NoteNextra-origin/pages/Math4121/Exam_reviews/Math4121_Final.md

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<b\in \mathbb{R}	b-a	b-a	b-a
[a,b),a<b\in \mathbb{R}	b-a	b-a	b-a
(a,b],a<b\in \mathbb{R}	b-a	b-a	b-a
(a,b),a<b\in \mathbb{R}	b-a	b-a	b-a

Borel Measure

Our desired property of measures:

1. Measure of interval is the length of the interval. m([a,b])=m((a,b))=m([a,b))=m((a,b])=b-a

2. Countable additivity: If S_1, S_2, \cdots, S_N are pairwise disjoint Borel measurable sets, then m(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N m(S_i)

3. Closure under set minus: If S is Borel measurable and T is Borel measurable, then S\setminus T is Borel measurable with m(S\setminus T)=m(S)-m(T)

Borel Measurable Sets

\mathcal{B} is the smallest $\sigma$-algebra that contains all closed intervals.

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<b\in \mathbb{R}	b-a
[a,b),a<b\in \mathbb{R}	b-a
(a,b],a<b\in \mathbb{R}	b-a
(a,b),a<b\in \mathbb{R}	b-a

Lebesgue Measure

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<l_1<\cdots<l_n=L, and S_i=\{x|l_i<f(x)<l_{i+1}\}

The Lebesgue integral of f over [a,b] is bounded by:

\sum_{i=0}^{n-1} l_i m(S_i)\leq \int_a^b f(x) \, dx\leq \sum_{i=0}^{n-1} l_{i+1} m(S_i)

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)<c\} is Lebesgue measurable.
4. For all c\in \mathbb{R}, the set \{x\in [a,b]|f(x)\leq c\} is Lebesgue measurable.
5. For all c<d\in \mathbb{R}, the set \{x\in [a,b]|c\leq f(x)<d\} is Lebesgue measurable.

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)<r, x\in I}|A_I f(x)-f(x)|=0. For x almost every x\in[a,b].