---- We define the integral first for nonnegative real-valued functions, then for signed functions, then complex-valued functions. Each is built from, and in turn subsumes, its predecessor. > [!definition] Definition. ($\Sigma$-partition) > Let $\Sigma$ is a [[σ-algebra]] on a set $X$. A **$\Sigma$-partition** of $X$ is a finite collection of disjoint subsets $A_{1},...A_{m} \in \Sigma$ such that $A_{1} \sqcup\dots \sqcup A_{m}=X$. ^definition > [!definition] Definition. ([[integral]] of a nonnegative extended-real function ) > > > > Suppose $(X, \Sigma, \mu)$ is a [[measure|measure space]], $f:X \to [0, \infty]$ is a [[measurable function]], and $P=\{ A_{1},\dots,A_{m} \}$ is a $\Sigma$-partition of $X$. The **lower Lebesgue sum** $L(f, P)$ is defined $L(f, P):=\sum_{j=1}^{m} \mu(A_{j}) \inf_{A_{j}}f.$ > The **(Lebesgue) integral** of the nonnegative function $f:X \to [0, \infty]$ is [[supremum|then]] $ \int f \, d\mu := \sup \{L(f, P): P \text{ is a $\Sigma$-partition of }X\}.$ >For a subset $E \in \Sigma$, the notation $\int _{E} f \, d\mu$ stands for $\int \chi_{E} f \, d\mu$, should the latter be defined. > > [!equivalence] Equivalence > > An equivalent construction of $\int \_ \, d\mu$ arises by defining *first* the integral $\int s \, d\mu$ of a [[simple function]] $s=\sum_{j=1}^{m}c_{j} \chi_{E_{j}}$, $E_{j} \in \Sigma$ with $E_{i} \cap E_{j}=\emptyset$ for $i \neq j$ and $c_{j} \geq 0$, by taking $\int \chi_{E_{j}} \, d\mu:=\mu(E_{j})$ and linearly extending; then defining $\int f \, d\mu := \sup \left\{ \int s \, d\mu : 0 \leq s \leq f, s \text{ simple} \right\}. $ > > Assuming the $E_{j}$ are disjoint is nicest from a definition standpoint, but in general one can show that the integral of a simple function is unaffected by the way in which it is represented as a linear combination of [[characteristic function|characteristic functions]] (Axler 3.13-3.15).[^1] > > > > > [!proof]- Proof of Equivalence. > > > > > > We want to show that $\sup \{ L(f,P) :P \text{ is a }\Sigma \text{-partition of }X\} \ (1)$ > > > equals $\begin{align} > > > \sup \left\{ \sum_{j=1}^{m} c_{j}\mu(A_{j}): A_{j} \in \Sigma \text{ are disjoint}, \right. \\ > > > \left. c_{j} \in [0,\infty), f(x) \geq \sum_{j=1}^{m}c_j \chi_{A_{j}}(x) \ \forall x \in X \right\} \ (2). > > > \end{align}$ > > > The fact that $(1) \geq (2)$ follows from the 'integral of a simple function' and 'monotonicity' properties we proved for our original integral definition. > > > > > > To prove $(2) \geq (1)$, we'll examine two cases. First, let's assume $\inf_{A}f<\infty$ for every $A \in \Sigma$ with $\mu(A)>0$. Then, for $P$ a $\Sigma$-partition $A_{1},\dots,A_{m}$ of nonempty subsets of $X$, take $c_{j}= \inf_{A_{j}}f$ to witness $(2) \geq L(f,P)$ (since $L(f,P)$ is in the set whose [[supremum]] we are taking). Second, let's assume there exists a set $A \in \Sigma$ such that $\mu(A) >0$ and $\inf_{A}f=\infty$ (which implies $f(x)=\infty$ for all $x \in A$). In this case, for arbitrary $t \in (0, \infty)$ we can take $m=1$, $A_{1}=A$, and $c_{1}=t$. Then $(2)$ is at least $t \mu (A)$. Because $t$ is arbitrary positive number, this means $(2)$ equals $\infty$, which is of course greater than or equal to $(1)$. > > > ^equivalence > > > > > [!basicproperties] Basic Properties of the integral of nonnegative functions > > > > - (Integral of simple function)[^3] If $E_{1},\dots,E_{n}$ are disjoint sets in $\Sigma$ and $c_{1},\dots,c_{n} \in [0,\infty]$, then $\int \left( \sum_{k=1}^{n} c_{k} \chi_{E_{k}} \right) \, d\mu =\sum_{k=1}^{n} c_{k} \mu(E_{k})$ > > > > > [!proof]- Proof. > > > WLOG we can assume $E_{1},\dots,E_{n}$ forms a $\Sigma$-partition $P$ of $X$ (by replacing $n$ by $n+1$, setting $E_{n+1}=X-(E_{1} \sqcup \dots \sqcup E_{n})$, and setting $c_{n+1}=0$). In this case, $L\left( \sum_{k=1}^{n} c_{k} \chi_{E_{k}} , P \right)=\sum_{k=1}^{n}c_{k} \mu(E_{k})$. Thus $\int \left( \sum_{k=1}^{n} c_{k} \chi_{E_{k}} \right) \, d\mu \geq \sum_{k=1}^{n} c_{k} \mu(E_{k})$ > > > For the reverse inequality, let $P$ be any $\Sigma$-partition $A_{1},\dots,A_{m}$ of $X$. Then $\begin{align} > > > L\left( \sum_{k=1}^{n} c_{k} \chi_{E_{k}}, P \right)&=\sum_{j=1}^{m} \overbrace{ \mu(A_{j}) }^{ =\sum_{k=1}^{n} \mu(A_{j} \cap E_{k}) } \min_{\{ i : A_{i} \cap A_{j} \neq \emptyset\}}c_{i} \\ > > > &= \sum_{j=1}^{m} \sum_{k=1}^{n} \mu(A_{j} \cap E_{k}) \underbrace{ \min_{\{ i : A_{i} \cap A_{j} \neq \emptyset\}}c_{i} }_{ \leq c_{k} } \\ > > > & \leq\sum_{k=1}^{n} c_{k}\underbrace{ \sum_{j=1}^{m} \mu(A_{j} \cap E_{k}) }_{ \mu(E_{k}) } \\ > > > &= \sum_{k=1}^{n} c_{k} \mu(E_{k}). > > > \end{align}$ > > > The result follows. > > > > > > - (Monotonicity) If $f,g:X \to [0,\infty]$ are [[measurable function|measurable functions]] such that $f(x) \leq g(x)$ for all $x \in X$, then $\int f \, d\mu \leq \int g \, d\mu$. > > > > > [!proof]- Proof. > > > > > > Suppose $P=\{ A_{1},\dots,A_{m} \}$ is a $\Sigma$-partition of $X$. Then $\inf_{A_{j}}f \leq \inf_{A_{j}}g$ > > > and hence $L(f,P) \leq L(g,P)$ for all $\Sigma$-partitions $P$ of $X$. The result follows. > > - [[monotone convergence theorem for nonnegative measurable functions|monotone convergence theorem]] > > - (linearity) For $c \geq 0$, $f, g:X \to [0, \infty]$ measurable, $\int (cf + g) \, d\mu= c\int f \, d\mu+\int b \, d\mu$. (This takes some machinery to prove!) > > > > > > > [!proof]- Proof. > > > **Linearity.** If $P$ is a $\Sigma$-partition of $X$, then clearly $L(cf, P)=cL(f,P)$. Since $c \geq0$, homogeneity follows. > > > **Additivity.** The desired result holds for simple nonnegative measurable functions (Axler 3.13-3.15). Thus we approximate by such functions. So let $f_{1},f_{2},\dots$ and $g_{1},g_{2},\dots$ be increasing sequences of simple nonnegative measurable functions such that $\lim_{k \to \infty}f_{k}(x)=f(x)$ and $\lim_{k \to \infty}g_{k}(x)=g(x)$ for all $x \in X$; see [[approximation by simple functions]]. Then $\begin{align} > > > \int (f+g) \, d\mu &= \int (\lim_{k\to \infty} f_{k} + \lim_{k \to \infty} g_{k}) \, d\mu \\ > > > &= \int \left( \lim_{k\to \infty} (f_{k} + g_{k}) \right) d \mu \\ > > > &= \lim_{k \to \infty} \int\underbrace{ f_{k} + g_{k} }_{ \text{simple} } \, d\mu \ \ (\text{MCT}) \\ > > > &= \lim_{k \to \infty} \left( \int f_{k} \, d\mu + \int g_{k} \, d\mu \right) \\ > > > &= \lim_{k \to \infty} \int f_{k} \, d\mu + \lim_{k \to \infty} \int g_{k} \, d\mu \\ > > > &= \int f\, d\mu + \int g \, d\mu \ \ (\text{MCT}). > > > \end{align}$ > > > > > > > > > > > > > > [^1]: There is something to be proven here: for us, this fact doesn't follow from linearity because we use it to *prove* linearity. [^3]: These first two integration properties are used to prove the others, which is why they must be proved straight from the definition. ^definition > [!definition] Definition. ([[integral]] of a real-valued function) > Suppose $(X, \Sigma, \mu)$ is a [[measure|measure space]] and $f:X \to [-\infty, \infty]$ is a [[measurable function]] such that at least one of $\int f^{+} \, d\mu$, $\int f^{-} \, d\mu$ is finite.[^2] The **(Lebesgue) integral** of $f$ with respect to $\mu$, denoted $\int f \, d\mu$, is defined by $\int f \, d\mu =\int f^{+} \, d\mu - \int f^{-} \, d\mu .$ > > > > [!definition]- Definition. ($f^{+};f^{-}$) > > Suppose $f:X \to [-\infty, \infty]$ is a function. Define functions $f^{+}$ and $f^{-}$ from $X$ to $[0,\infty]$ by $f^{+}(x)=\begin{cases} > f(x) & f(x) \geq 0 \\ > 0 & f(x)<0 > \end{cases} \text{ and } f^{-}(x)=\begin{cases} > 0& f(x) \geq 0 \\ > -f(x) & f(x)<0. > \end{cases}$ > >Then $f=f^{+}-f^{-}$, while $|f|=f^{+}+f^{-}$. > > > >This definition allows us to extend our definition of integration to signed functions. > > > [!basicproperties] > - (Linearity) For $c \in \mathbb{R}$ and $f,g:X \to [-\infty, \infty]$ [[measurable function|measurable functions]] we have $\int cf \, d\mu =c\int f \, d\mu .$ > If$\int |f| \, d\mu<\infty$ and $\int |g| \, d\mu <\infty$, then $\int (f+g) \, d\mu =\int f \, d\mu + \int g \, d\mu .$ > > > > [!proof]- Proof. > > **Homogeneity.** We have already shown homogeneity in the case that $c \geq 0$ and $f$ is nonnegative. Next suppose that still $c \geq 0$, but $f$ is valued in $[-\infty, \infty]$. In this case, $\begin{align} > > \int cf \, d\mu &= \int (cf)^{+} \, d\mu - \int (cf)^{-} \, d\mu \\ > > &= \int cf^{+} \, d\mu - \int cf^{-} \, d\mu \\ > > &= c \int f^{+} \, d\mu - c \int f^{-} \, d\mu \text{ (Homogeneity when }f \geq ) \\ > > &= c \int f \, d\mu . > > \end{align}$ > > Finally, suppose $c<0$. Then $-c>0$ and $\begin{align} > > \int cf \, d\mu &= \int (cf)^{+} \, d\mu - \int (cf)- \, d\mu \\ > > &= \int (-c)f^{-} \, d\mu - \int (-c)f^{+} \, d\mu \\ > > &= (-c) \left( \int f^{-} \, d\mu - \int f^{+} \, d\mu \right) \\ > > &= c \int f \, d\mu . > > \end{align}$ > > > > > > **Additivity.** Clearly $(f+g)^{+} - (f+g)^{-}=f+g=f^{+}-f^{-}+ g^{+}-g^{-}.$ > > Thus $(f+g)^{+}+f^{-}+g^{-}=(f+g)^{-}+f^{+}+g^{+}.$ > > Both sides of this equation are sums of nonnegative functions. Thus integrating both sides wrt $\mu$ and using the additivity of the integral of nonnegative functions which we have already shown: $\int (f+g)^{+} \, d\mu + \int f^{-} \, d\mu + \int g^{-} \, d\mu =\int (f+g)^{-} \, d\mu +\int f^{+} \, d\mu + \int g^{+} \, d\mu .$ > > Rearranging gives $\int (f+g)^{+} \, d\mu -\int (f+g)^{-} \, d\mu =\int f^{+} \, d\mu -\int f^{-} \, d\mu +\int g^{+} \, d\mu -\int g^{-} \, d\mu ,$ > > where the left side is not of the form $\infty-\infty$ because $(f+g)^{+}\leq f^{+}+g^{+}$ and $(f+g)^{-} \leq f^{-}+g^{-}$. The equation above can be rewritten as $\int (f+g) \, d\mu =\int f \, d\mu +\int g \, d\mu ,$ > > completing the proof. > > > > >- (Monotonicity) If $f,g:X \to \mathbb{R}$ are measurable functions such that $\int f \, d\mu$ and $\int g \, d\mu$ are defined. Then if $f(x) \leq g(x)$ for all $x \in X$, then $\int f \, d\mu \leq \int g \, d\mu$. >- [[triangle inequality for integrals]] > - (bounding an integral) For all $E \in \Sigma$, $|\int _{E} f \, d\mu|\leq \mu(E) \sup_{E} |f|$. > > > [!proof]- Proof. > > > > We have $\begin{align} > > |\int _{E} f\, d\mu | &= |\int _{E} f\, d\mu | \\ > > &\leq \int \chi_{E} |f| \, d\mu \\ > > &\leq \int \chi_{E} \sup_{E} |f| \, d\mu \\ > > &= \sup_{E} |f| \underbrace{ \int \chi_{E} \, d\mu }_{ \mu(E) }, > > \end{align}$ > where we have used monotonicity, linearity, and the [[triangle inequality for integrals]]. > > - If $\mu\{x \in X: h(x) \neq 0\}=0$, then $\int h \, d\mu=0$. Taking $h:=f-g$, we obtain $\mu(\{ x \in X: f(x) \neq g(x)\})=0 \implies \int f \, d\mu=\int g \, d\mu$ as a corollary. > > > > > [!proof]- Proof. > > Suppose $\mu \{ x \in X: h(x) \neq 0 \}=0$. Then for every $E \in \Sigma$, $\inf_{E} h^{+}=0=\inf_{E}h^{-}$. Hence $L(h^{+}, P)=0=L(h^{-},P)$ for every $\Sigma$-partition $P$ of $X$. It follows that $\int h^{+} \, d\mu=0=\int h^{-} \, d\mu$. Hence $\int h \, d\mu=\int h^{+} \, d\mu-\int h^{-} \, d\mu=0$. > > > - In fact, a converse holds: if $\int |h| \, d\mu=0$, then $h=0 \text{ a.e.}$ > > > > [!proof]- Proof. > > Suppose $\int |h| \, d\mu=0$. First note that $\mu\left( \left\{ x \in X: |h(x)| \geq \frac{1}{n} \right\} \right)=0$ for each $n$, as follows e.g. from [[Markov's inequality]].[^5] Now, $h^{-1}(\mathbb{R} - \{ 0 \})=\bigcup_{n=1}^{\infty} \left\{ x \in X : |h(x)| \geq \frac{1}{n} \right\},$ > > and applying $\mu$ to both sides and using [[measure|countable subadditivity of measure]] we find $\mu\big( h ^{-1}(\mathbb{R} - \{ 0 \}) \big) \leq \sum_{n=1}^{\infty} \mu\left( \left\{ x \in X: |h(x)| \geq \frac{1}{n} \right\} \right)=0.$ > > Hence $h=0$ [[almost-everywhere]]. [^5]: To see this: [[Markov's inequality]] gives us that $\mu\left( \left\{ x \in X : |h(x)| \geq \frac{1}{n} \right\} \right) \leq n \underbrace{ \int |h| \, d\mu }_{ =0 }$, and since since $\mu$ is a nonnegative [[measure]] equality holds. We can also get there without Markov, by observing $0 = \int |h| \, d\mu \geq \int _{\left\{ x : |h(x) |\geq \frac{1}{n} \right\}} |h| \, d\mu \geq \frac{1}{n} \mu\left( \left\{ x : |h(x)| \geq \frac{1}{n} \right\} \right).$ > [!definition] Definition. ([[integral]] of a complex-valued function) > Suppose $(X, \Sigma, \mu)$ is a [[measure|measure space]] and $f:X \to \mathbb{C}$ is in $\mathcal{L}^{1}(\mu)$. Then $\int f \, d\mu$ is defined by $\int f \, d\mu =\int (\operatorname{re}f) \, d\mu + i \int \operatorname{\operatorname{im }}f \, d\mu .$ ^definition > [!basicproperties] > - $\int \overline{f} \, d\mu=\overline{\int f \, d\mu}$ > - - [ ] other properties we want (note: the integral triangle inequality can't just transfer over the proof from the real case, but is not hard; see 6.22) ^properties [^2]: The finiteness assumption is needed to avoid undefined expressions of the form $\infty-\infty$. > [!basicexample] > The integral of a [[characteristic function]] $\chi_{E}$, $E \in \Sigma$, is $\int \chi_{E} \, d\mu = \mu(E). $ > > > > [!proof]- Proof. > > If $P$ is the $\Sigma$-partition of $X$ given by $\{ E, X-E \}$, then $L(\chi_{E},P)=\mu(E)$. Hence $\int \chi_{E} \, d\mu \geq \mu(E)$. For the other direction, let $P=\{ A_{1},\dots,A_{m} \}$ be an arbitrary $\Sigma$-partition of $X$. WTS $L(\chi_{E}, P) \leq \mu(E)$. Indeed, > $L(\chi_{E}, P)=\sum_{\{ j: E \supset A_{j} \}} \mu(A_{j})=\mu(\bigsqcup_{\{ j: E \supset A_{j} \}} A_{j}) \leq \mu(E),$ > > > where the second equality used additivity of [[measure]] and the inequality used monotonicity. > ^proof > > [!basicexample] > *([[series|Series]] of nonnegative numbers as integrals)* If $\mu$ is the [[measure|counting measure]] on $\mathbb{N}$ and $a:\mathbb{N} \to [0, \infty]$ is a [[sequence]], $a=(a_{n})$, [[series|then]] $\int a \, d\mu =\sum_{n=1}^{\infty}a_{n}.$ > > > [!proof]- Proof. > > There are two [[supremum|suprema]] at play. First, [[series|recall]] that $\sum_{n=1}^{\infty}a_{n}:=\sup\underbrace{ \left\{ \sum_{n \in F} a_{n} : F \subset \mathbb{N}\text{ finite}\right\} }_{=: A}.$ > > Second, recall the characterization $\int a \, d\mu =\sup \underbrace{ \left\{ \int s \, d\mu : 0 \leq s \leq a \text{ simple}\right\} }_{ =:B }.$ > > $\geq$. A finite subset $F \subset \mathbb{N}$ determines a [[simple function]] $s_{F}:= \sum_{n \in F} a_{n} \chi_{\{ n \}} \leq a$, for which $\int s_{F} \, d\mu=\sum_{n \in F}a_{n}$. Thus $A \subset B$. Now recall that $\sup(\cdot)$ is [[monotonic map|monotone]] wrt set inclusion. > > > > > > $\leq.$ Let $\int s \, d\mu \in B$. If $\int s \, d\mu=\infty$, then for each $M$ there exists a finite set $F_{M} \subset \mathbb{N}$ with $\sum_{n \in F_{M}}s(n)>M$. Since $s \leq a$, this entails $\sum_{n=1}^{\infty}a_{n}=\infty$. So assume $\int s \, d\mu<\infty$. In this case it suffices to check that every element of $B$ satisfying this assumption has an upper bound in $A$.[^1] Since $\int s \, d\mu < \infty$, $s$ has finite support; write $s=\sum_{j=1}^{m} c_{j} \chi_{F_{j}}$ for each $F_{j} \subset \mathbb{N}$ finite. Then $\int s \, d\mu= \underbrace{ c_{1} + \dots + c_{1} }_{ \#F_{1} \text{ times} }+ \dots + \underbrace{ c_{m}+\dots+c_{m} }_{ \#F_{m} \text{ times} }$ > > where each $c_{j} \leq a |_{F_{j}}$. The result follows. > [^1]: In general for two sets $A$ and $B$ satisfying the property that for all $a \in A$ there exists $b \in B$ such that $a \leq b$, one has $\sup A \leq \sup B$ provided each sup exists. Indeed, for all $a \in A$ we have $a \leq b \leq \sup B$ for some $b \in B$, hence $\sup B$ is an upper bound of $A$; then $\sup A \leq \sup B$ by definition of [[supremum]]. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` Suppose $\mu$ is the [[measure|counting measure]] on $(\mathbb{N}, 2^{\mathbb{N}})$ and $(b_{n})$ is a [[sequence]] of nonnegative numbers, thought of as a function $b:(n \in \mathbb{N}) \mapsto (b_{n} \in [0, \infty))$. Then $\int b \, d\mu =\sum_{n=1}^{\infty}b_{n}.$ For $P=\{ A_{1},\dots,A_{m} \}$ be a $2^{\mathbb{N}}$-partition of $\mathbb{N}$, clearly $L(f,P)=\sum_{j=1}^{m}\#(A_{j}) \ \inf_{n \in A_{j}}b_{n} \leq \sum_{n=1}^{\infty} \overbrace{ (\#\{ n \}) }^{ 1 } b_{n}=\sum_{n=1}^{\infty}b_{n}.$ Hence $\int b \, d\mu \leq \sum_{n=1}^{\infty}b_{n}$. The partition $Q=\{ 1 \} \sqcup \{ \mathbb{N} - \{ 1 \} \}$ witnesses the reverse inequality: $L(f, Q)=b_{1} + \sum $ We can always obtain a finer partition by choosing a non-singleton $A_{j}=\{ a_{j,1}, a_{j,2},\dots \}$ and (WLOG) segmenting into $\{ a_{j, 1} \} \sqcup \{ a_{j,2},\dots \}$. This yields a $2^{\mathbb{N}}$-partition $P'$ of $\mathbb{N}$, with $L(b, P) \leq L(b, P')$ since $\inf_{A_{j}}b \leq \inf_{\{ a_{j, 1} \}}b + \inf_{\{ a_{j,2},\dots \}}b$. $\begin{align} L(b,P)&= \sum_{j=1}^{m}\big(\#A_{j} \big) \inf_{n \in A_{j}}b_{n} \\ &\leq \sum_{j=1}^{m} \big( \# A_{j} - \{ a_{j} \} \big) \inf_{n \in (A_{j} - \{ a_{j} \})}b_{n}+\sum_{j=1}^{m}\inf_{n \in \{ a_{j} \}_{j=1}^{m}}b_{n}= L(b, P') \end{align}$ where $P'$ is the partition obtained by $P=\{ A_{1}, \dots, A_{m} \}$ by 'splitting off' an element from each $A_{j}$.