# Math4121 Final Review ## Guidelines There is one question from Exam 2 material. 3 T/F from Exam 1 material. The remaining questions cover the material since Exam 2 (Chapters 5 and 6 of Bressoud and my lecture notes for the final week). The format of the exam is quite similar to Exam 2, maybe a tad longer (but not twice as long, don't worry). ## Chapter 5: Measure Theory ### Jordan Measure > Content > > Let $\mathcal{C}_S^e$ be the set of all finite covers of $S$ by closed intervals ($S\subset C$, where $C$ is a finite union of closed intervals). > > Let $\mathcal{C}_S^i$ be the set of disjoint intervals that contained in $S$ ($\bigcup_{i=1}^n I_i\subset S$, where $I_i$ are disjoint intervals). > > Let $c_e(S)=\sup_{C\in\mathcal{C}_S^e} \sum_{i=1}^n |I_i|$ be the outer content of $S$. > > Let $c_i(S)=\inf_{I\in\mathcal{C}_S^i} \sum_{i=1}^n |I_i|$ be the inner content of $S$. > > _Here we use $|I|$ to denote the length of the interval $I$, in book we use volume but that's not important here._ > > The content of $S$ is defined if $c(S)=c_e(S)=c_i(S)$ Note that from this definition, **for any pairwise disjoint collection of sets** $S_1, S_2, \cdots, S_N$, we have $$ \sum_{i=1}^N c_i(S_i)\leq c_i(\bigcup_{i=1}^N S_i)\leq c_e(\bigcup_{i=1}^N S_i)\leq \sum_{i=1}^N c_e(S_i) $$ by $\sup$ and $\inf$ in the definition of $c_e(S)$ and $c_i(S)$. #### Proposition 5.1 $$ c_e(S)=c_i(S)+c_e(\partial S) $$ Note the boundary of $S$ is defined as $\partial S=\overline{S}\setminus S^\circ$ (corrected by Nathan Zhou). > Some common notations for sets: > > $S^\circ$ is the interior of $S$. $S^\circ=\{x\in S| \exists \epsilon>0, B(x,\epsilon)\subset S\}$ (largest open set contained in $S$) > > $S'$ is the set of limit points of $S$ (derived set of $S$). $S'=\{x\in \mathbb{R}^n|\forall \epsilon>0, B(x,\epsilon)\setminus \{x\}\cap S\neq \emptyset\}$ (Topological definition of limit point). > > $\overline{S}$ is the closure of $S$. $\overline{S}=S\cup S'$ (smallest closed set containing $S$) Equivalently, $\forall x\in \partial S$, $\forall \epsilon>0$, $\exists p\notin S$ and $q\notin S$ s.t. $d(x,p)<\epsilon$ and $d(x,q)<\epsilon$. So the content of $S$ is defined if and only if $c_e(\partial S)=0$. > Jordan Measurable > > A set $S$ is Jordan measurable if and only if $c_e(\partial S)=0$, ($c(S)=c_e(S)=c_i(S)$) #### Proposition 5.2 Finite additivity of content: Let $S_1, S_2, \cdots, S_N$ be a finite collection of pairwise disjoint Jordan measurable sets. $$ c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i) $$ Example for Jordan measure of sets | Set | Inner Content | Outer Content | Content | | --- | --- | --- | --- | | $\emptyset$ | 0 | 0 | 0 | | $\{q\},q\in \mathbb{R}$ | 0 | 0 | 0 | | $\{\frac{1}{n}\}_{n=1}^\infty$ | 0 | 0 | 0 | | $\{[n,n+\frac{1}{2^n}]\}_{n=1}^\infty$ | 1 | 1 | 1 | | $SVC(3)$ | 0 | 1 | Undefined | | $SVC(4)$ | 0 | $\frac{1}{2}$ | Undefined | | $Q\cap [0,1]$ | 0 | 1 | Undefined | | $[0,1]\setminus Q$ | 0 | 1 | Undefined | | $[a,b], a Sigma algebra: A $\sigma$-algebra is a collection of sets that is closed under **countable** union, intersection, and complement. > > That is: > > 1. $\emptyset\in \mathcal{B}$ > 2. If $A\in \mathcal{B}$, then $A^c\in \mathcal{B}$ > 3. If $A_1, A_2, \cdots, A_N\in \mathcal{B}$, then $\bigcup_{i=1}^N A_i\in \mathcal{B}$ #### Proposition 5.3 Borel measurable sets does not contain all Jordan measurable sets. Proof by cardinality of sets. Example for Borel measure of sets | Set | Borel Measure | | --- | --- | | $\emptyset$ | 0 | | $\{q\},q\in \mathbb{R}$ | 0 | | $\{\frac{1}{n}\}_{n=1}^\infty$ | 0 | | $\{[n,n+\frac{1}{2^n}]\}_{n=1}^\infty$ | 1 | | $SVC(3)$ | 0 | | $SVC(4)$ | 0 | | $Q\cap [0,1]$ | 0 | | $[0,1]\setminus Q$ | 1 | | $[a,b], a Lebesgue measure > > Let $\mathcal{C}$ be the set of all countable covers of $S$. > > The Lebesgue outer measure of $S$ is defined as: > > $$m_e(S)=\inf_{C\in\mathcal{C}} \sum_{i=1}^\infty |I_i|$$ > > If $S\subset[a,b]$, then the inner measure of $S$ is defined as: > > $$m_i(S)=(b-a)-m_e([a,b]\setminus S)$$ > > If $m_i(S)=m_e(S)$, then $S$ is Lebesgue measurable. #### Proposition 5.4 Subadditivity of Lebesgue outer measure: For any collection of sets $S_1, S_2, \cdots, S_N$, $$m_e(\bigcup_{i=1}^N S_i)\leq \sum_{i=1}^N m_e(S_i)$$ #### Theorem 5.5 If $S$ is bounded, then any of the following conditions imply that $S$ is Lebesgue measurable: 1. $m_e(S)=0$ 2. $S$ is countable (measure of countable set is 0) 3. $S$ is an interval > Alternative definition of Lebesgue measure > > The outer measure of $S$ is defined as the infimum of all the open sets that contain $S$. > > The inner measure of $S$ is defined as the supremum of all the closed sets that are contained in $S$. #### Theorem 5.6 Caratheodory's criterion: A set $S$ is Lebesgue measurable if and only if for any set $X$ with finite outer measure, $$m_e(X-S)=m_e(X)-m_e(X\cap S)$$ #### Lemma 5.7 Local additivity of Lebesgue outer measure: If $I_1, I_2, \cdots, I_N$ are any countable collection of **pairwise disjoint intervals** and $S$ is a bounded set, then $$ m_e\left(S\cup \bigcup_{i=1}^N I_i\right)=\sum_{i=1}^N m_e(S\cap I_i) $$ #### Theorem 5.8 Countable additivity of Lebesgue outer measure: If $S_1, S_2, \cdots, S_N$ are any countable collection of pairwise disjoint Lebesgue measurable sets, **whose union has a finite outer measure,** then $$ m_e\left(\bigcup_{i=1}^N S_i\right)=\sum_{i=1}^N m_e(S_i) $$ #### Theorem 5.9 Any finite union or intersection of Lebesgue measurable sets is Lebesgue measurable. #### Theorem 5.10 Any countable union or intersection of Lebesgue measurable sets is Lebesgue measurable. #### Corollary 5.12 Limit of a monotone sequence of Lebesgue measurable sets is Lebesgue measurable. If $S_1\subseteq S_2\subseteq S_3\subseteq \cdots$ are Lebesgue measurable sets, then $\bigcup_{i=1}^\infty S_i$ is Lebesgue measurable. And $m(\bigcup_{i=1}^\infty S_i)=\lim_{i\to\infty} m(S_i)$ If $S_1\supseteq S_2\supseteq S_3\supseteq \cdots$ are Lebesgue measurable sets, **and $S_1$ has finite measure**, then $\bigcap_{i=1}^\infty S_i$ is Lebesgue measurable. And $m(\bigcap_{i=1}^\infty S_i)=\lim_{i\to\infty} m(S_i)$ #### Theorem 5.13 Non-measurable sets (under axiom of choice) Note that $(0,1)\subseteq \bigcup_{q\in \mathbb{Q}\cap (-1,1)}(\mathcal{N}+q)\subseteq (-1,2)$ $$ \bigcup_{q\in \mathbb{Q}\cap (-1,1)}(\mathcal{N}+q) $$ is not Lebesgue measurable. ## Chapter 6: Lebesgue Integration ### Lebesgue Integral Let the partition on y-axis be $l=l_0 Definition of measurable function: > > A function $f$ is measurable if for all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)>c\}$ is Lebesgue measurable. > > Equivalently, a function $f$ is measurable if any of the following conditions hold: > > 1. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)>c\}$ is Lebesgue measurable. > 2. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)\geq c\}$ is Lebesgue measurable. > 3. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x) 4. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)\leq c\}$ is Lebesgue measurable. > 5. For all $c > Prove by using the fact$\{x\in [a,b]|f(x)\geq c\}=\bigcap_{n=1}^\infty \{x\in [a,b]|f(x)>c-\frac{1}{n}\}$ #### Proposition 6.3 If $f,g$ is a measurable function, and $k\in \mathbb{R}$, then $f+g,kf,f^2,fg,|f|$ is measurable. > Definition of almost everywhere: > > A property holds almost everywhere if it holds everywhere except for a set of Lebesgue measure 0. #### Proposition 6.4 If $f_n$ is a sequence of measurable functions, then $\limsup_{n\to\infty} f_n, \liminf_{n\to\infty} f_n$ is measurable. #### Theorem 6.5 Limit of measurable functions is measurable. > Definition of simple function: > > A simple function is a linear combination of indicator functions of Lebesgue measurable sets. #### Theorem 6.6 Measurable function as limit of simple functions. $f$ is a measurable function if and only if ffthere exists a sequence of simple functions $f_n$ s.t. $f_n\to f$ almost everywhere. ### Integration #### Proposition 6.10 Let $\phi,\psi$ be simple functions, $c\in \mathbb{R}$ and $E=E_1\cup E_2$ where $E_1\cap E_2=\emptyset$. Then 1. $\int_E \phi(x) \, dx=\int_{E_1} \phi(x) \, dx+\int_{E_2} \phi(x) \, dx$ 2. $\int_E (c\phi)(x) \, dx=c\int_E \phi(x) \, dx$ 3. $\int_E (\phi+\psi)(x) \, dx=\int_E \phi(x) \, dx+\int_E \psi(x) \, dx$ 4. If $\phi\leq \psi$ for all $x\in E$, then $\int_E \phi(x) \, dx\leq \int_E \psi(x) \, dx$ > Definition of Lebesgue integral of simple function: > > Let $\phi$ be a simple function, $\phi=\sum_{i=1}^n l_i \chi_{S_i}$ > > $$\int_E \phi(x) \, dx=\sum_{i=1}^n l_i m(S_i\cap E)$$ > Definition of Lebesgue integral of measurable function: > > Let $f$ be a nonnegative measurable function, then > > $$\int_E f(x) \, dx=\sup_{\phi\leq f} \int_E \phi(x) \, dx$$ > > If $f$ is not nonnegative, then > > $$\int_E f(x) \, dx=\int_E f^+(x) \, dx-\int_E f^-(x) \, dx$$ > > where $f^+(x)=\max(f(x),0)$ and $f^-(x)=\max(-f(x),0)$ #### Proposition 6.12 Integral over a set of measure 0 is 0. #### Theorem 6.13 If a nonnegative measurable function $f$ has integral 0 on a set $E$, then $f(x)=0$ almost everywhere on $E$. #### Theorem 6.14 Monotone convergence theorem: If $f_n$ is a sequence of monotone increasing measurable functions and $f_n\to f$ almost everywhere, and $\exists A>0$ s.t. $|\int_E f_n(x) \, dx|\leq A$ for all $n$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists almost everywhere and it's integrable on $E$ with $$ \int_E f(x) \, dx=\lim_{n\to\infty} \int_E f_n(x) \, dx $$ #### Theorem 6.19 Dominated convergence theorem: If $f_n$ is a sequence of integrable functions and $f_n\to f$ almost everywhere, and there exists a nonnegative integrable function $g$ s.t. $|f_n(x)|\leq g(x)$ for all $x\in E$ and all $n$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists almost everywhere and it's integrable on $E$ with $$ \int_E f(x) \, dx=\lim_{n\to\infty} \int_E f_n(x) \, dx $$ #### Theorem 6.20 Fatou's lemma: If $f_n$ is a sequence of nonnegative integrable functions, then $$ \int_E \liminf_{n\to\infty} f_n(x) \, dx\leq \liminf_{n\to\infty} \int_E f_n(x) \, dx $$ > Definition of Hardy-Littlewood maximal function > > Given integrable $f$m and an interval $I$, look at the averaging operator $A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(y)dy$. > > The maximal function is defined as > > $$f^*(x)=\sup_{I \text{ is an open interval}} A_I f(x)$$ ### Lebesgue's Fundamental theorem of calculus If $f$ is Lebesgue integrable on $[a,b]$, then $F(x) = \int_a^x f(t)dt$ is differentiable **almost everywhere** and $F'(x) = f(x)$ **almost everywhere**. Outline: Let $\lambda,\epsilon > 0$. Find $g$ continuous such that $\int_{\mathbb{R}}|f-g|dm < \frac{\lambda \epsilon}{5}$. To control $A_I f(x)-f(x)=(A_I(f-g)(x))+(A_I g(x)-g(x))+(g(x)-f(x))$, we need to estimate the three terms separately. Our goal is to show that $\lim_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)