+ \, d\mu - \int_X f^- \, d\mu.$$

(iii) If ${E \in \mathcal{M}}$ and $f$ is integrable, we define the real number

$$\int_E f \, d\mu := \int_X f \chi_E \, d\mu.$$

Figure 4. The idea of Definition 11: a function ${f,}$ its positive part ${f^+,}$ and its negative part ${f^-,}$ dashed, which is the part of $f$ below the axis reflected above it.

Notation. The integrable functions form the set

$$\mathcal{L}^1(X, \mathcal{M}, \mu) := \{ f \colon X \to \mathbb{R} : f \text{ integrable} \},$$

also written ${\mathcal{L}^1(X)}$ or ${\mathcal{L}^1.}$ The calligraphic $\mathcal{L}^1$ is different from ${L^1.}$

Remark 12. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f \colon X \to \mathbb{R}}$ be measurable.
(1) ${f \in \mathcal{L}^1}$ if and only if ${|f| \in \mathcal{L}^1,}$ if and only if $f^+$ and $f^-$ belong to ${\mathcal{L}^1.}$
(2) If ${f \in \mathcal{L}^1,}$ then ${\bigl| \int_X f \, d\mu \bigr| \le \int_X |f| \, d\mu.}$

Proof (sketch). Only (2) is proved:

$$\begin{aligned} \biggl| \int_X f \, d\mu \biggr| &= \biggl| \int_X f^+ \, d\mu - \int_X f^- \, d\mu \biggr| \\ &\le \biggl| \int_X f^+ \, d\mu \biggr| + \biggl| \int_X f^- \, d\mu \biggr| \\ &= \int_X f^+ \, d\mu + \int_X f^- \, d\mu = \int_X (f^+ + f^-) \, d\mu = \int_X |f| \, d\mu, \end{aligned}$$

since ${f^+ + f^- = |f|.}$ $\blacksquare$

Remark 13. The set ${\mathcal{L}^1(X, \mathcal{M}, \mu)}$ is a vector space, and the map

$$\begin{aligned} \int_X \cdot \, d\mu \colon \mathcal{L}^1 &\to \mathbb{R} \\ f &\mapsto \int_X f \, d\mu, \end{aligned}$$

is linear, that is,

$$\int_X (\alpha f + \beta g) \, d\mu = \alpha \int_X f \, d\mu + \beta \int_X g \, d\mu$$

for all ${\alpha, \beta \in \mathbb{R}}$ and ${f, g \in \mathcal{L}^1.}$

Proposition 14 (Vanishing lemma for real valued functions). Let ${(X, \mathcal{M}, \mu)}$ be a complete measure space and let ${f, g \in \mathcal{L}^1(X, \mathcal{M}, \mu).}$ The following are equivalent:
(1) ${f = g}$ almost everywhere;
(2) ${\int_X |f - g| \, d\mu = 0;}$
(3) ${\int_E f \, d\mu = \int_E g \, d\mu}$ for every ${E \in \mathcal{M}.}$
In particular, if ${\int_E f \, d\mu = 0}$ for every ${E \in \mathcal{M},}$ that is, if ${\int_X f \chi_E \, d\mu = 0}$ for every ${E \in \mathcal{M},}$ then ${f = 0}$ almost everywhere.

Proof (complete). If (1) holds, then (2) holds.

To prove that (2) implies ${(3),}$ let ${E \in \mathcal{M}.}$ Then

$$\begin{aligned} \biggl| \int_E f \, d\mu - \int_E g \, d\mu \biggr| &= \biggl| \int_E (f - g) \, d\mu \biggr| \\ &\le \int_E |f - g| \, d\mu \le \int_X |f - g| \, d\mu = 0, \end{aligned}$$

where the first inequality is (2) of Remark 12, the second is monotonicity and the last equality is the hypothesis (2).

To prove that (3) implies ${(1),}$ let ${E = \{ x \in X : f(x) \ge g(x) \}.}$ We prove that ${f = g}$ almost everywhere on ${E;}$ in the same way one gets ${f = g}$ almost everywhere on ${\{ f \le g \}.}$ By ${(3),}$

$$0 = \int_E f \, d\mu - \int_E g \, d\mu = \int_E (f - g) \, d\mu,$$

where ${f - g \ge 0}$ on ${E.}$ Hence ${f - g = 0}$ almost everywhere on $E$ by the vanishing lemma for nonnegative functions, (1) of Corollary 10. $\blacksquare$

Notation. The set of the continuous functions with compact support, the continuous functions from $\mathbb{R}^N$ to $\mathbb{R}$ whose support is compact, is denoted ${C_c(\mathbb{R}^N).}$

Theorem 15 (Vanishing for $\lambda$ with respect to other classes of functions). Let ${f \colon \mathbb{R}^N \to \mathbb{R}}$ with ${f \in \mathcal{L}^1(\mathbb{R}^N, \mathcal{L}(\mathbb{R}^N), \lambda).}$ If

$$\int_{\mathbb{R}^N} f g \, d\lambda = 0 \qquad \text{for every } g \in C_c(\mathbb{R}^N),$$

then ${f = 0}$ almost everywhere.

Notation. For an interval of ${\mathbb{R},}$ such as $(a, b)$ or ${[a, b],}$ the family of the sets of $\mathcal{L}(\mathbb{R})$ contained in it is written ${\mathcal{L}((a, b))}$ or ${\mathcal{L}([a, b]).}$ The set $C_c((a, b))$ is defined as $C_c(\mathbb{R}^N)$ with $(a, b)$ in place of ${\mathbb{R}^N,}$ the set $C^\infty((a, b))$ is the set of the infinitely differentiable functions on ${(a, b),}$ and

$$C_c^\infty(a, b) := C_c((a, b)) \cap C^\infty((a, b)).$$

Corollary 16 (Vanishing and constancy on intervals). Let ${f \in \mathcal{L}^1((a, b), \mathcal{L}((a, b)), \lambda).}$
(1) If ${\int_{(a, b)} f g \, d\lambda = 0}$ for every ${g \in C_c^\infty(a, b),}$ then ${f = 0}$ almost everywhere.
(2) If ${\int_{(a, b)} f g' \, d\lambda = 0}$ for every ${g \in C_c^\infty(a, b),}$ then there exists ${c \in \mathbb{R}}$ such that ${f = c}$ almost everywhere.
(3) If ${\int_{(a, b)} f g \, d\lambda = 0}$ for every ${g \in C_c^\infty(a, b)}$ such that ${\int_{(a, b)} g \, d\lambda = 0,}$ then there exists ${c \in \mathbb{R}}$ such that ${f = c}$ almost everywhere.

Lebesgue versus Riemann integral#

Notation. In this part, ${\int_{(a, b)} f \, d\lambda}$ denotes the Lebesgue integral and ${\int_a^b f \, dx}$ the Riemann integral.

Theorem 17 (Comparison of Lebesgue and Riemann integrals). (1) Let ${a, b \in \mathbb{R}}$ with ${a < b,}$ and let ${f \colon [a, b] \to \mathbb{R}}$ be Riemann integrable. Then ${f \in \mathcal{L}^1([a, b], \mathcal{L}([a, b]), \lambda)}$ and

$$\int_{[a, b]} f \, d\lambda = \int_a^b f \, dx.$$

(2) Let ${-\infty \le a < b \le +\infty,}$ and let ${f \colon (a, b) \to \mathbb{R}}$ be Riemann integrable on every closed bounded interval contained in ${(a, b),}$ with $|f|$ Riemann integrable in the generalized sense. Then ${f \in \mathcal{L}^1((a, b), \mathcal{L}((a, b)), \lambda)}$ and

$$\int_{(a, b)} f \, d\lambda = \int_a^b f \, dx,$$

where the right-hand side is the Riemann integral in the generalized sense.

The function $(\sin x)/x$#

Example 18. Let ${f \colon [\pi, +\infty) \to \mathbb{R}}$ be given by ${f(x) = (\sin x)/x.}$
(i) The function $f$ is Riemann integrable in the generalized sense. Integration by parts gives, for every ${R > \pi,}$

$$\begin{aligned} \int_\pi^R \frac{\sin x}{x} \, dx &= -\frac{\cos x}{x} \biggr|_{\pi}^{R} - \int_\pi^R \frac{-\cos x}{-x^2} \, dx \\ &= -\frac{\cos R}{R} - \frac{1}{\pi} - \int_\pi^R \frac{\cos x}{x^2} \, dx, \end{aligned}$$

where ${(\cos R)/R \to 0}$ as ${R \to +\infty}$ and ${|(\cos x)/x^2| \le 1/x^2.}$ Hence the limit

$$\lim_{R \to +\infty} \int_\pi^R \frac{\sin x}{x} \, dx$$

exists and is finite.

(ii) The function $f$ is not Lebesgue integrable on ${[\pi, +\infty).}$ In fact,

$$\int_{[\pi, +\infty)} \biggl| \frac{\sin x}{x} \biggr| \, d\lambda = +\infty.$$

Let ${N \in \mathbb{N}}$ with ${N \ge 1.}$ For ${1 \le m \le N,}$ on ${[m\pi, (m+1)\pi)}$ we have ${x \le (m+1)\pi;}$ moreover, the function $\lvert \sin x \rvert$ is ${\pi}$-periodic and ${\int_{[0, \pi]} \lvert \sin x \rvert \, d\lambda = 2.}$ Hence

$$\begin{aligned} \int_{[\pi, +\infty)} \biggl| \frac{\sin x}{x} \biggr| \, d\lambda &\ge \int_{[\pi, (N+1)\pi]} \biggl| \frac{\sin x}{x} \biggr| \, d\lambda \\ &= \sum_{m=1}^{N} \int_{[m\pi, (m+1)\pi)} \frac{\lvert \sin x \rvert}{x} \, d\lambda \\ &\ge \sum_{m=1}^{N} \frac{1}{(m+1)\pi} \int_{[m\pi, (m+1)\pi)} \lvert \sin x \rvert \, d\lambda \\ &= \sum_{m=1}^{N} \frac{1}{(m+1)\pi} \int_{[0, \pi]} \lvert \sin x \rvert \, d\lambda \\ &= \sum_{m=1}^{N} \frac{2}{\pi(m+1)}. \end{aligned}$$

This holds for every ${N \ge 1,}$ hence

$$\int_{[\pi, +\infty)} \biggl| \frac{\sin x}{x} \biggr| \, d\lambda \ge \sum_{m=1}^{\infty} \frac{2}{\pi(m+1)} = +\infty,$$

since the last series behaves like the harmonic series.

N

3

Figure 5. Above, the function $(\sin x)/x$ on ${[\pi, (N+1)\pi],}$ with the regions above and below the axis. Below, its absolute value and, under it on each ${[m\pi, (m+1)\pi),}$ the function ${\lvert \sin x \rvert/((m+1)\pi)}$ used for (ii), whose integral ${2/(\pi(m+1))}$ is the shaded region. The slider sets ${N.}$

From now on, ${\int_a^b f}$ and ${\int_a^b f \, d\lambda}$ denote Lebesgue integrals over intervals, and every integral is understood in the sense of Lebesgue.