> Notation: > $\chi_{A}$ is the indicator function of the set $A$. > $I_{i}$ is an interval, unless indicated otherwise. > $l(I_{i})$ is its length. > $(I_{i})$ is a sequence of intervals, in particular a countable set. > $\Lambda(J)=\sum_{i}l(I_{i})$ is the total length of the sequence $J=(I_{i})$. ```dataview table without id File as "Topics", dateformat(file.mtime, "yyyy-MM-dd") as "Last Modified" from "" FLATTEN "[[" + file.path + "|" + truncate(file.name, 30) + "]]" as File FLATTEN course as course where (course and (course = this.file.link) and (file.name != this.file.name)) and !contains(file.path, "2 - Snippets") sort file.mtime desc ``` ```dataview table without id File as "Snippets", dateformat(file.mtime, "yyyy-MM-dd") as "Last Modified" from "2 - Snippets" FLATTEN "[[" + file.path + "|" + truncate(file.name, 30) + "]]" as File FLATTEN course as course where (course and (course = this.file.link) and (file.name != this.file.name)) sort file.mtime desc ``` ## Measurable Spaces * The Lebesgue measure $m$ in $\mathbb{R}$ can be generalized to other spaces, giving measure functions with similar properties. ### General Measurable Spaces * Given a set $\Omega$, a $\sigma$**-algebra** (or $\sigma$-field) is a set $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ if: - (1) $\Omega \in \mathcal{F}$, * (2) closed to complements: if $E \in \mathcal{F}$, then $E^{c}\in \mathcal{F}$, * (3) closed to countable unions: if $(E_{n})_{n=1}^{\infty} \subseteq \mathcal{F}$, then $\cup_{n}E_{n } \in \mathcal{F}$. * Then the pair $(\Omega, \mathcal{F})$ is called a **measurable space**. * A **measure** $\mu:\mathcal{F} \to [0, \infty]$ in the space $(\Omega,\mathcal{F})$ satisfies: * $\mu(\emptyset)=0$, * countable additivity: pairwise disjoint sets $(E_{n})_{n=0}^{\infty}\subseteq \mathcal{F}$ have $\mu(\cup_{n}E_{n})=\sum_{n}\mu(E_{n})$. * A measure is **finite** if $\mu(\Omega) < \infty$, and a **probability measure** if $\mu(\Omega)=1$. * The triple $(\Omega, \mathcal{F}, \mu)$ is a **measure space**. * For example, $(\mathbb{R}, \mathcal{M},m)$ is a measure space. * For any set $\Omega$, $(\Omega, \mathcal{P}(\Omega),\mu)$ is a measure space, where $\mu(E)=\text{No. of elements of }E$, the counting measure. * If $E \subseteq \mathbb{R}$, the "subspace" $(E, \mathcal{M}_{Leb}|_{E},m|_{E})$ is also a measurable space, where $m|_{E}$ is the restriction of $m$ to $E$, and $\mathcal{M}_{Leb}|_{E}=\{ F \subseteq E:F \in \mathcal{M}_{Leb} \}=\{ F \cap E :F \in \mathcal{M}_{Leb}\}$ ### Properties of General Measures *Big idea: general measures have the same properties that the Lebesgue measure has.* * Monotonocity: if $A,B \in \mathcal{F},A \subseteq B$, then $\mu(A) \le \mu(B)$. * Continuity: if $A_{1} \subseteq A_{2} \subseteq \dots$, then $\mu(\cup_{n}A_{n})=\lim_{ n \to \infty }\mu(A_{n})$. > Proof similar to the result for the Lebesgue measure. * if $A_{1} \supseteq A_{2} \supseteq \dots$, and $\mu(A_{1})< \infty$, then $\mu(\cap_{n=0}^{\infty}A_{n})=\lim_{ n \to \infty }\mu(A_{n})$. > Proof: take their complements $C_{1}\subseteq C_{2} \subseteq\dots$, and apply the previous result. ### Generation and Borel Sets * An arbitrary intersection of $\sigma$-fields is also a $\sigma$-field. > Proof: just a routine check of the definition. * Given $B \subseteq \Omega$, it **generates** a $\sigma$-algebra $\mathcal{F_{B}}$, the smallest $\sigma$-algebra containing $B$. Equivalently, the $\sigma$-algebra generated by $B$ is$\mathcal{F}_{B}=\bigcap\{\mathcal{F}\subset \mathbb{R}: \mathcal{F} \text{ is a }\sigma \text{-field containing }B \}$ * **Borel sets** are members of the **Borel $\sigma$-algebra** $\mathcal{M}_{Bor}$, which is generated by the set of all intervals. * All open sets, being countable union of intervals, are Borel sets. * All countable sets, being countable union of singleton intervals $[a,a]$, are Borel sets. * The same $\mathcal{M}_{Bor}$ can be generated by any of the following families: * All intervals. * All finite open intervals $(a,b)$. * All finite closed intervals $[a,b]$. * All unbounded intervals $(a, \infty)$. * Etc... * In particular, since all intervals are in $\mathcal{M}$, $\mathcal{M}_{Bor} \subset \mathcal{M}_{Leb}$. The inclusion is strict, and in particular some of the null sets are not in $\mathcal{M}_{Bor}$. ## Measurable Functions * On a measurable set $E \subseteq \mathbb{R}$, a function $f: E \to \mathbb{R}$ is **(Lebesgue) measurable** if any interval $I \subseteq \mathbb{R}$ has a measurable preimage: $f^{-1}(I)=\{ x \in E: f(x) \in I \} \in \mathcal{M}_{Leb}$ * More generally, in a measurable space $(\Omega, \mathcal{F})$, a function $f$ is $\mathcal{F}$**-measurable** if any interval $I \subseteq \mathbb{R}$ is in $\mathcal{F}$: $f^{-1}(I)=\{ x \in \Omega:f(x) \in I \} \in \mathcal{F}$ * In particular, functions measurable in $(\mathbb{R},\mathcal{M}_{Bor})$ are called **Borel measurable**. * A function $f$ being measurable is equivalent to one of the following families having measurable preimages: * $\forall a: (a,\infty)$ * $\forall a:[a,\infty)$ * $\forall a:(-\infty,a)$ * $\forall a:(-\infty, a]$ > Proof (for $(a,\infty)$): > $(-\infty,a]$: by taking complements $f^{-1}((-\infty,a])=E\setminus f^{-1}((a,\infty))$. > $(-\infty, a)$: by countable union $f^{-1}((-\infty, a))=\cup_{n}f^{-1}((-\infty,a-1 / n])$ > $[a,\infty)$: by taking complement of $f^{-1}((-\infty, a))$ > $(a,b)$: intersect $f^{-1}((-\infty, b)) \cap f^{-1}((a, \infty))$ * Generally, $f: \Omega \to \mathbb{R}$ is $\mathcal{F}$-measurable if $\forall G \in B:f^{-1}(G) \in \mathcal{F}$, where $B$ generates the Borel set $\mathcal{B}$. > Proof: define the measurable preimages $f^{*}(\mathcal{F})=\{ G \subseteq \mathbb{R}:f^{-1}(G) \in \mathcal{F} \}$, which form a $\sigma$-algebra in $\mathbb{R}$. Hence $B \subseteq f^{*}(\mathcal{F}) \iff \mathcal{B}\subseteq f^{*}(\mathcal{F}) \iff I \in f^{*}(\mathcal{F})$ for all intervals $I$. * Constant functions and continuous functions are measurable. * Indicator functions of measurable sets are measurable. ### Deriving Measurability * If $f$ is Lebesgue-mesurable, $g$ is $\mathcal{M}_{Bor}$-measurable, then $g \circ f$ is Lebesgue measurable. > Proof: for any interval $I$, $g^{-1}(I) \in \mathcal{M}_{Bor}$, so it is a countable union of intervals $I_{i}$. Then the preimage $f^{-1}(\cup_{i}I_{i})=\cup_{i}f^{-1}(I_{i}) \in \mathcal{M}_{Leb}$. * If $f,g$ are measurable functions, their combination via a continuous function $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, given by $h(x)=F(f(x),g(x))$ is also measurable. * In particular, sums and products of measurable functions are also measurable. > Proof: for $a \in \mathbb{R}$, define $G_{a}=\{ (u,v) \in \mathbb{R}^{2}:F(u,v)>a \}=F^{-1}((a,\infty))$ > Since $F$ is continuous, $G_{a}$ is open in $\mathbb{R}^{2}$, hence it can be decomposed into a countable union of open rectangles: $G_{a}=\bigcup_{i=1}^{\infty}R_{i},\, \text{ where }R_{i}=(a_{i},b_{i}) \times (c_{i},d_{i}) \in \mathbb{R}^2$(a rough proof of this theorem can be found [here](https://math.stackexchange.com/q/1007436)). > > Then for any $(a, \infty)$, its preimage is $\begin{align*} h^{-1}((a,\infty))&=\{ x:h(x)>a \}\\ &=\{ x:(f(x),g(x)) \in G_{a} \}\\ &= \bigcup_{i}\{ x: (f(x),g(x)) \in R_{i} \}\\ &= \bigcup_{i} \{ x:f(x) \in (a_{i},b_{i}) \} \cap \{ x: g(x) \in (c_{i},d_{i}) \}\\ &= \bigcup_{i} f^{-1}[(a_{i},b_{i})] \cap g^{-1}[(c_{i},d_{i})] \\ &\in \mathcal{M}_{Leb} \end{align*}$so $h=F(f,g)$ is measurable. * For a sequence of measurable functions $(f_{n}) \subset \mathcal{M}$, the following are all measurable: $\sup_{n \in \mathbb{N}}f_{n},\,\, \inf_{n \in \mathbb{N}} f_{n},\,\, \limsup_{n \to \infty} f_{n},\,\, \liminf_{n \to \infty} f_{n} $ * As a result, if $f_{n} \to f$ pointwise, then $f=\lim_{ n \to \infty }f_{n}=\limsup_{ n \to \infty }f_{n}$ is also measurable. > Proof: it's sufficient to show that the preimage of $(a,\infty)$ is measurable. Under $\sup_{n \in \mathbb{N}}f_{n}$, the preimage is $\{ x: \sup_{n \in \mathbb{N}}f_{n}(x)>a \}=\bigcup_{n}\{ x: f_{n}(x)>a \}=\bigcup_{n}f^{-1}_{n}((a, \infty)) \in \mathcal{M}_{Leb}$$\inf_{n \in \mathbb{N}} f_{n}=-\sup_{n \in \mathbb{N}}(-f_{n})$, hence also measurable. > $\limsup_{ n \to \infty }f_{n} = \inf_{n \ge 1}\{ \sup_{m \ge n} f_{m} \}$, hence measurable measurable; similarly for $\liminf$. * If $f$ is a measurable function, and $g=f\,\,\mathrm{a.e.}$, then $g$ is also measurable. * As a result, if the measurable sequence $(f_{n}) \to f\,\,\mathrm{a.e.}$ , $f$ is also measurable. > Proof: their difference $d=f-g$ is measurable. This is because $d=0\,\,\mathrm{a.e.}$, except on the null set $A$. > It's sufficient to show that all $(a,\infty)$ has a measurable preimage: $d^{-1}((a,\infty))=\begin{cases} A^{>a} \subseteq A &\text{if }a \ge 0\\ B \supset A^{c} & \text{if }a<0 \end{cases}$where $A^{>a}$ (null, being a subset of $A$) and $B$ (has a null, hence measurable complement) are both measurable. Hence $g=f+h$ also measurable. ## Lebesgue Integration - Introduction ### Simple Functions * A **simple function** $\phi: \mathbb{R} \to \mathbb{R}$ is a measurable function with a finite image: $\begin{align*} \phi(\mathbb{R})&=\{ a_{1},\dots,a_{n} \} \text{ finite,}\\ \phi^{-1}(a_{i})&=\{ x \in \mathbb{R}:\phi(x)=a_{i} \} \equiv A_{i} \in \mathcal{M}_{Leb}. \end{align*}$so in its **canonical/standard form**, $\phi=\sum_{i}a_{i}\chi_{A_{i}}$ (assuming $a_{i}$ are all distinct). * If $f:\mathbb{R} \to [0,\infty]$ is a non-negative measurable function, then it is the limit of some increasing sequence of simple functions: $\exists (\phi_{n})\text{ simple}:\phi_{n} \nearrow f$ > Proof: define $\phi_{n}$ to be $f$ rounded down to the nearest $2^{-n}$, capped at $2^{n}$. > To do so, split $[0,2^n)$ in the image equally into $4^{n}$ intervals, namely $\{[0,2^{-n})+k2^{-n}: k =0,\dots,4^{n}-1 \}$Then for each interval, take its preimage: $B_{k,n}=f^{-1}\left([0,2^{-n})+k 2^{-n}\right)$and define $\phi_{n}$ to be $k2^{-n}$ on this set. Otherwise, $\phi_{n}=2^{n}$. > Then for points $x:f(x)<\infty$, $f-\phi_{n}<2^{-n} \to 0$. For points $x: f(x)=\infty$, $\phi_{n}=2^{n} \to f(x)$. ### Lebesgue Integral of Non-Negative Functions * The (Lebesgue) integral of a simple function $\phi=\sum_{i}a_{i}\mathbb{1}_{A_{i}}$ is $\int \phi = \sum_{i}a_{i}\,m(A_{i})$with the convention that if $a_{i}=0,\,m(A_{i})=\infty$, their product is $0$. * Any indicator of null sets have integral $0$, for example the rationals in $[0,1]$: $\int _{[0,1]} \mathbb{1_{Q}} \, dm=m(\mathbb{Q}\cap[0,1])=0$. * The **(Lebesgue) integral** of a non-negative function $f$ is defined to be: $\int f = \sup \left\{ \int \phi : 0 \le \phi \le f,\,\,\phi \text{ simple} \right\}$and equivalently denoted $\int f \, dm$ to make the measure explicit. * The integral of $f$ over a measurable set $E \subseteq \mathbb{R}$ is defined to be $\int _{E} f \equiv \int f \cdot \chi_{E}$ ### Properties of the Integral (Simple Functions) *Big idea: the integral of simple functions have the expected properties. These will carry over to general non-negative functions.* * Let $\phi, \psi$ be simple functions, and $D,E \in \mathcal{M}$, $a \in \mathbb{R}$, then their integrals are: * Order-preserving: $0 \le \phi \le \psi \Longrightarrow \int \phi \le \int \psi$. * Set-additive: $D \cap E=\emptyset \Longrightarrow \int_{D \cup E} \phi=\int _{D} \phi + \int _{E} \phi$. * Linear: $\int (\phi+\psi)=\int \phi+\int \psi$, and $\int a\phi=a\int\phi$. > Proof: > <u>Order</u>: $Y(D, \phi) \subseteq Y(D, \psi)$, so their suprema follow the inequality. > <u>Additivity</u>: write $\phi=\sum_{i}a_{i}\mathbb{1}_{A_{i}}$ as in the definition, then use the countable additivity of the measure in its integral: $\begin{align*} \int _{D \cup E} \phi \, dm&=\sum_{i} a_{i}\,m(A_{i} \cap (D \cup E))\\ &= \sum_{i} a_{i}\,m\left((A_{i}\cap D) \cup (A_{i} \cap E)\right) \\ &= \sum_{i} a_{i}\,m(A_{i} \cap D) + \sum_{i} a_{i}\,m(A_{i} \cap E)\\ &= \int _{D} \phi \, dm + \int _{E} \phi \, dm \end{align*}$ > <u>Linearity</u>: just grinding set notations. ### Monotone Convergence Theorem *Big idea: the MCT allows swapping the integral and the limit. Hence properties of increasing simple functions $(\phi_{n})$ carry over to their limit $f$.* * The **monotone convergence theorem (MCT)** states that if $(f_{n})$ are non-negative, measurable functions, and $(f_{n}) \nearrow f$, then $\lim_{ n \to \infty } \int f_{n} = \int \lim_{ n \to \infty } f_{n} = \int f $ > Proof: ($\le$) is immediate from monotonocity as $f_{n} \le f \Rightarrow \int f_{n} \le \int f$. > ($\ge$): take any simple function $\phi:0 \le \phi \le f$, and it's sufficient to show that $\lim_{ n \to \infty }\int f_{n} \, dm \ge \int \phi \, dm$. > Scale down $\phi$ by any $\alpha \in (0,1)$ to get $\alpha \phi$, and define $B_{n} \equiv \{ x: f_{n} \ge \alpha \phi(x) \}$so that $B_{n}$ is increasing, with $B=\cup_{n} B_{n} = \mathbb{R}$. Then $\int _{\mathbb{R}} f_{n} \ge \int_{B_{n}} f_{n} \ge \alpha\int _{B_{n}} \phi $Letting $n \to \infty$, $\lim_{ n \to \infty } \int _{\mathbb{R}}f_{n} \, dx \ge \lim_{ n \to \infty }\alpha\int _{B_{n}} \phi= \alpha\int _{\mathbb{R}} \phi\, d$where convergence in $\mathrm{RHS}$ can be shown with the standard form of simple functions, and that $m(B_{n} \cap E) \to m(\mathbb{R} \cap E)$ when $B_{n}$ increases to $\mathbb{R}$. > Then taking the limit as $\alpha \nearrow 1$ gives the result. * The **baby MCT** states that if $(E_{n}) \subseteq \mathcal{M}$ is an increasing sequence, and $E=\cup_{n}E_{n}$, $f \ge 0$ is measurable, then $\int _{E} f = \sup_{n} \int_{E_{n}} f = \lim_{ n \to \infty } \int _{E_{n}}f $which is just the special case of the MCT on $(f\cdot \chi_{E_{n}})\nearrow f$. * It's known by this name only at Oxford. * **MCT for series**: for a sequence $(f_{n}) \ge 0\,\,\mathrm{a.e.}$, $\int \sum_{n=0}^{\infty} f_{n} = \sum_{n=0}^{\infty}\int f_{n}$which is the MCT applied to $\sum_{n=0}^{k} f_{n} \nearrow \sum_{n=0}^{\infty} f_{n}$. ### Properties of the Integral (Non-Negative Functions) *Big idea: since non-negative functions is the limit of some increasing sequence of simple functions, many properties carry over via the MCT.* - If $f$ is non-negative and measurable, then there is a sequence of increasing, non-negative simple functions $(\phi_{n}) \nearrow f$, so by the MCT, $\int f \, dm = \lim_{ n \to \infty } \int \phi_{n}$ * Additivity: if $f,g \ge 0$ are measurable, then $\int (f+g) \, dm = \int f \, dm + \int g \, dm$ > Proof: approximate $f,g$ with $(\phi_{n})\nearrow f$, and $(\psi_{n}) \nearrow g$. Then $(\phi_{n}+\psi_{n}) \nearrow (f+g)$, so $\begin{align*} \int (f+g) \, dx \overset{\mathrm{MCT}}{=} &\lim_{ n \to \infty } \int (\phi_{n} + \psi_{n}) \\ =\,\,\, &\lim_{ n \to \infty } \left( \int \phi_{n} +\int \psi_{n} \right)\\ \overset{\mathrm{ MCT}}{=} &\int f + \int g \end{align*}$ * Null sets are negligible: if a measurable function $f \ge 0$, then $\int f=0 \iff f=0\,\, \mathrm{a.e.} $and hence $g=h\,\,\mathrm{a.e.}\iff \int g =\int h \, dx$ for $f,g \ge0$. > Proof: find non-negative simple functions $(\phi_{n}) \nearrow f$. > ($\Leftarrow$) is trivial: any simple function $\phi:0\le\phi \le f$ mush be $0\,\,\mathrm{a.e.}$, hence having an integral of $0$. > ($\Rightarrow$) consider $E_{n}=\{ x:f(x)>1 / n \}$, and we want to show that $\cup_{n}E_{n}=\{ x: f(x)>0\}\equiv E$ is null. > It's then enough to show that $E_{n}$ is null for all $n$. But if for contradiction $m(E_{n})>0$, then $\frac{1}{n}\chi_{E_{n}} \le f$ has integral $0<\frac{m(E_{n})}{n}=\int \frac{1}{n}\chi_{E_{n}} \le \int f =0 $a contradiction. > ($g=h$): take $f=g-h$ and apply the above. * On the interval $[a,b]$, for function $f \in C[a,b]$, its Riemann and Lebesgue integrals are the same. > Proof: let $(\phi_{n})$ by the optimal minorant step functions obtained by bisecting $[a,b]$ $n$ times, then $(\phi_{n}) \nearrow f$. > Their integrals converge to the Riemann integral by definition, and to the Lebesgue integral by the MCT. ## Lebesgue Integration - General Functions ### Definitions * The positive and negative parts of a function $f$ are: $f^{+}\equiv \max(f,0),\,\,f^{-}\equiv \max(-f,0)$and note that $f=f^{+}-f^{-}$, and $|f|=f^{+}+f^{-}$. * A measuable function $f$ is **integrable** if both its positive and negative parts have a finite integral: $\int _{E} f^{\pm} \, dm \in (-\infty,\infty)$; in that case its integral is $\int f \equiv \int f^{+} -\int f^{-} $ * Over a set $E \in \mathcal{M}_{Leb}$, a measurable function $f$ is integrable if $f\cdot\chi_{E}$ is. The set of integrable functions is denoted $\mathcal{L}^{1}(E)$. ### Properties of the Integral (General Functions) *Big idea: most properties carry over to general functions $f$ via their positive/negative parts $f^{\pm}$.* * Integrals are linear to addition: $\int (f+g) \, dm=\int f \, dm + \int g \, dm$. > Proof: note that $\begin{align*} (f+g)&= (f+g)^{+}-(f+g)^{-}\\ &= f^{+}-f^{-}+g^{+}-g^{-} \end{align*}$so rearrange to get $(f+g)^{+}+f^{-}+g^{-}=(f+g)^{-}+f^{+}+g^{+}$, all of which are non-negative functions, and apply the linearity property on integrals of non-negative functions. * Integrals are linear to scaling: for any $c \in \mathbb{R}, f \in \mathcal{L}^{1}(A)$, $\int _{A} (cf) \, dx=c\int _{A} f \, dx$ > Proof: approximate both sides with simple functions, for which the identity is trivial. * Integrals preserve orders: if $f \le g$, then $\int f \, dm \le \int g \, dm$. > Proof: if $f \le g$, then $f^{+} \le g^{+}$, and $f^{-} \ge g^{-}$. Then apply the result for non-negative functions to their integrals. * Bounds on the integrand: if $f$ is integrable, then * $|f| \ne \infty\,\,\mathrm{a.e.}$; * If $\int |f| =0$, then $f=0\,\,\mathrm{a.e.}$ * Bounds on the integral: if $f$ is integrable, then * $|\int f |\le \int |f|$. * Over a measurable set $A$, $m(A)\inf _{x \in A} f \le \int _{A} f\, dm \le m(A)\sup _{x \in A} f$ > Hint for the first property: write $f$ and $|f|$ in terms of $f^{\pm}$. ### Deducing Integrability * <u>Absolute values</u>: if $f$ is integrable, so is $|f|$; the converse is true if $f$ is measurable. * The extra condition to the converse is just because $|f|$ measurable $\not \Rightarrow f$ measurable. > Proof/Hint: again, write $f$ and $|f|$ in terms of $f^{\pm}$, then use linearity. * The **Comparison Test**: if $f,g$ are measurable, * $|f| \le g$, $g$ integrable, then $f$ is also integrable; * $|f| \ge g \ge 0$, $g$ not integrable, then neither is $f$. > Proof/Hint: use the order-preserving property to find $|f|$ integrable/not integrable, then use the previous result. * <u>Products</u>: if $f$ is integrable, $g:|g| \le c$ is measurable and bounded, then $f \cdot g$ is integrable. > Proof: $fg$ is measurable, and $|fg| \le c|f|$, so the Comparison Test gives integrability. * <u>Changing on a null set</u>: if measurable functions $f=g \,\,\mathrm{a.e.}$, then $f$ integrable $\iff g$ integrable, and if they are, $\int f = \int g$. > Proof: $f^{\pm}=g^{\pm}\,\,\mathrm{a.e.}$, so $\int f^{\pm} =\int g^{\pm}$ by the result for non-negative functions. > WLOG assume $f$ integrable, then $\int g^{\pm}=\int f^{\pm} <\infty$, $g$ is integrable, and their integrals are the same. ### Lebesgue and Riemann Integrals of General Functions * Any Riemann-integrable function $f: [a,b] \to \mathbb{R}$ is Lebesgue-integrable. > Proof: if $f$ is Riemann-integrable on $[a,b]$, it is bounded. So it remains to show that $f$ is measurable. > Since $f$ is integrable, take a sequence of minorants $(\phi_{n}) \nearrow f$ and majorants $(\psi_{n})\searrow f$. Let $g=\sup_{n}(\phi_{n})$ and $h=\inf_{n}(\psi_{n})$ be their pointwise limits, and $g,h$ are measurable since $\phi_{n},\psi_{n}$ are. > Then $g \le f \le h$, and $0 \le h-g \le \psi_{n}-\phi_{n}$, so $0 \le\int_{[a,b]} (h-g) \le \int _{[a,b]} (\psi_{n}-\phi_{n}) \to 0 $hence $h=g\,\,\mathrm{a.e.}$, and so $f=g\,\,\mathrm{a.e.}$, and $g$ measurable implies $f$ measurable. * Furthermore, in that case Lebesgue and Riemann integrals agree: $\underset{\text{Lebesgue}}{\vphantom{\underline{\Bigg\uparrow}}\int _{[a,b]} f }=\underset{\text{Riemann}}{\vphantom{\underline{\Bigg\uparrow}}\int _{a}^{b}f \, dx}$ > Proof: obviously Riemann and Lebesgue integrals agree for step functions. > Denote $I(S)=\left\{ \int_{a}^{b} \phi | \phi \in S \right\}$, then $\underset{\text{Riemann}}{\vphantom{\underline{\Bigg\uparrow}}\int _{a}^{b}f \, dx} = \sup I(\mathrm{minorants}) \le \underset{\text{Lebesgue}}{\vphantom{\underline{\Bigg\uparrow}}\int _{[a,b]} f } \le \inf I(\mathrm{majorants}) = \underset{\text{Riemann}}{\vphantom{\underline{\Bigg\uparrow}}\int _{a}^{b}f \, dx}$where the inequalities are justified by the integral preserving the orders in $\text{minorant}\le f \le \text{majorant}$. * If the improper Riemann integral of $f \ge 0$ exists, so does the Lebesgue integral, and they are equal: $\lim_{ a,b \to \infty }\int _{-a}^{b}f \, dx \equiv \int _{-\infty}^{\infty}f \, dx=\int _{\mathbb{R}} f $ * $f \ge 0$ is necessary: otherwise the positive and negative components of $f$ can have unbounded integrals (hence Lebesgue non-integrable) that cancel out (giving finite Riemann integral). E.g. $\sin(x) / x$ over $\mathbb{R}$. > Proof: $f_{n}\equiv f \chi_{[-n,n]}\nearrow f$ is Riemann (hence Lebesgue) integrable, so apply the previous result for finite intervals to get $\int _{-n}^{n}f \, dx =\int _{[-n,n]} f $then $\mathrm{LHS} \to \int _{-\infty}^{\infty}f \, dx$, and $\mathrm{RHS} \to\int _{\mathbb{R}} f \, dx$ by the baby MCT; hence these are equal. ## Dominated Convergence Theorem ### The DCT * **Fatou's Lemma** states that if $(f_{n})$ is a sequence of non-negative measurable functions, then $\int \liminf_{ n \to \infty } f_{n} \le \liminf_{ n \to \infty } \int f_{n} $ > Proof: denote $g_{k}=\inf_{n \ge k}f_{n}$, then $(1)\,g_{n}\nearrow\liminf_{n \to \infty}f_{n}$, and $(2)\, g_{n} \le f_{n}$, > Hence $\int \liminf_{ n \to \infty }f_{n} \, \overset{\mathrm{MCT}+(1)}{=} \lim_{ n \to \infty } \int g_{n} = \liminf_{ n \to \infty } \int g_{n} \overset{(2)}{\le} \liminf_{ n \to \infty } \int f_{n} $ * The **Dominated Convergence Theorem (DCT)** states that for a sequence of integrable functions $(f_{n})$, if: * <u>Domination</u>: $(f_{n}) \to f\,\,\mathrm{a.e.}$, and * <u>Convergence</u>: $\exists g$ integrable, such that $\forall n \ge 1:|f_{n}| \le g \,\,\mathrm{a.e.}$, * Then the limit $f$ is integrable, and $\int f =\lim_{ n \to \infty }\int f_{n}$. * *That is, if a sequence is dominated by an integrable function, then the integral of its limit equals the limit of its integrals.* > Proof: $f$ is measurable as the $\mathrm{a.e.}$ limit of measurable functions. > The Comparison Test with $g \ge |f|\,\, \mathrm{a.e.}$ gives integrability of $|f|$, hence $f$. > Consider $(g-f_{n})$ and $(g+f_{n})$, both non-negative and measurable. > Apply Fatou's Lemma to each to get the inequalities in $\begin{align*} (1) &\int (g-f) = \int \liminf_{ n \to \infty } (g-f_{n}) \le \liminf_{ n \to \infty } \int (g-f_{n}) = \int g - \limsup_{ n \to \infty } \int f_{n}\\ (2) &\int (g+f) = \int \liminf_{ n \to \infty } (g+f_{n}) \le \liminf_{ n \to \infty } \int (g+f_{n}) = \int g + \liminf_{ n \to \infty } \int f_{n} \end{align*}$Hence $\limsup_{ n \to \infty } \int f_{n} \overset{(1)}{\le} \int f\, \overset{(2)}{\le} \liminf_{ n \to \infty } \int f_{n}$which is only possible if they are all equal (and hence equal to $\lim_{ n \to \infty } \int f_{n}$). ### Derivatives of the DCT * The **Bounded Convergence Theorem (BCT)**: on a bounded interval $I$, if $(f_{n}) \subseteq \mathcal{L}^{1}(I)$ is bounded by some constant $c$ almost everywhere, and $(f_{n}) \to f\,\,\mathrm{a.e.}$; then $f \in \mathcal{L}^{1}(I)$, and $\int f =\lim_{ n \to \infty }\int f_{n}$. > Proof: constants are integrable on bounded intervals, so this is just the special case of the DCT where $g \equiv c$. * **Beppo-Levi Theorem** states that if $(g_{n})$ are integrable, $\sum_{n} \int |g_{n}| < \infty$, then $\sum_{n} g$ converges $\mathrm{a.e.}$ to an integrable limit, and $\int \sum_{n=0}^{\infty} g_{n} = \sum_{n=0}^{\infty} \int g_{n}$ > Proof: > <u>Integrability</u>: define $\gamma(x)=\sum_{n}|g_{n}(x)|$, a series of non-negative, measurable functions, so the MCT of series gives $\int \gamma =\lim_{ k \to \infty }\sum_{n=0}^{k}\int |g_{n}| < \infty $so $\gamma$ is integrable, hence finite $\mathrm{a.e.}$. So the series $\sum_{n}g_{n}$ converges absolutely $\mathrm{a.e.}$. > > <u>Equality</u>: since partial sums $\sum_{n=0}^{k}g_{n}$ converge and are dominated by the integrable function $\gamma$, the DCT guarantees $\sum_{n} g_{n}$ to be integrable, and: $\int \sum_{n=0}^{\infty} g_{n} =\int \lim_{ k \to \infty } \sum_{n=0}^{k} g_{n}\overset{\mathrm{DCT}}{=} \lim_{ k \to \infty } \int \sum_{n=0}^{k}g_{n} = \lim_{ k \to \infty } \sum_{n=0}^{k} \int g_{n}=\sum_{n=0}^{\infty}\int g_{n}$ ### DCT of Integrals with a Parameter - This section focuses on the integrals of the form $F(y)=\int f(x,y) \, dx $and their differentiability and continuity. - The **continuous parameter DCT** states that for a function $f:I \times J \to \mathbb{R}$, where $I,J$ are intervals, then the function $F(y)=\int _{I}f(x,y) \, dx$ is continuous over $J$ if: - $(1)$ integrability: fixing any $y$, $x \mapsto f(x,y)$ is integrable over $I$. - $(2)$ continuity: fixing any $y_{0}$ and for almost all $x$, mapping $y \mapsto f(x,y)$ is continuous at $y_{0}$. - $(3)$ domination: $\exists g \in \mathcal{L}^{1}(I)$ such that for $\forall y \in J$, $|f(x,y)|\le g(x)$ for almost all $x$. > Proof: fix any $y$, and we shall show that for any $(y_{n}) \to y$, there is $F(y_{n})\to F(y)$. > Define $f_{n}(x)=f(x,y_{n})$, where $|f_{n}(x)|\le g(x)\,\,\mathrm{a.e.}$ by condition $(3)$. By $(2)$, $f_{n}(x) \to f(x,y)$ for almost all $x$. Then the (single-variable) DCT applies, giving $F(y_{n})=\int f_{n}(x) \, dx \to \int f \, dx=F(y) $and hence $F$ is continuous at any $y$. - <u>Differentiability</u>: with $f:I \times J \to \mathbb{R}$, its integral $F(y)=\int _{I}f(x,y) \, dx$ is differentiable with $F'(y)=\int _{I}f_{y} \, dx$ if: - $(1)$ integrability: fixing any $y$, $x \mapsto f(x,y)$ is integrable over $I$. - $(2)$ differentiability: for almost all $x$, $f_{y}$ exists for all $y$. - $(3)$ domination: $\exists h(x) \in \mathcal{L}^{1}(I)$ such that $|f_{y}(x,y)|\le h(x)$ for almost all $x$, for $\forall y \in J$. > Proof: we shall show that $F$ is differentiable at an arbitrary $y$. Find any sequence $(y_{n}) \to y$ with $y_{n} \ne y$, $\begin{align*} \lim_{ n \to \infty } \frac{F(y_{n})-F(y)}{y_{n}-y}&=\lim_{ n \to \infty } \int_{I} \frac{f(x,y_{n})-f(x,y)}{y_{n}-y} \,dx\\ &= \int _{I} \lim_{ n \to \infty } \frac{f(y_{n})-f(y)}{y_{n}-y} \,dx && (3),\,\text{DCT}\\ &= \int _{I} f_{y}(x,y) \,dx && (2) \end{align*} $where the DCT applies because $x \mapsto\frac{f(x,y_{n})-f(x,y)}{y_{n}-y}$ is integrable for all $n$, and they are dominated by $h(x)$ for almost all $x$. ## Product Measures ### General Product Measures Given measure spaces $(\Omega_{1},\mathcal{F}_{1},m_{1})$ and $(\Omega_{2},\mathcal{F}_{2},m_{2})$, their **product measure space** is $(\Omega,\mathcal{F},m)$ where $\Omega=\Omega_{1}\times \Omega_{2}$ (the usual Cartesian product). - The simplest sets in $\mathcal{F}$ are the "rectangles" $A=A_{1} \times A_{2}$, where $A_{i} \in F_{i}$. - For a general set $A \in \mathcal{F}$, its section for $\omega_{1}\in \Omega_{1}$ is $A_{\omega_{1}}=\{ \omega_{2} \in \Omega_{2}:(\omega_{1},\omega_{2}) \in A \}$and similarly for $\omega_{2} \in \Omega_{2}$. In particular for rectangles, $(A_{1}\times A_{2})_{\omega_{2}}=\begin{cases} A_{1} & \text{if }\omega _{2} \in A_{2} \\ \emptyset &\text{if } \omega_{2} \not\in A_{2} \end{cases}$ - In the product measure space, the measure $m$ is called the **product measure**, given by: $m(A)=\int _{\Omega_{2}} m_{1}(A_{\omega_{2}})\, dm_{2} $note that the integrand $\omega_{2} \mapsto m_{1}(A_{\omega_{2}})$ is just a function $\Omega_{2} \to \mathbb{R}$. ### Fubini's and Tonelli's Theorems - From now on, we only work with $(\mathbb{R},\mathcal{M}_{Leb},m)^{2}$, and for clarity, the first space is labeled $(\mathbb{R}_{x},\mathcal{M}_{x},m(x))$, and the same for the second space with $y$. - Let $f: \mathbb{R}^{2} \to [0,\infty]$ be measurable in the product measure, then: - For almost all $y$, mapping $x \mapsto f(x,y)$ is measurable in $m(x)$. - Mapping $y \mapsto \int_{\mathbb{R}_{x}} f(x,y) \, dm(x)$ is measurable in $m(y)$. - Integration in $\mathbb{R}^{2}$ equals a double integral: $\int _{\mathbb{R}^{2}}f(x,y) \, d(x,y) =\int _{\mathbb{R}} \left( \int _{\mathbb{R}}f(x,y) \, dx \right) \, dy $where the inner integral can be seen as a function $y \mapsto \int _{\mathbb{R}} f(x,y) \, dx$. - *Note that $f$ is non-negative.* - **Fubini's Theorem**: let $f: \mathbb{R}^{2} \mapsto \mathbb{R}$ be integrable in the product measure. Then $\int _{\mathbb{R}}\left( \int_{\mathbb{R}} f(x,y) \, dx \right) \, dy= \int _{\mathbb{R}^{2}}f(x,y) \, d(x,y)=\int _{\mathbb{R}}\left( \int_{\mathbb{R}} f(x,y) \, dy \right) \, dx $and the double integrals exist, in the sense that (e.g. for $\mathrm{LHS}$) - The inner integrand $x \mapsto f(x,y)$ is integrable in $m(x)$, with integral $F(y) \equiv \int _{\mathbb{R}}f(x,y) \, dm(x)$. - Integrating again, $y \mapsto F(y)$ is integrable in $m(y)$. - **Tonelli's Theorem**: $f: \mathbb{R}^{2}\to \mathbb{R}$ is integrable if either of the double integrals is finite: $\begin{align*} \int _{\mathbb{R}}\left( \int_{\mathbb{R}} |f| \, dx \right) \, dy<\infty &&\text{OR}&&\int _{\mathbb{R}}\left( \int_{\mathbb{R}} |f| \, dy \right) \, dx< \infty \end{align*}$so Fubini's theorem applies to $f$. ### Worked Examples in Notes - $(8.4)$ and Sheet 1: contrapositive of Fubini. - $(8.5)$: Tonelli + Fubini (basic application and well-explained). - $(8.8)$: comparison test + Tonelli. - $(8.10,11, 12)$: change-of-variables to polar coordinates. ## Spaces of Integrable Functions ### Integrals as Norms - Given $f \in \mathcal{L}^{1}(\mathbb{R})$, $\|f\|_{1} \equiv\int |f|$ has the desired properties of a norm, except that $\|f\|_{1}=0 \not \Rightarrow f\equiv 0$, but only $\mathrm{a.e.}$. - The solution is to define the norm $\|*\|_{1}$ on space of equivalence classes of the relation $\sim$, where $f \sim g \iff f=g\,\, \mathrm{a.e.}$and the quotient space $L^{1}=\mathcal{L}^{1} / \{ f:f=0\,\,\mathrm{a.e.} \}$ forms a normed space with $\|*\|_{1}$. - The same can be done with $\|f\|_{2}\equiv \sqrt{ \int |f|^{2} }$ and $\mathcal{L}^{2}=\left\{ f: \|f\|_{2}<\infty \right\}$ by defining $L^{2}=\mathcal{L}^{2} / \{ f:f=0\,\,\mathrm{a.e.} \}$. - The difference between $\mathcal{L}^{p}$ and $L^{p}$ is not important, so it will be dropped. ### General $L^{p}$ Spaces and Inequalities - In fact, for general $p>0$, we can define $\|f\|_{p}=\left( \int |f|^p \, dx \right)^{1 / p}$and it forms a norm on the space $\mathcal{L}^{p}=\{f:\|f\|_{p}<\infty \}$. - Its linearity is straightforward; the triangular inequality less so. - **Minkowski's inequality**: for $p\ge1$, $f,g \in \mathcal{L}^{p}$, there is $\|f+g\|_{p}<\| f\|_{p}+\| g \|_{p}$. - **Hölder's Inequality**: for $1<p,q<\infty$ such that $p^{-1}+q^{-1}=1$, and $f \in \mathcal{L}^{p},g \in \mathcal{L}^{q}$, there is $fg \in \mathcal{L}^{1}$, and that $\|fg\|_{1}\le \|f\|_{p}\|g\|_{q}$ > Both proofs are algebraic manipulations and use convexity. See notes or the textbook. - The space $(\mathcal{L}^{p},\| \ast \|)_{p}$ is complete: - Any Cauchy sequence $(f_{n}) \subseteq \mathcal{L}^{p}$ converges in $\| \ast \|_{p}$ to a function $f \in \mathcal{L}^{p}$. - Also, there is a subsequence $(f_{n_{k}}) \to f\,\,\mathrm{a.e.}$ in pointwise convergence. ### Maximum Norm - Instead of $p<\infty$, letting $p=\infty$ gives the norm $\|f\|_{\infty}=\mathrm{ess\,sup}f=\inf\{ c:f<c\,\,\mathrm{a.e.} \}$on the space $\mathcal{L}^{\infty}=\{ f:\|f\|_{\infty}<\infty \}$, the set of essentially bounded functions.