The Integration Theory of Lebesgue

--- tags: analysis, measure theory --- # The Integration Theory of Lebesgue [TOC] ## Properties of Measurable Functions **Def** Let $(X, \mathcal m)$ be a measurable set. An exntended real valued function $f: D =\text{Dom}(f) \rightarrow \overline{\mathbb R}$ is ***measurable with respect to*** $\mathcal m$ if (1) $D \in \mathcal m$ (2) $f$ is $(\mathcal m, \mathscr{B(\overline{\mathbb R})}$ )-measurable. Proposition $(X, \mathcal m)$: measuarable space A function $f: D \rightarrow \overline {\mathbb R}$ is a measurable with respect to $\mathcal m$ if and only if $D \in \mathcal m$ and $\{ x \in D | f(x) \le a \} = f^{-1}([-\infty, a]) \in \mathcal m$ for every $a \in \mathbb R$. This also holds when the interval is right-opened. *proof.* (sketch) We only need to show the fact that $$ \mathscr B(\overline {\mathbb R}) = \mathscr A(\{ [-\infty, a) | a \in \mathbb R\}) $$ It's can be done by proving that $\{[ -\infty, a)| a\ \in \mathbb R\}$ can generate all the connected open interval $(x, y), (x,y \in \overline{\mathbb R})$. Then, using the fact that any open set in $\overline {\mathbb R}$ is a countable union of disjoint open interval to complete the proof. Example Let $(X, \mathcal m)$ be a measurable space and $f, g$ be two measurable function with respect to $\mathcal m$. Show that $f + g$ is also measurable with respect to $\mathcal m$. *proof*. By the proposition, we only need to check $S_a = \{ x \in D | (f + g)(x) < a \}$ is measurable for $a \in \overline {\mathbb R}$. Since $(f + g)(x) < a$ is equivalent to $f(x) < g(x) - a$ and thus $\exists q \in \mathbb Q$ such that $f(x) < q < g(x) - a$, we have $$ S_a = \bigcup_{q \in \mathbb Q} \left( \{ x | f(x) < q\} \cap \{x | g(x) > q + a\} \right) $$ ,which is countable union of measurable sets and is thus also measurable. Example Define $$ (\sup_n f_n)(x) := \sup_n f_n(x), $$ $$ (\inf_n f_n)(x) := \inf_n f_n(x) $$ and $$ \text{Dom} (\sup_n f_n) = \text{Dom} (\inf_n f_n) = \cap_n \text{Dom}(f_n) $$ Let $f_n: \text{Dom}(f_n) \rightarrow \overline{\mathbb R} (n \in \mathbb N)$ be a sequence of functions which are measurable with respect to $\mathcal m$. Then, if $f_n (n \in \mathbb N)$ is measurable with respect to $\mathcal m$, then $\sup f_n, \inf f_n, \overline{\lim}_n f_n, \underline{\lim}_n f_n$ are all measurable with respect to $\mathcal m$. *proof* We only show that $\sup f_n$ and $\overline{\lim}_n f_n(x)$ is measurable with respect to $\mathcal m$. For any $a \in \mathbb R$, we have $$ \begin{align*} & \left\{ x \in \text{Dom}(\sup f_n) \middle| \sup_n f_n(x) \le a \right\} \\ = \ & \left\{ x \in \text{Dom}(\sup f_n) \middle| \forall n \in \mathbb N, f_n(x) \le a \right\} \\ = \ & \bigcap_{n}^{\infty} \left\{ x \in \text{Dom}(f_n) \middle| f_n(x) \le a \right\} \in \mathcal m \end{align*} $$ The proof is done by applying the proposition. For $\overline{\lim}_n f_n(x)$, we have $$ \overline{\lim}_n f_n = \inf_m \sup_{n \ge m} f_n $$ and we have proved that $sup_{n \ge m} f_n$ is also measurable with respect to $\mathcal m$. Therefore, $\inf_m \sup_{n \ge m} f_n$ is also measurable with respect to $\mathcal m$. Remark Measurable function的良好性質：經過許多種操作之後仍為measurable。 ## Construction of Lebesgue Integral **Def** For any $c \in \overline{\mathbb R}$, we define $c^+ := \max\{c, 0 \}$ and $c^- := \max\{-c, 0\}$, Remark * $c^+$ and $c^-$ cannot be $\infty$ simultaneously. * $c = c^+ - c^-$, $|c| = c^+ + c^-$. * If $c = a - b$ for some $a, b \in \overline{\mathbb R}_+$ then $c^+ \le a$ and $c^- \le b$. We can extend this definition to a extended real-valued function $f$ and the with the properties above preserved. Moreover, we have $f = f^+ - f^-$. We use this definition to simplify some discussons (e.g. avoid $\infty - \infty$ when integrating). Note that since $f^+ = \max\{f, 0\}$, we can use the previous example to show that $f^\pm$ is measurable with respect to $\mathcal m$ if $f$ is measurable with respect to $\mathcal m$. For $E \in \mathcal m$ and $f: D \rightarrow \overline{\mathbb R}$ measurable with respect to $\mathcal m$, we want to discuss "$\int_E f d\mu$" (sometimes written as $\int_E f(x) d \mu(x)$ or $\int_E f(x) \mu(dx)$). Now, we consider the integration of $f: D \rightarrow \overline{\mathbb R}_+ = [0, \infty]$ which are measurable with respect to $\mathcal m$ on $E$. **Def** A function or map is ***simple*** if it takes only finitely many values (Its shape is just like steps). If $\alpha_1, \cdots, \alpha_n$ are the distinct values of a simple function $s$ and if we set $A_i = \{x: s(x) = \alpha_i \}$, then clearly $$ s = \sum_{i=1}^n \alpha_i \chi_{A_i} $$ , where $\chi_{A_i}$ (the characteristic function of $A_i$) is defined as $$ \chi_E(x) = \begin{cases} 1 & \text{ if } x \in E \\ 0 & \text{ if } x \notin E \end{cases} $$ For a measure space $(X, \mathcal m, \mu)$ (fixed if no particular mention). we aim to defining $$ \int_E f(x) d\mu(x) = \int_E f(x) \mu(dx) $$ (or $\int_E f d\mu$ for short) for suitable measurable functions $f:X \rightarrow \overline{\mathbb R} = [-\infty, \infty]$ and for any $E \in \mathcal m$ in a systemetic manner. Steps: 1. Reduction to the $[0, \infty]$-valued case. We do this using $f = f^+ - f^-$ where $f^\pm = \max\{ \pm f, 0 \}$. ($f^\pm$ is measurable if $f$ is measurable.) 2. We say that $\int_X f d\mu$ is defined if $\int_X f^+ d\mu$ and $\int_X f^- d\mu$ have already been defined (as elements in $\overline{\mathbb R}$) and are not $\infty$ simultaneously. If this is the case, $\int_X f d\mu := \int_X f^+ d\mu - \int_X f d\mu \in \overline{\mathbb R}$. If furthermore $\int_X f d \mu \in \mathbb R$, we say that $f$ is ***$\mu$-integrable*** (The whole space $X$ can be replaced by a measurable set $E$, and we say that $f$ is $\mu$-integrable on $E$ if $\int_E f d\mu \in \mathbb R$). 3. Reduction to the case of measurable $[0, \infty)$-valued simple functions (those which takes only finitely many values). For any measurable function $f: X \rightarrow [0, \infty]$, we define $$ \int_E f d\mu := \sup \left\{ \int_E s(x) d\mu(x) \middle | 0 \le s \le f, s: \text{simple}, [0, \infty) \text{-valued, measurable with respect to } \mathcal m \right\} $$ , where all the distinct value of $s$ is denoted by $\{\alpha_1, \cdots \alpha_k \} \subseteq [0, \infty)$ and the integral $\int_E s(x) d\mu (x)$ is defined as $$ \int_E s d\mu = \sum_{i = 1}^k \alpha_i \mu(A_i \cap E) $$ where $$ A_i = s^{-1}(\alpha_i) \forall i \in \{1, \ldots, k\} $$ ## Basic Properties of Lebesgue Integral **Proposition** Let $f, g: X \rightarrow [0, \infty]$ be measurable functions, $A, B \in \mathcal m$ ,and $c \in [0, \infty]$, (1) ***monotonicity*** $f \le g \Rightarrow \int_A f d\mu \le \int_A g d\mu$ and $A \in B \Rightarrow \int_A f d\mu \le \int_B f d\mu$. (2) ***positivie homogeneity*** $\int_A c f d\mu = c \int_A f d\mu$. (Check the case $c = \infty$) (3) ***zeroness*** $f = 0 \Rightarrow \int_A f d \mu = 0$ (even if $\mu(A) = \infty$) and $\mu(A) = 0 \Rightarrow \int_A f d\mu = 0$ (even if $f = \infty$). (4) $\int_A f d \mu = \int_X \chi_A \cdot f d\mu$ Question: Does the additivity hold? That is, for $A \cap B = \phi$, does $\int_{A \cup B} f d\,u$ equals to $\int_A fd\mu + \int_B f d\mu$? Moreover, is $\int_A (f + g) d \mu$ equal to $\int_A f d\mu + \int_A g d\mu$? **Lemma** Let $s, t: X \rightarrow [0, \infty)$ be two measurable simple functions. Consider $\nu: \mathcal m \rightarrow [0, \infty]$ defined by $$ A \overset{\nu}\mapsto \int_A s d\mu $$ Then (1) $\nu$ is a positive measure. (2) $\int_X (s+t) d\mu = \int_X s d\mu + \int_X t d\mu$. *proof* (1) Suppose that $A_n \in \mathcal m (n \in \mathbb N)$ be a disjoint family. We can express $s$ as $$ s = \sum_{\alpha \in [0, \infty)} \alpha \cdot \chi_{_{s^{-1}(\alpha)}} $$ Then $$ \begin{align*} \nu\left( \bigcup_{n=1}^\infty A_n \right) &= \int_{\bigcup_{n=1}^\infty A_n} s d\mu \\ &= \int_{\bigcup_{n=1}^\infty A_n} \left( \underbrace{\sum_{\alpha \in [0, \infty)} \alpha \chi_{s^{-1}(\alpha)}}_{\text{finite sum}} d\mu \right) \end{align*} $$ Since the summation is a sum of finite terms, we can rearrange the expression as$$ \sum_{\alpha \in [0, \infty)} \int_{\bigcup_{n=1}^\infty A_n} \alpha \chi_{s^{-1}(\alpha}) d\mu = \sum_{\alpha \in [0, \infty)} \alpha \cdot \int_{\bigcup_{n=1}^\infty A_n} \chi_{s^{-1}(\alpha}) d\mu $$ Since $s$ is simple and measurable, the integral inside the summation is well defined and thus the expression is equal to $$ \begin{align*} \sum_{\alpha \in [0, \infty)} \alpha \cdot \mu \left( s^{-1}({\alpha}) \cap \bigcup_{n=1}^\infty A_n ) \right) =& \sum_{\alpha\in[0,\infty)} \alpha \cdot \sum_{n=1}^\infty \mu\left( s^{-1}(\alpha) \cap A_n \right) \\=& \sum_{n=1}^\infty \sum_{\alpha\in[0,\infty)} \alpha \mu \left( s^{-1}(\alpha) \cap A_n \right) \\ =& \sum_{n=1}^\infty \int_{A_n} s d\mu = \sum_{n=1}^\infty \nu(A_n) \end{align*} $$ Besides, it's obvious that $\nu(\phi) = 0$. Therefore, we have proved that $nu$ is a positive measure. (2) From (1) we can write $$ \int_X (s+t) d\mu = \nu_{s+t} (X) $$ , where $\nu_{s+t}$ is a positive measure. The expression above is also equal to the following (actually) finite summation: $$ \begin{align*} \sum_{\alpha, \beta \in [0, \infty)} \nu_{s+t} (s^{-1}(\alpha) \cap t^{-1}(\beta)) =& \sum_{\alpha, \beta \in [0, \infty)} \int_{s^{-1}(\alpha) \cap t^{-1}(\beta)} (s+t) d\mu \\ =& \sum_{\alpha, \beta \in [0, \infty)} \int_X \chi_{s^{-1}(\alpha) \cap t^{-1}(\beta)}(s+t) d\mu \\ =& \sum_{\alpha, \beta \in [0, \infty)} \int_X \chi_{s^{-1}(\alpha) \cap t^{-1}(\beta)}(\alpha + \beta) d\mu \\ =& \sum_{\alpha, \beta \in [0, \infty)} (\alpha + \beta) \nu(s^{-1}(\alpha) \cap t^{-1}(\beta)) \\ =& \sum_\alpha \alpha \sum_\beta \mu(s^{-1}(\alpha) \cap t^{-1} (\beta)) + \sum_\beta \beta \sum_\alpha \mu(s^{-1}(\alpha) \cap t^{-1} (\beta)) \\ =& \sum_{\alpha} \alpha \cdot \mu(s^{-1} (\alpha)) +\sum_{\beta} \beta \cdot \mu(s^{-1} (\beta) ) \\ =& \nu_s(X) + \nu_t(X) = \int_X s d\mu + \int_X t d\mu \end{align*} $$ ### Lebesgue's monotone convergence theorem **Theorem** Lebesgue's monotone convergence theorem Let $f_n: X \rightarrow [0, \infty] (n \in \mathbb N)$ be a sequence of measurable functions such that $$ \forall x \in X, f_n(x) \uparrow \text{ as } n \rightarrow \infty $$ (We have shown that $\lim_{n\rightarrow \infty} f_n$ is measurable). Then we have $$ \int_X (\lim_{n \rightarrow \infty} f_n) d\mu = \lim_{n \rightarrow \infty} \int_X f_n d\mu $$ *proof* Let $f := \lim_{n \rightarrow \infty} f_n$. Since $f_n \le f_{n+1} \le f (n \in \mathbb N)$ by the monotonicity, we have $$ \alpha = \lim_{n\rightarrow \infty} \int_X f_n d\mu \le \int_X f d\mu $$ , where $\alpha \in \overline{\mathbb R}_+$. It remains to show that $\alpha \ge \int_X f d\mu$. By the definition of $\int_X f d\mu$, it suffices to prove that $$ \alpha \ge \int_X s d\mu $$ for all measurable $[0, \infty)$-valued simple function $s \le f$. Now we fix $s(x)$. Let $c \in (0, 1)$ and consider $E_n := \left\{ x \in X \middle | f_n(x) \ge c \cdot s(x) \right\}, n \in \mathbb N$. Obviously $E_n \uparrow \cup_{n=1}^\infty E_n$. > Claim: $\bigcup_{n=1}^\infty E_n = X$ > > *proof* > > We divide $x \in X$ into two cases and show that $x \in \bigcup_{n=1}^\infty E_n$ for both cases > > case(1) $f(x) = 0$ > > Obviously, $x \in \bigcup_{n=1}^\infty E_n$. > > case(2) $f(x) > 0$ > > Since $c < 1$ and $f_n \uparrow f$, $\exists k \in \mathbb N$ such that $f_k(x) \ge c \cdot s(x)$ and thus $x \in E_k$. Therefore, $$ \int_X f_n d\mu \ge \int_{E_n} f_n d\mu \ge \int_{E_n} c \cdot s d\mu = c \int_{E_n} s d\mu $$ Since $\int_{E} sd \mu$ is a positvie measure by previous lemma and $E_n$ monotonically converges to $X$, we can use the [property of measurable space](https://hackmd.io/UdqdcU29RL6DHbuAtCrlZw#Basic-Properties-of-Measurable-Spaces) to get $$ \lim_{n \rightarrow \infty} \int_{E_n} s d\mu = \int_{X} s d\mu $$ Therefore, we have $$ \alpha = \lim_{n\rightarrow \infty} \int_{E_n} f_n d\mu \ge c \cdot \lim_{n \rightarrow \infty} \int_{E_n} s d\mu = c\int_{X} s d\mu, \forall c \in (0, 1) $$ Let $c \rightarrow 1$ then we obtain $$ \alpha \ge \int_X s d \mu $$ for any $s$. Example Approximating Measurable Arbitrary Measurable Function by Simple Functions For any measurable function $f: X \rightarrow [0, \infty]$, we let $$ \begin{align*} s_n(x) := \begin{cases} \frac{k-1}{2^n} & \text{if } \frac{k-1}{2^n} \le f(x) < \frac{k}{2^n}, k \in \{1,2,...,n \cdot 2^n\} \\ n & \text{if } f(x) \ge n \end{cases} \end{align*} $$ Then, * $s_n: X \rightarrow [0, \infty)$ is a measurable simple function. * $s_n \uparrow f$ as $n \rightarrow \infty$. **Proposition** (generalization of previous lemma) For any measurable functions $f, g: X \rightarrow [0, \infty]$, we have $$ \int_X (f+g) d\mu = \int_X f d\mu + \int_X g d\mu $$ *proof* We use the same rule in previous example to construct $s^f_n \uparrow f$ and $s_n^g \uparrow g$ as $n \rightarrow \infty$. Then, we apply Lebesgue’s monotone convergence theorem to $\int_X (s_n^f + s_n^g) d\mu = \int_X s_n^f d\mu + \int_X s_n^g d\mu$, which complete the proof. Corollary Let $g_m: X \rightarrow [0, \infty] (m \in \mathbb N)$ be measurable. Then $$ \sum_{m=1}^\infty g_m $$ is measurable and $$ \int_X \left( \sum_{m=1}^\infty d\mu\right) = \sum_{m=1}^\infty \int_X g_m d\mu $$ *proof* Let $f_n = \sum_{m=1}^n g_m (n \in \mathbb N)$, then $f_n \uparrow \sum_{m=1}^\infty g_m$. The proposition tells that $$ \int_X f_n = \int_X \left(\sum_{m=1}^n g_m \right)d \mu = \sum_{m=1}^n \int_X g_m d\mu $$ Then, applying Lebesgue’s monotone convergence theorem completes the proof. From this corollary we can prove that for $$ a_{i,j} \in [0, \infty], (i, j) \in \mathbb N^2 $$ , we have $$ \sum_i \sum_j a_{i,j} = \sum_j \sum_i a_{i, j} $$ and then use counting measure as the measure (or refer to [this](https://math.stackexchange.com/questions/466757/why-can-we-interchange-summations)). ### Use Integration to Construct Measure **Theorem** (A generalization of previous lemma) Let $f: X \rightarrow [0, \infty]$ be a measurable function and $\nu: m \rightarrow [0, \infty]$ define by $$ A \mapsto \int_A f d\mu $$ 1. Then $\nu$ is a positive measure on $(X, \mathcal m)$ 2. for any measurable function $g: X \rightarrow [0, \infty]$, we have $$ \int_X g d\nu = \int_X g\cdot f d\mu $$ *proof* Let $A_n \in m (n \in \mathbb N)$, we want to show that $$ \nu(\bigcup_{n=1}^\infty A_n) = \sum_{n=1}^\infty \nu(A_n) $$ By definition of $\nu$, $$ \begin{align} \nu(\bigcup_{n=1}^\infty A_n ) &= \int_{\bigcup_{n=1}^\infty A_n} f d\nu \\ &= \int_X \chi_{\bigcup_{n=1}^\infty A_n} f d\mu \\ &= \int_X \sum_{n-1}^\infty \chi_{A_n} f d\mu \end{align} $$ Apply previous corollary, the expression above is equal to $$ \sum_{n=1}^\infty \int_X \chi_{A_n} \cdot f d\mu = \sum_{n=1}^\infty \int_{A_n} f d\mu $$ Thus, we have completed the first part of the proof. To prove the second part, we construct $s_n^g \uparrow g$ as before. We have $$ \begin{align} \int_X g d\nu &= \int_X \left( \lim_{n \rightarrow \infty} s_n^g \right) d\nu \\ &= \lim_{n\rightarrow \infty}\int_X s_n^g d \nu \end{align} $$ On the other hand $$ \begin{align} \int g\cdot f d\mu &= \int_X \lim_{n \rightarrow \infty} (s_n^g \cdot f) d\mu \\ &= \lim_{n \rightarrow \infty} \int_X s_n^g \cdot f d \mu \end{align} $$ and $s_n \cdot f \uparrow g \cdot f$ as $n \rightarrow \infty$. Therefore, it suffices to show that $\int_X s_n^g d\nu = \int_X s_n^g\cdot f d\mu$. For any $A \in \mathcal m$, we have $$ \int_X \chi_A d\nu = \nu(A) = \int_A f d\mu = \int_X \chi_A \cdot f d\mu $$ Since any simple function is a linear combination of characteristic function, the equality above completes the proof of the second part. ### Fatou's Lemma **Fatou's Lemma** Let $f_n: X \rightarrow [0, \infty] (n \in \mathbb N)$ be measurable. Then $$ \int_X \left( \liminf_n f_n \right) d\mu \le \liminf_n \int_X f_n d\mu $$ *proof* Note that $\liminf_k f_k = \sup_n (\inf_{m\ge n} f_m) := g_n$ ($\ge f_m$ if $m\ge n$). To be completed... Note that $f$ is $\mu$-integrable if and only if $|f|$ is $\mu$-integrable. ## Properties of Measurable Complex-Valued Functions **Def** $f: X \rightarrow \mathbb C$, where $u :=\operatorname{Re}(f)$ and $v := \operatorname{Im}(f)$ (both of which are measurable). Define 1. $f$ is $\mu$-integrable if $u$ and $v$ are $\mu$-integrable or, equivalently, $|f|$ is $\mu$-integrable ($|u|, |v| \le |f| \le |u| + |v|$). If $f$ is $\mu$-integrale, we define $$ \int_X f d\mu := \int_X u d\mu + i \int_X v d\mu $$ 2. $\mathcal L^p (\mu) := \left\{ f: X\rightarrow \mathbb C \middle| | f |^p \text{ is } \mu \text{-integrable} \right\}$ and $\mathcal L^p (\mu)_{\mathbb R} := \left\{ f: X\rightarrow \mathbb R \middle| | f |^p \text{ is } \mu \text{-integrable} \right\}$ **Proposition** If $f, g \in \mathcal L^1(\mu)$ and $\alpha, \beta \in \mathbb C$, then $\alpha \cdot f + \beta \cdot g \in \mathcal L^1 (\mu)$ and $\int_X (\alpha \cdot f + \beta \cdot g) d\mu = \alpha \int_X f d\mu + \beta \int_X g d\mu$. *proof* First we need to show that $\alpha \cdot f + \beta \cdot g$ is integrable: $$ \begin{align} \int_X |\alpha f + \beta g| d\mu &\le \int_X \left( |\alpha| |f| + |\beta||g| \right) d\mu \\ &\le |\alpha| \int_X |f| d\mu + |\beta| \int_X |g| d\mu \end{align} $$ It suffices to show that $\int_X (f + g) d\mu = \int_X f d\mu + \int_X g d \mu$ if $f, g \in \mathcal L^1 (X)_{\mathbb R}$ and $\int_X (\alpha f) d\mu = \alpha \int_X f d\mu$ if $f\in \mathcal L^1(X)_{\mathbb R}$ and $\alpha \in \mathbb R$. ## Lebesgue's Dominant Convergence Theorem Let $f_n \in \mathcal L^1 (\mu)_{\mathbb R} (n \in \mathbb N)$ be a sequence of $\mu$-integrable functions which converge pointwise to a function $f$. Suppose that there exists $g \in \mathcal L^1 (\mu)$ such that $\forall n \in \mathbb N, | f_n | \le g$. Then $$ \lim_{n \rightarrow \infty} \int_X | f_n -f| d\mu = 0 $$ **Lemma** If $f \in \mathcal L^1 (\mu)$, then $|\int_X f d\mu| \le \int_X |f| d\mu$. *proof* There exist $\alpha \in \mathbb C$ such that $|\alpha| = 1$ and $\alpha \int_X f d\mu \in [0, \infty)$. $$ \begin{align} \left | \int_X f d\mu\right | &= \left | \alpha \int_X f d\mu\right| = \left | \int_X \alpha f d\mu \right | \\ &= \int_X \alpha f d\mu = \int_X \operatorname{Re} (\alpha f) d\mu \end{align} $$ , which is bounded by $\int_X f d\mu$. Corollary Lebesgue's dominant convergence theorem impise that $f \in \mathcal L^1(\mu)$ and $\lim_{n\rightarrow \infty} \int_X f_n d\mu = \int_X f d\mu$. *proof* Skipped (application of the lemma).