diff --git a/pages/Math4121/Exam_reviews/Math4121_E1.md b/pages/Math4121/Exam_reviews/Math4121_E1.md new file mode 100644 index 0000000..81de67c --- /dev/null +++ b/pages/Math4121/Exam_reviews/Math4121_E1.md @@ -0,0 +1,264 @@ +# Math4121 Exam 1 Review + +Range: Chapter 5 and 6 of Rudin. We skipped (and so you will not be tested on) + +- Differentiation of Vector Valued Functions (pp. 111-113) +- Integration of Vector-Valued Function and Rectifiable Curves (pp.135-137) + +You will also not be tested on Uniform Convergence and Integration, which we cover in class on Monday 2/10. + +## Chapter 5: Differentiation + +### Definition of the Derivative + +Let $f$ be a real function defined on an closed interval $[a,b]$. We say that $f$ is differentiable at a point $x \in [a,b]$ if the following limit exists: + +$$ +f'(x) = \lim_{t\to x} \frac{f(t) - f(x)}{t - x} +$$ + +If the limit exists, we call it the derivative of $f$ at $x$ and denote it by $f'(x)$. + +#### Theorem 5.2 + +Every differentiable function is [continuous](https://notenextra.trance-0.com/Math4111/Math4111_L22#definition-45). + +The converse is not true, consider $f(x) = |x|$. + +#### Theorem 5.3 + +If $f,g$ are differentiable at $x$, then + +1. $(f+g)'(x) = f'(x) + g'(x)$ +2. $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$ +3. If $g(x) \neq 0$, then $(f/g)'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)^2$ + +#### Theorem 5.4 + +Constant function is differentiable and its derivative is $0$. + +#### Theorem 5.5 + +Chain rule: If $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then the composite function $g\circ f$ is differentiable at $x$ and + +$$ +(g\circ f)'(x) = g'(f(x))f'(x) +$$ + +#### Theorem 5.8 + +The derivative of local extremum ($\exists \delta > 0$ s.t. $f(x)\geq f(y)$ or $f(x)\leq f(y)$ for all $y\in (x-\delta,x+\delta)$) is $0$. + +#### Theorem 5.9 + +Generalized mean value theorem: If $f,g$ are differentiable on $(a,b)$, then there exists a point $x\in (a,b)$ such that + +$$ +(f(b)-f(a))g'(x) = (g(b)-g(a))f'(x) +$$ + +If we put $g(x) = x$, we get the mean value theorem. + +$$ +f(b)-f(a) = f'(x)(b-a) +$$ + +for some $x\in (a,b)$. + +#### Theorem 5.12 + +Intermediate value theorem: + +If $f$ is differentiable on $[a,b]$, for all $\lambda$ between $f'(a)$ and $f'(b)$, there exists a $c\in (a,b)$ such that $f'(x) = \lambda$. + +#### Theorem 5.13 + +L'Hôpital's rule: If $f,g$ are differentiable in $(a,b)$ and $g'(x) \neq 0$ for all $x\in (a,b)$, where $-\infty \leq a < b \leq \infty$, + +Suppose +$$ +\frac{f'(x)}{g'(x)} \to A \text{ as } x\to a +$$ + +If +$$ +f(x) \to 0, g(x) \to 0 \text{ as } x\to a +$$ + +or if + +$$ +g(x) \to \infty \text{ as } x\to a +$$ + +then + +$$ +\lim_{x\to a} \frac{f(x)}{g(x)} = A +$$ + +#### Theorem 5.15 + +Taylor's theorem: If $f$ is $n$ times differentiable on $[a,b]$, $f^{(n-1)}$ is continuous on $[a,b]$, and $f^{(n)}$ exists on $(a,b)$, for any distinct points $\alpha, \beta \in [a,b]$, there exists a point $x\in (\alpha, \beta)$ such that + +$$ +f(\beta) =\left(\sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!}(\beta-\alpha)^k\right) + \frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n +$$ + +## Chapter 6: Riemann-Stieltjes Integration + +### Definition of the Integral + +Let $\alpha$ be a monotonically increasing function on $[a,b]$. + +A partition of $[a,b]$ is a set of points $P = \{x_0, x_1, \cdots, x_n\}$ such that + +$$ +a = x_0 < x_1 < \cdots < x_n = b +$$ + +Let $\Delta \alpha_i = \alpha(x_{i}) - \alpha(x_{i-1})$ for $i = 1, \cdots, n$. + +Let $m_i = \inf \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ and $M_i = \sup \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ for $i = 1, \cdots, n$. + +The lower sum of $f$ with respect to $\alpha$ is + +$$L(f,P,\alpha) = \sum_{i=1}^{n} m_i \Delta \alpha_i$$ + +The upper sum of $f$ with respect to $\alpha$ is + +$$U(f,P,\alpha) = \sum_{i=1}^{n} M_i \Delta \alpha_i$$ + +Let $\overline{\int_a^b} f(x) d\alpha(x)=\sup_P L(f,P,\alpha)$ and $\underline{\int_a^b} f(x) d\alpha(x)=\inf_P U(f,P,\alpha)$. + +If $\overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x)$, we say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ and we write + +$$ +\int_a^b f(x) d\alpha(x) = \overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x) +$$ + +#### Theorem 6.4 + +Refinement of partition will never make the lower sum smaller or the upper sum larger. + +$$ +L(f,P,\alpha) \leq L(f,P^*,\alpha) \leq U(f,P^*,\alpha) \leq U(f,P,\alpha) +$$ + +#### Theorem 6.5 + +$\underline{\int_a^b} f(x) d\alpha(x) \leq \overline{\int_a^b} f(x) d\alpha(x)$ + +#### Theorem 6.6 + +$f\in \mathscr{R}(\alpha)$ on $[a,b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a,b]$ such that + +$$ +U(f,P,\alpha) - L(f,P,\alpha) < \epsilon +$$ + +#### Theorem 6.8 + +Every continuous function on a closed interval is Riemann-Stieltjes integrable with respect to any monotonically increasing function. + +#### Theorem 6.9 + +If $f$ is monotonically increasing on $[a,b]$ and **$\alpha$ is continuous on $[a,b]$**, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$. + +Key: We can repartition the interval $[a,b]$ using $f$. + +#### Theorem 6.10 + +If $f$ is bounded on $[a,b]$ and has only **finitely many discontinuities** on $[a,b]$, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$. + +Key: We can use the bound and partition around the points of discontinuity to make the error arbitrary small. + +#### Theorem 6.11 + +If $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $m\leq f(x) \leq M$ for all $x\in [a,b]$, and $\phi$ is a continuous function on $[m,M]$, then $\phi\circ f\in \mathscr{R}(\alpha)$ on $[a,b]$. + +_Composition of bounded integrable functions and continuous functions is integrable._ + +#### Theorem 6.12 + +Properties of the integral: + +Let $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, and $c$ be a constant. Then + +1. $f+g\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b (f(x) + g(x)) d\alpha(x) = \int_a^b f(x) d\alpha(x) + \int_a^b g(x) d\alpha(x)$ +2. $cf\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b cf(x) d\alpha(x) = c\int_a^b f(x) d\alpha(x)$ +3. $f\in \mathscr{R}(\alpha)$ on $[a,b]$ and $c\in [a,b]$, then $\int_a^b f(x) d\alpha(x) = \int_a^c f(x) d\alpha(x) + \int_c^b f(x) d\alpha(x)$. +4. **Favorite Estimate**: If $|f(x)| \leq M$ for all $x\in [a,b]$, then $\left|\int_a^b f(x) d\alpha(x)\right| \leq M(\alpha(b)-\alpha(a))$. +5. If $f\in \mathscr{R}(\beta)$ on $[a,b]$, then $\int_a^b f(x) d(\alpha+\beta) = \int_a^b f(x) d\alpha + \int_a^b f(x) d\beta$. + +#### Theorem 6.13 + +If $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, then + +1. $fg\in \mathscr{R}(\alpha)$ on $[a,b]$ +2. $|f|\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\left|\int_a^b f(x) d\alpha(x)\right| \leq \int_a^b |f(x)| d\alpha(x)$ + +Key: (1), use Theorem 6.12, 6.11 to build up $fg$ from $(f+g)^2-f^2-g^2$. (2), take $\phi(x) = |x|$ in Theorem 6.11. + +#### Theorem 6.14 + +Integration over indicator functions: + +If $a 0$, there exists a positive integer $N$ such that + +$$ +|f_n(x) - f(x)| < \epsilon \text{ for all } x\in E \text{ and } n\geq N +$$ + +If $E$ is a point, then that's the common definition of convergence. + +If we have uniform convergence, then we can swap the order of limits. + +#### Theorem 7.16 + +If $\{f_n\}\in \mathscr{R}(\alpha)$ on $[a,b]$, and $\{f_n\}$ converges uniformly to $f$ on $[a,b]$, then + +$$ +\int_a^b f(x) d\alpha(x) = \lim_{n\to \infty} \int_a^b f_n(x) d\alpha(x) +$$ + +Key: Use the definition of uniform convergence to bound the difference between the integral of the limit and the limit of the integral. $\int_a^b (f-f_n)d\alpha \leq |f-f_n| \int_a^b d\alpha = |f-f_n| (\alpha(b)-\alpha(a))$. diff --git a/pages/Math4121/Exam_reviews/Math4121_E2.md b/pages/Math4121/Exam_reviews/Math4121_E2.md index 7a4eeaf..4e2991d 100644 --- a/pages/Math4121/Exam_reviews/Math4121_E2.md +++ b/pages/Math4121/Exam_reviews/Math4121_E2.md @@ -44,6 +44,45 @@ where $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x)$ and $\alpha_i = \inf_{x \in [x_{i ### Fail of Riemann-Stieltjes Integration +Consider the function + +$$ +((x)) = \begin{cases} +x-\lfloor x \rfloor & x \in [\lfloor x \rfloor, \lfloor x \rfloor + \frac{1}{2}) \\ +0 & x=\lfloor x \rfloor + \frac{1}{2}\\ +x-\lfloor x \rfloor - 1 & x \in (\lfloor x \rfloor + \frac{1}{2}, \lfloor x \rfloor + 1] \end{cases} +$$ + +![Graph of y=((x))](https://notenextra.trance-0.com/Math4121/y=((x)).png) + +We define + +$$ +f(x) = \sum_{n=1}^{\infty} \frac{((nx))}{n^2}=\lim_{N\to\infty}\sum_{n=1}^{N} \frac{((nx))}{n^2} +$$ + +![Graph of y=f(x)](https://notenextra.trance-0.com/Math4121/sum_y=((x)).png) + +(i) The series converges uniformly over $x\in[0,1]$. + +$$ +\left|f(x)-\sum_{n=1}^{N} \frac{((nx))}{n^2}\right|\leq \sum_{n=N+1}^{\infty}\frac{|((nx))|}{n^2}\leq \sum_{n=N+1}^{\infty} \frac{1}{n^2}<\epsilon +$$ + +As a consequence, $f(x)\in \mathscr{R}$. + +(ii) $f$ has a discontinuity at every rational number with even denominator. + +$$ +\begin{aligned} +\lim_{h\to 0^+}f(\frac{a}{2b}+h)-f(\frac{a}{2b})&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))}{n^2}-\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}))}{n^2}\\ +&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\ +&=\sum_{n=1}^{\infty}\lim_{h\to 0^+}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\ +&>0 +\end{aligned} +$$ + + #### Some integrable functions are not differentiable (violates the fundamental theorem of calculus) Solve: @@ -70,7 +109,7 @@ And we claim that the function is integrable on $[a,b]$ if and only if the outer > Outer content: > -> The **outer content** of a set $S$ is the infimum of the lengths of all finite covers of $S$. $c_e(S) = \inf_{C\in \mathcal{C}_S}\ell(C)$. (e denotes "exterior") +> The **outer content** of a set $S$ is the infimum of the lengths of all **finite covers** of $S$. $c_e(S) = \inf_{C\in \mathcal{C}_S}\ell(C)$. (e denotes "exterior") Homework question: You cannot cover an interval $[a,b]$ with length $k$ with a finite cover of length strictly less than $k$. @@ -150,7 +189,7 @@ $\mathbb{R}$ is not first species. > > The **boundary** of a set $S$ is the set of all points in $S$ that are not in the interior of $S$. $\partial S = \overline{S} \setminus S^\circ$. -#### Missing Thoerem 3.4 +#### Theorem 3.4 Bolzano-Weierstrass Theorem: @@ -190,12 +229,51 @@ For any open cover of a compact set, there exists a finite subcover. A set $S$ is **nowhere dense** if there are no open intervals in which $S$ is dense. -That is equivalent to $S'$ contains no open intervals. +That is equivalent to **$S'$ contains no open intervals**. + +Note: If $S$ is nowhere dense, then $S^c$ is dense. But if $S$ is dense, $S^c$ is not necessarily nowhere dense. (Consider $\mathbb{Q}$) ### Perfect Set A set $S$ is **perfect** if $S'=S$. +Example: open intervals, Cantor set. + +#### Cantor set + +The Cantor set ($SVC(3)$) is the set of all real numbers in $[0,1]$ that can be represented in base 3 using only the digits 0 and 2. + +The outer content of the Cantor set is 0. + +#### Generalized Cantor set (SVC(n)) + +The outer content of $SVC(n)$ is $\frac{n-3}{n-2}$. + +#### Lemma 4.4 + +Osgood's Lemma: + +Let $G$ be a closed, bounded set and Let $G_1\subseteq G_2\subseteq \ldots$ and $G=\bigcup_{n=1}^{\infty} G_n$. Then $\lim_{n\to\infty} c_e(G_n)=c_e(G)$. + +Key: Using Heine-Borel Theorem. + +#### Theorem 4.5 + +Arzela-Osgood Theorem: + +Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of continuous, uniformly bounded functions on $[0,1]$ that converges pointwise to $0$. It follows that + +$$ +\lim_{n\to\infty}\int_0^1 f_n(x) \, dx = \int_0^1 \lim_{n\to\infty} f_n(x) \, dx=0 +$$ + +Key: Using Osgood's Lemma and do case analysis on bounded and unbounded parts of the Riemann-Stieltjes integral. + +#### Theorem 4.7 + +Baire Category Theorem: + +An open interval cannot be covered by a countable union of nowhere dense sets. diff --git a/pages/Math4121/Exam_reviews/Math4121_Final.md b/pages/Math4121/Exam_reviews/Math4121_Final.md new file mode 100644 index 0000000..1dc4f88 --- /dev/null +++ b/pages/Math4121/Exam_reviews/Math4121_Final.md @@ -0,0 +1,362 @@ +# 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 $\partial S=\overline{S}\setminus S$. + +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) Any set or subset of a set with $c_e(S)=0$ is Jordan measurable. -3. SVC(4) +3. $SVC(4)$ At each step, we remove $2^n$ intervals of length $\frac{1}{4^n}$. diff --git a/pages/Math4121/Math4121_L9.md b/pages/Math4121/Math4121_L9.md index 1fd0e6f..45a09be 100644 --- a/pages/Math4121/Math4121_L9.md +++ b/pages/Math4121/Math4121_L9.md @@ -75,30 +75,9 @@ Let $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$. Proof: -Property (aa), (b), (e) holds for Riemann Sums themselves. +**Property (aa), (b), (e) holds for Riemann Sums themselves.** -$$ -\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R} -$$ - -$$ -U(P,cf, \alpha) = cU(P,f,\alpha) -$$ - -For (b), notice that if $f(x)\leq g(x)$, then $\sup f(x)\leq \sup g(x)$, $U(P,f,\alpha)\leq U(P,g,\alpha)$. and $L(P,f,\alpha)\leq L(P,g,\alpha)$. - -For (e), notice that - -$$ -\begin{aligned} -\Delta (\alpha+\beta)_i &= \alpha(x_i)-\alpha(x_{i-1})+\beta(x_i)-\beta(x_{i-1}) \\ -&= \Delta \alpha_i + \Delta \beta_i -\end{aligned} -$$ - -(c),(d) are left as homework. - -For (a), Set $h(x)=f(x)+g(x)$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$ and we will show $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$. +**For (a)**, Set $h(x)=f(x)+g(x)$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$ and we will show $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$. Since $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, for any $\epsilon > 0$, there exists a partition $P_1,P_2$ of $[a, b]$ such that $U(f,P_1,\alpha)-L(f,P_1,\alpha) < \epsilon$ and $U(g,P_2,\alpha)-L(g,P_2,\alpha) < \epsilon$. @@ -108,4 +87,76 @@ So $U(P,h,\alpha)\leq U(P,f,\alpha)+U(P,g,\alpha)\leq \int_a^b f d\alpha + \int_ Since $\epsilon$ is arbitrary, $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$. + +$$ +\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R} +$$ + +$$ +U(P,cf, \alpha) = cU(P,f,\alpha) +$$ + +**For (b)**, notice that if $f(x)\leq g(x)$, then $\sup f(x)\leq \sup g(x)$, $U(P,f,\alpha)\leq U(P,g,\alpha)$. and $L(P,f,\alpha)\leq L(P,g,\alpha)$. + +**For (c)**, if $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and if $a