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# 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 $a<s<b$, $f$ is bounded on $[a,b]$, and $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then
$$
\int_a^b f(x) d\alpha(x) = f(s)
$$
Key: Note the max difference can be made only occurs at $s$.
#### Theorem 6.15
Integration over step functions:
If $\alpha(x) = \sum_{i=1}^{n} c_i I(x-x_i)$ for $x\in [a,b]$, then
$$
\int_a^b f(x) d\alpha(x) = \sum_{i=1}^{n} c_i f(x_i)
$$
#### Theorem 6.21
Fundamental theorem of calculus:
Let $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $F(x) = \int_a^x f(t) d\alpha(t)$. Then
1. $F$ is continuous on $[a,b]$
2. If $f$ is continuous at $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x) = f(x)$
## Chapter 7: Sequence and Series of Functions
### Example of non-Riemann integrable function
$\lim_{m\to \infty} \lim_{n\to \infty} (\cos(m!\pi x))^{2n}=\begin{cases} 1 & x\in \mathbb{Q} \\ 0 & x\notin \mathbb{Q} \end{cases}$
This function is everywhere discontinuous and not Riemann integrable.
### Uniform Convergence
#### Definition 7.7
A sequence of functions $\{f_n\}$ converges uniformly to $f$ on $E$ if for every $\epsilon > 0$, 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. $\int_a^b (f-f_n)d\alpha \leq |f-f_n| \int_a^b d\alpha = |f-f_n| (\alpha(b)-\alpha(a))$.

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# Math4121 Exam 2 Review
Range: Chapter 2-4 of Bressoud's A Radical Approach to Lebesgue's Theory of Integration
## Chapter 2
### The Riemann-Stieltjes Integral
#### Definition of the Riemann-Stieltjes Integral
Let $f$ be a bounded function on $[a,b]$ and $\alpha$ be a bounded function on $[a,b]$.
We say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ if there exists a number $I$ such that for every $\epsilon > 0$, there exists a $\delta > 0$ such that for every partition $P = \{a = x_0, x_1, \ldots, x_n = b\}$ of $[a,b]$ with $||P|| < \delta$, we have
$$
\left| \int_a^b f \, d\alpha - I \right| < \epsilon
$$
If $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$, we write
$$
\int_a^b f \, d\alpha = I
$$
#### Darboux Sums
Let $P = \{a = x_0, x_1, \ldots, x_n = b\}$ be a partition of $[a,b]$.
The upper Darboux sum of $f$ with respect to $\alpha$ is
$$
U(f, \alpha, P) = \sum_{i=1}^n M_i (x_i - x_{i-1})
$$
where $M_i = \sup_{x \in [x_{i-1}, x_i]} f(x)$ and $\alpha_i = \sup_{x \in [x_{i-1}, x_i]} \alpha(x)$.
The lower Darboux sum of $f$ with respect to $\alpha$ is
$$
L(f, \alpha, P) = \sum_{i=1}^n m_i (x_i - x_{i-1})
$$
where $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x)$ and $\alpha_i = \inf_{x \in [x_{i-1}, x_i]} \alpha(x)$.
### 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:
Define the oscilation of $f$ on $[x_{i-1}, x_i]$ as
$$
\omega(f, [x_{i-1}, x_i]) = \sup_{x,y \in [x_{i-1}, x_i]} |f(x) - f(y)|-\inf_{x,y \in [x_{i-1}, x_i]} |f(x) - f(y)|
$$
And define continuous functions as those functions that have oscilation 0 on every subinterval of their domain.
that is, the function $f$ is continuous at $c$ if $\omega(f,c) = 0$.
And we claim that the function is integrable on $[a,b]$ if and only if the outer measure of the set of discontinuities of $f$ is 0.
> Finite cover:
>
> Given a set $S$, an **finite cover** of $S$ is a **finite** collection of open/ or closed/ or half-open intervals $\{I_1, I_2, \ldots, I_n\}$ such that $S \subseteq \bigcup_{i=1}^n I_i$. The set of all finite covers of $S$ is denoted by $\mathcal{C}_S$.
> Length of a cover:
>
> The **length** of a cover $\ell(C)$ is the sum of the lengths of the intervals in the cover. (open/closed/half-open doesn't matter.)
> 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")
Homework question: You cannot cover an interval $[a,b]$ with length $k$ with a finite cover of length strictly less than $k$.
Proceed by counting the intervals $I_i = [l_i, r_i]$ in the cover, and $r_n-l_0$ is less than or equal to $c_e(S)$ and $l_0\leq a$ and $r_n\leq b$.
#### Theorem 2.5
Given a bounded function $f$ defined on the interval $[a,b]$, let $S_\sigma$ be the points in $[a,b]$ with oscilation greater than $\sigma$.
The function $f$ is Riemann-Stieltjes integrable over $[a,b]$ if and only if $\lim_{\sigma \to 0} |S_\sigma| = 0$. That is, for every $\sigma > 0$, the outer content of $S_\sigma$ is 0.
Extra terminology:
> Dense:
>
> A set $S$ is **dense** in the interval $I$ is every open subinterval of $I$ contains a point of $S$.
>
> This is equivalent to saying that $S$ is dense in $I$ if every point of $I$ is a limit point of $S$ or a point of $S$. (proved in homework)
> Totally discontinuous:
>
> A discontinuous function is **totally discontinuous** in an interval if the set of points of continuity is not dense in that interval.
>
> In other words, there exists an open interval $I$ such that the set of points of continuity of $f$ in $I$ is empty.
> Pointwise discontinuity:
>
> A discontinuous function is **pointwise discontinuous** if the set of points of discontinuity is dense in the domain of $f$.
> Accumulation point (limit point):
>
> A point $p$ is an **accumulation point** of a set $S$ if every neighborhood of $p$ contains a point of $S$ other than $p$ itself. (That is, there exists a convergent sequence $\{p_n\}_{n=1}^\infty$ in $S$ such that $\lim_{n\to\infty} p_n = p$ and $p_n \neq p$ for all $n \in \mathbb{N}$. Proved in Rudin)
> Derived set:
>
> The **derived set** of a set $S$ is the set of all accumulation points of $S$. $S' = \{p \in \mathbb{R} \mid \forall \epsilon > 0, \exists x \in S \text{ s.t. } 0 < |x-p| < \epsilon\}$.
> Type 1 set:
>
> A set $S$ is a **type 1 set** if $S'\neq \emptyset$ and $S''=\emptyset$.
> Type $n$ set:
>
> A set $S$ is a **type $n$ set** if $S'$ is a type $n-1$ set.
> First species:
>
> A set $S$ is of **first species** if it is type $n$ for some $n\geq 0$, otherwise it is of **second species**.
$\mathbb{Q}$ is not first species since it is dense in $\mathbb{R}$ and $\mathbb{Q}' = \mathbb{R}$.
$\mathbb{R}$ is not first species.
## Chapter 3
### Topology of $\mathbb{R}$
> Open set:
>
> A set $S$ is **open** if for every $x \in S$, there exists an $\epsilon > 0$ such that $B_\epsilon(x) \subseteq S$.
> Closed set:
>
> A set $S$ is **closed** if its complement is open.
>
> Equivalently, a set $S$ is closed if it contains all of its limit points. That is $S' \subseteq S$.
> Interior of a set:
>
> The **interior** of a set $S$ is the set of all points in $S$ such that there exists an $\epsilon > 0$ such that $B_\epsilon(x) \subseteq S$. $S^\circ = \{x \in S \mid \exists \epsilon > 0 \text{ s.t. } B_\epsilon(x) \subseteq S\}$. (It is also the union of all open sets contained in $S$.)
> Closure of a set:
>
> The **closure** of a set $S$ is the set of all points that for every $\epsilon > 0$, $B_\epsilon(x) \cap S \neq \emptyset$. $\overline{S} = \{x \in \mathbb{R} \mid \forall \epsilon > 0, B_\epsilon(x) \cap S \neq \emptyset\}$.
> Boundary of a set:
>
> The **boundary** of a set $S$ is the set of all points in $S$ that are not in the interior of $S$. $\partial S = \overline{S} \setminus S^\circ$.
#### Theorem 3.4
Bolzano-Weierstrass Theorem:
Every bounded infinite set has an accumulation point.
Proof:
Let $S$ be a bounded infinite set. Cut the interval $[a,b]$ into two halves, and let $I_1$ be one with infinitely many points of $S$. (such set exists since $S$ is infinite.)
Let $I_2$ be the one half with infinitely many points of $I_1$.
By induction, we can cut the interval into two halves, and let $I_{n+1}$ be the one half with infinitely many points of $I_n$.
By the nested interval property, there exists a point $c$ that is in all $I_n$.
$c$ is an accumulation point of $S$.
QED
#### Theorem 3.6 (Heine-Borel Theorem)
For any open cover of a compact set, there exists a finite subcover.
> Compact set:
>
> A set $S$ is **compact** if every open cover of $S$ has a finite subcover. In $\mathbb{R}$, this is equivalent to being closed and bounded.
> Cardinality:
>
> The **cardinality** of $\mathbb{R}$ is $\mathfrak{c}$.
>
> The **cardinality** of $\mathbb{N}$, $\mathbb{Z}$, and $\mathbb{Q}$ is $\aleph_0$.
## Chapter 4
### Nowhere Dense set
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**.
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.

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# 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 |
| $[a,b], a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
| $[a,b),a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
| $(a,b],a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
| $(a,b),a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
### 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 $x\in[a,b]$.

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# Math4121 Lecture 1
## Chapter 5: Differentiation
### The derivative of a real function
#### Definition 5.1
Let $f$ be a real-valued function on an interval $[a,b]$ ($f: [a,b] \to \mathbb{R}$).
We say that $f$ is _differentiable_ at a point $x\in [a,b]$ if the limit
$$
\lim_{t\to x} \frac{f(t)-f(x)}{t-x}
$$
exists.
Then we defined the derivative of $f$, $f'$, a function whose domain is the set of all $x\in [a,b]$ at which $f$ is differentiable, by
$$
f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}
$$
#### Theorem 5.2
Let $f:[a,b]\to \mathbb{R}$. If $f$ is differentiable at $x\in [a,b]$, then $f$ is continuous at $x$.
Proof:
> Recall [Definition 4.5](https://notenextra.trance-0.com/Math4111/Math4111_L22#definition-45)
>
> $f$ is continuous at $x$ if $\forall \epsilon > 0, \exists \delta > 0$ such that if $|t-x| < \delta$, then $|f(t)-f(x)| < \epsilon$.
>
> Whenever you see a limit, you should think of this definition.
We need to show that $\lim_{t\to x} f(t) = f(x)$.
Equivalently, we need to show that
$$
\lim_{t\to x} (f(t)-f(x)) = 0
$$
So for $t\ne x$, since $f$ is differentiable at $x$, we have
$$
\begin{aligned}
\lim_{t\to x} (f(t)-f(x)) &= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right)(t-x) \\
&= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right) \lim_{t\to x} (t-x) \\
&= f'(x) \cdot 0 \\
&= 0
\end{aligned}
$$
Therefore, differentiable is a stronger condition than continuous.
> There exists some function that is continuous but not differentiable.
>
> For example, $f(x) = |x|$ is continuous at $x=0$, but not differentiable at $x=0$.
>
> We can see that the left-hand limit and the right-hand limit are not the same.
>
> $$ \lim_{t\to 0^-} \frac{|t|-|0|}{t-0} = -1 \quad \text{and} \quad \lim_{t\to 0^+} \frac{|t|-|0|}{t-0} = 1 $$
>
> Therefore, the limit does not exist. for $f(x) = |x|$ at $x=0$.
#### Theorem 5.3
Suppose $f$ is differentiable at $x\in [a,b]$ and $g$ is differentiable at a point $x\in [a,b]$. Then $f+g$, $fg$ and $f/g$ are differentiable at $x$, and
(a) $(f+g)'(x) = f'(x) + g'(x)$
(b) $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
(c) $\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$, provided $g(x)\ne 0$
Proof:
Since the limit of product is the product of the limits, we can use the definition of the derivative to prove the theorem.
(a)
$$
\begin{aligned}
(f+g)'(x) &= \lim_{t\to x} \frac{(f+g)(t)-(f+g)(x)}{t-x} \\
&= \lim_{t\to x} \frac{f(t)-f(x)}{t-x} + \lim_{t\to x} \frac{g(t)-g(x)}{t-x} \\
&= f'(x) + g'(x)
\end{aligned}
$$
(b)
Since $f$ is differentiable at $x$, we have $\lim_{t\to x} f(t) = f(x)$.
$$
\begin{aligned}
(fg)'(x) &= \lim_{t\to x} \left(\frac{f(t)g(t)-f(x)g(x)}{t-x}\right) \\
&= \lim_{t\to x} \left(f(t)\frac{g(t)-g(x)}{t-x} + g(x)\frac{f(t)-f(x)}{t-x}\right) \\
&= f(t) \lim_{t\to x} \frac{g(t)-g(x)}{t-x} + g(x) \lim_{t\to x} \frac{f(t)-f(x)}{t-x} \\
&= f(x)g'(x) + g(x)f'(x)
\end{aligned}
$$
(c)
$$
\begin{aligned}
\left(\frac{f}{g}\right)'(x) &= \lim_{t\to x}\left(\frac{f(t)g(x)}{g(t)g(x)} - \frac{f(x)g(x)}{g(t)g(x)}\right) \\
&= \frac{1}{g(t)g(x)}\left(\lim_{t\to x} (f(t)g(x)-f(x)g(t))\right) \\
\end{aligned}
$$

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# Math4121
Riemann integration; measurable functions; Measures; the Lebesgue integral; integrable functions; $L^p$ spaces.
## Textbook
Principles of Mathematical Analysis by Walter Rudin
A radical Approach to Lebesgue's Theory of Integration by David
## Grade
| item | percentage |
| --- | --- |
| Homework | 40% |
| Midterm 1 | 15% |
| Midterm 2 | 15% |
| Final | 30% |
## Homework
Due every Monday.
## Office Hour
Monday 1pm-2pm
Wednesday 12pm-1pm
Friday 11:30am-12:30pm