[Measure Theory] The Lebesgue Integral
Definitions
Let $(X, \mathcal{A}, \mu)$ be a measure space.
- If $s = \sum_{i=1}^n a_i \chi_{E_i}$ is a non-negative measurable simple function, the Lebesgue integral of $s$ is defined as $$ \begin{aligned} \int s \; d\mu := \sum_{i=1}^n a_i \mu(E_i). \end{aligned} $$ If $a_i = 0$ and $\mu (E_i) = \infty$, we use the convention that $a_i \mu (E_i) = 0$.
- If $f \geq 0$ is measurable function, we define $$ \begin{aligned} \int f \; d\mu = \text{sup} \left\{ \int s \; d\mu : 0 \leq s \leq f, s \text{ simple }\right\} \end{aligned} $$
- Let $f$ be measurable and let $f^+ = \text{max}(f, 0)$ and $f^- = \text{max}(-f, 0)$. Provided $\int f^+ \; d\mu$ and $\int f^- \; d\mu$ are not both infinite, define $$ \begin{aligned} \int f \; d\mu = \int f^+ \; d\mu - \int f^- \; d\mu. \end{aligned} $$ Here, we often omit the $\mu$ and write $\int f$ if it is clear which measure is being used. Or, we often write $\int f(x) \; \mu (dx)$, or $\int f(x) \; d\mu(x)$
- If $f = u + iv$ is complex-valued and $\int (|u | + | v |) \; d\mu$ is finite, define $$ \begin{aligned} \int f \; d\mu = \int u \; d\mu + i \int v \; d\mu. \end{aligned} $$
- The integral $\int f \chi_A d\mu$ is often written $\int_A f \; d\mu$.
And when we are integrating a function $f$ with respect to Lebesgue measure $m$, it is usual to write $\int f(x) \; dx$ for $\int f(x) m (dx)$ and to define
\[\begin{aligned} \int_{a}^b f(x) \; dx = \int_{[a, b]} f(x) m (dx). \end{aligned}\]If $f$ is measurable and $\int | f | \; d\mu < \infty$, we say $f$ is integrable.
From the definitions, the following properties of Lebesgue integral are naturally derived:
- If $0 \leq a \leq f(x) \leq b$ for all $x$ and $\mu (X) < \infty$, then $$ \begin{aligned} a \mu(X) \leq \int f \; d\mu \leq b\mu(X); \end{aligned} $$
- If $f$ and $g$ are integrable, and $0 \leq f(x) \leq g(x)$ for all $x$, then $$ \begin{aligned} \int f \; d\mu \leq \int g \; d\mu; \end{aligned} $$
- If $f$ is integrable, non-negative, and $c$ is a non-negative real number, then $$ \begin{aligned} \int cf \; d\mu = c \int f \; d\mu; \end{aligned} $$
- If $\mu (A) = 0$ and $f$ is non-negative and measurable, then $$ \begin{aligned} \int f \chi_A \; d\mu = 0. \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
(1) Let $s_n = a \chi_X$. Then, it is obvious to see that $s_n$ is a simple function that $0 \leq s_n (x) \leq f(x)$ for all $x \in X$. Thus,
\[\begin{aligned} \int s \; d\mu = a \mu(X) \leq \text{sup} \{ \int s \; d\mu : 0 \leq g \leq f, $g$ \text{ simple } \} = \int f \; d\mu \end{aligned}\]The upper bound case can be shown in the analogous way.
(2) Let $s$ be any simple function such that $0 \leq s \leq f$. Since $s$ is also simple function with $0 \leq s \leq g$ for all $x$, it follows
\[\begin{aligned} \int s \; d\mu \leq \int g \; d\mu. \end{aligned}\]Since $s$ is arbitrary, by the definition of supremum:
\[\begin{aligned} \int f \; d\mu \leq \int g \; d\mu. \end{aligned}\](3) If $c = 0$, it is trivial.
For $c > 0$, let $s$ be any simple function that $0 \leq s \leq cf$ for all $x$. As $s / c$ is a simple function that $0 \leq s \ c \leq f$, $\int s / c \; d\mu \leq \int f \; d\mu$. But, from the definition of integral of simple function,
\[\begin{aligned} \int s / c \; d\mu = \frac{1}{c} \int s \; d\mu \leq \int f \; d \mu \end{aligned}\]so that $\int s \; d \mu \leq c \int f \; d \mu$. Thus, by definition of supremum we see that
\[\begin{aligned} \int cf \; d \mu \leq c \int f \; d \mu. \end{aligned}\]$\int cf \; d \mu \geq c \int f \; d \mu$ case can be proved in the analogous way.
(4) For any simple function $s$ with $0 \leq s \leq f$, if we write $s = \sum_{i=1}^n a_i \chi_{E_i} (x)$
\[\begin{aligned} \int_{A} s \; d\mu = \sum_{i=1}^n a_i \mu (E_i \cap A) \leq \sum_{i=1}^n a_i \mu (A) = 0 \end{aligned}\]Hence, supremum must be $0$.
\[\tag*{$\blacksquare$}\]Criteria for a function to be zero a.e.
Non-negative functions
$\color{red}{\mathbf{Proof.}}$
If not, there exists an $n$ such that $\mu (A_n) > 0$ where
\[\begin{aligned} A_n = \left\{ x \; \middle| \; f(x) > \frac{1}{n} \right\}. \end{aligned}\]since
\[\begin{aligned} \{ x \mid f(x) > 0 \} = \bigcup_{n=1}^\infty \left\{ x \; \middle| \; f(x) > \frac{1}{n} \right\}. \end{aligned}\]But since $f$ is non-negative,
\[\begin{aligned} 0 = \int f \geq \int_{A_n} f \geq \frac{1}{n} \mu (A_n), \end{aligned}\]a contradiction. \(\tag*{$\blacksquare$}\)
Integrable functions
$\color{red}{\mathbf{Proof.}}$
Let $A = f^{-1} ((\varepsilon, \infty))$ for any $\varepsilon > 0$. Then
\[\begin{aligned} 0 = \int_A f \geq \int_A \varepsilon = \varepsilon \mu (A) \end{aligned}\]since $f \chi_A \geq \varepsilon \chi_A$. Hence $\mu (A) = 0$. Then
\[\begin{aligned} \mu( \left\{ x \; \middle| \; f(x) > 0 \right\} ) & = \mu \left( \bigcup_{n=1}^\infty \left\{ x \; \middle| \; f(x) > \frac{1}{n} \right\} \right) \\ & \leq \sum_{n=1}^\infty \mu \left( \left\{ x \; \middle| \; f(x) > \frac{1}{n} \right\} \right) = 0. \end{aligned}\]Similarly, we can show that \(\mu \{ x \mid f(x) < 0 \} = 0\).
\[\tag*{$\blacksquare$}\]$\color{blue}{\mathbf{Proof.}}$
Step 1. $\int_G f = 0$ if $G$ is the finite union of disjoint intervals
For any interval $[c, d]$,
\[\begin{aligned} \int_c^d f = \int_a^d f - \int_a^c f = 0. \end{aligned}\]Therefore, if $G$ is the finite union of disjoint intervals, by linearity $\int_G f = 0$.
Step 2. $\int_G f = 0$ if $G$ is an open set
Also, as any open set $G \in \mathbb{R}$ can be written as the countable union of disjoint open interval, let $G = \bigcup_{i=1}^\infty (a_i, b_i)$.
Let $f_n = f \cdot \chi_{\bigcup_{i=1}^n (a_i, b_i)}$. Then,
\[\begin{aligned} | f_n | \leq | f | \text{ for all } n \quad \text{ and } \quad f_n \to f \cdot \chi_G \end{aligned}\]As $| f |$ is integrable, by the dominated convergence theorem,
\[\begin{aligned} 0 = \int_{\bigcup_{i=1}^n (a_i, b_i)} f = \int f_n \to \int f \cdot \chi_G = \int_G f \end{aligned}\]thus we have $\int_G f = 0$ for all open $G$.
Step 3. $\int_E f = 0$ for all Lebesgue measurable $E$
If $E$ is a Lebesgue measurable set, recall that there exists a decreasing sequence of open sets $G_n$ such that $E \subset G_{\delta} = \bigcap_{n=1}^\infty G_n$ and $m(G_{\delta} - E) = 0$. Since
\[\begin{aligned} | f \cdot \chi_{G_n} | \leq | f | \text{ for all } n \quad \text{ and } \quad f \cdot \chi_{G_n} \to f \cdot \chi_{G_{\delta}} \end{aligned}\]by the dominated convergence theorem, we get $\int f \cdot \chi_{G_{\delta}} = 0$:
\[\begin{aligned} 0 = \int_{G_n} f = \int f \cdot \chi_{G_n} \to \int f \cdot \chi_{G_{\delta}} \end{aligned}\]As $m(G_{\delta} - E) = 0$,
\[\begin{aligned} \int_E f = \int f \cdot \chi_E = \int f \cdot \chi_{G_{\delta}} = 0 \end{aligned}\]By the previous theorem, we conclude that $f = 0$ a.e.
\[\tag*{$\blacksquare$}\]Approximating a function
$\color{red}{\mathbf{Proof.}}$
Without loss of generality, we may assume $f$ is non-negative, i.e. $f \geq 0$. If we write $f = f^+ - f^-$, it is enough to find continuous functions $g_1$ and $g_2$ with compact support such that $\int | f^+ - g_1 | < \varepsilon / 2$ and $\int | f^- - g_2 | < \varepsilon / 2$ and to let $g = g_1 - g_2$.
Step 1. The existence of an integrable simple function $\phi$
Remind that by simple function approximation, there exists a sequence \(\{ \phi_n \}\) of non-negative measurable simple functions increasing to $f$. Since $| \phi_n - f | \leq 2 | f |$ and $f$ is integrable, by dominated convergence theorem we have
\[\begin{aligned} \int | \phi_n - f | < \varepsilon \end{aligned}\]for sufficiently large $n$.
For futher step, we show $\mu (E_j)$ is finite. Let $\phi_n = \sum_{j=1}^n a_j \chi_{E_j}$, where $E_j$ are pairwise disjoint and $a_j \neq 0$. Then
Step 2. When the measure is a Lebesgue-Stieltjes
As $\mu(E_j) < \infty$, we can approximate $E_j$ by finite union of open intervals $I_{j_k} = (a_{j_k}, b_{j_k})$ for $k = 1, \dots n_j$ with $$ \begin{aligned} \mu \left(E_j \; \Delta \; \left( \bigcup_{k=1}^{n_j} I_{j_k} \right) \right) < \frac{1}{n} \varepsilon \end{aligned} $$
$\color{#bf1100}{\mathbf{Proof.}}$
For Lebesgue measurable $E$ with finite measure, there exists an open set $A$ containing $E$ such that
\[\begin{aligned} \mu (A - E) \leq \frac{\varepsilon}{2} \end{aligned}\]Since $A$ is written as the countable union of open intervals $(a_j, b_j)$, we have
\[\begin{aligned} \mu (A) = \sum_{j=1}^\infty \mu ((a_j, b_j)) \leq \mu (E) + \frac{\varepsilon}{2} \end{aligned}\]Since $\mu (E) < \infty$, the sum is convergent and there exists an $N$ such that
\[\begin{aligned} \sum_{j = N}^\infty \mu (I_j) < \frac{\varepsilon}{2}. \end{aligned}\]Set $A = \bigcup_{j=1}^{N - 1} I_j$. It follows that $E - A \subset (\bigcup_{j=1}^\infty I_j) - A \subset \bigcup_{j = N}^\infty I_j$. Thus we have
\[\begin{aligned} \mu (E - A) \leq \mu \left( \bigcup_{j = N}^\infty I_j \right) < \frac{\varepsilon}{2}. \end{aligned}\]On the other hands, $A - E \subset \left(\bigcup_{j=1}^\infty I_j \right) - E$ so that
\[\begin{aligned} \mu ( A - E ) \leq \mu \left( \left(\bigcup_{j=1}^\infty I_j \right) - E \right) \end{aligned}\]From $E \subset \bigcup_{j=1}^\infty I_j$, we may split the above as follows
\[\begin{aligned} \mu \left( \left(\bigcup_{j=1}^\infty I_j \right) - E \right) = \mu \left( \bigcup_{j=1}^\infty I_j \right) - \mu (E) \leq \sum_{j=1}^\infty \mu (I_j) - \mu (E) \leq \mu (E) + \frac{\varepsilon}{2} - \mu (E) \end{aligned}\]and from $\mu (E) < \infty$ we obtain
\[\begin{aligned} \mu \left( \left(\bigcup_{j=1}^\infty I_j \right) - E \right) \leq \mu (E) + \frac{\varepsilon}{2} - \mu (E) = \frac{\varepsilon}{2} \end{aligned}\]Combining these two inequalities yields
\[\begin{aligned} \mu (E \; \Delta \; A) \leq \mu (E - A) + \mu (A - E) < \varepsilon. \end{aligned}\] \[\tag*{$\blacksquare$}\]In other words, we can approximate $\chi_{E_j}$ by $\sum_{k=1}^{n_j} \chi_{I_{j_k}}$:
\[\begin{aligned} \mu \left(E_j \; \Delta \; \left( \bigcup_{k=1}^{n_j} I_{j_k} \right) \right) = \int \left| \chi_{E_j} - \sum_{k=1}^{n_j} \chi_{I_{j_k}} \right| \; d\mu < \frac{1}{n} \varepsilon. \end{aligned}\]Finally, if $I = (a, b)$ we can approximate $\chi_I$ in the $L^1$-metric by continuous functions that vanish outside $(a, b)$. For example, for any given $\varepsilon > 0$, take $g$ to be the continuous function that equals $0$ on $(- \infty, a]$ and $[b, \infty)$, equals $1$ on $[a + \varepsilon, b - \varepsilon]$, and is linear on $[a, a + \varepsilon]$ and $[b - \varepsilon, b]$. Thus we can say that there exists a continuous $g$ such that $\int | \chi_I - g | < \varepsilon$.
Let’s putting these facts together. For $\phi_n = \sum_{j = 1}^n a_j \chi_{E_j}$, since $\chi_{E_j} = \sum_{k=1}^{n_j} \chi_{(a_{j_k}, b_{j_k})}$ where $(a_j, b_j)$ are disjoint, each $\chi_{E_j}$ can be approximated in $L^1$ metric by continuous functions that vanish outside $(a_{j_k}, b_{j_k})$, say $g_{(a_{j_k}, b_{j_k})} \geq \chi_{(a_{j_k}, b_{j_k})}$. Then $\phi_n$ can be approximated by $g = \sum_{j=1}^n a_j \sum_{k=1}^{n_j} g_{(a_{j_k}, b_{j_k})}$.
Therefore we have the following inequalities:
\[\begin{aligned} \int | \chi_{(a_{j_k}, b_{j_k})} - g_{(a_{j_k}, b_{j_k})} | < \frac{1}{n_j \cdot n} \varepsilon \end{aligned}\]hence
\[\begin{aligned} \int \left| \chi_{E_j} - \sum_{k=1}^{n_j} g_{(a_{j_k}, b_{j_k})} \right| & \leq \int \left| \chi_{E_j} - \sum_{k=1}^{n_j} \chi_{(a_{j_k}, b_{j_k})} \right| + \int \left| \sum_{k=1}^{n_j} \chi_{(a_{j_k}, b_{j_k})} - \sum_{k=1}^{n_j} \chi_{(a_{j_k}, b_{j_k})} \right| \\ & \leq \frac{1}{n} \varepsilon + \frac{1}{n} \varepsilon = \frac{2}{n} \varepsilon \end{aligned}\]And we approximate $\phi_n$:
\[\begin{aligned} \int | \phi_n - \sum_{j=1}^n a_j \sum_{k=1}^{n_j} g_{(a_{j_k}, b_{j_k})} | \leq \sum_{j=1}^n | a_j | \int | \chi_{E_j} - g_{(a_{j_k}, b_{j_k})} | < \sum_{j=1}^n | a_j | \frac{2}{n} \varepsilon \end{aligned}\]Since $\varepsilon > 0$ is arbitrary, it is fair to say $\int | \phi_n - g | < \varepsilon$. Finally we have
\[\begin{aligned} \int | f - g | \leq \int | f - \phi_n | + \int | \phi_n - g | < 2 \varepsilon. \end{aligned}\] \[\tag*{$\blacksquare$}\]Reference
[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
[3] Wikipedia, Measurable space
[4] Demystifying the Caratheodory Approach to Measurability
[5] Show that every point of cantor set is a limit point
[6] Wikiepedia, Pre-measure
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