# Random math: 05

Source of most content: Gunhee Cho

# MCT and NIP

## MCT

Monotone Convergence Theorem (MCT)

$\text{If } (a_n)^\infty_{n=1} \text{ is a monotone increasing (decreasing) sequence, i.e.,} \forall n \in \mathbb{N}, \ a_n \leq (\geq) a_{n+1}. \nonumber$ $\text{ Then, } \exists \lim a_n = \sup\{a_n \mid n \in \mathbb{N} \} \ (\inf \{a_n \mid n \in \mathbb{N} \}). \nonumber$

### Proof

1. Suppose $(a_n)$ is monotone increasing and bounded above, i.e., $\emptyset \ne \{ a_n \mid n \in \mathbb{N} \} \in \mathbb{R}$ is bounded above. By least upper bound property (LUP), $\exists \sup \{a_n \mid n\in \mathbb{N} \} =: \alpha \in \mathbb{R}$. Fix any $\epsilon >0$, then $\alpha - \epsilon$ is not an upper bound of $(a_n)$. Thus, $\exists N \in \mathbb{N}$ s.t. $\alpha - \epsilon < a_N$. Since $(a_n)$ is monotone increasing, $\forall n \geq N, \ \alpha - \epsilon < a_N \leq a_n \leq \alpha < \alpha + \epsilon \Leftrightarrow \forall n \geq N, \ |a_n - \alpha | < \epsilon$. This shows that $\lim a_n = \sup \{ a_n \}$ if $(a_n)$ is bounded above.
2. If $\{ a_n \mid n \in \mathbb{N}\}$ is not bounded above, we need to show $\lim a_n = \infty$. For any $M \in \mathbb{R}$, from the Asm., $\exists N \in \mathbb{N}$ s.t. $a_N \geq M$. Since $(a_n)$ is monotone increasing, $\forall n \geq N, \ a_n \geq a_N \geq M$. Thus, $\forall M \in \mathbb{R}, \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \ a_n \geq M \Leftrightarrow \lim a_n = \infty$. .

## NIP

Nested Interved Property (NIP)

$\text{If } (I_n)^\infty_{n=1} \text{ is a sequence of bounded closed intervals s.t. } I_n := [a_n, b_n] \geq I_{n+1} \ \forall n \in \mathbb{N}. \nonumber$ $\text{Then, } \cap^\infty_{n=1} I_n = [\sup a_n, \inf b_n] \ne \emptyset. \nonumber$

### Proof

Define $I_n = [a_n, b_n], \forall n \in \mathbb{N}$. Since $I_n \geq I_{n+1}, \forall n \in \mathbb{N}$, $a_n \leq a_{n+1}, \ b_n \geq b_{n+1} \forall n \in \mathbb{N} \leadsto \forall n,m \in \mathbb{N},$ $\ a_n \leq a_{n+1} \leq \cdots \leq a_{n+m} \leq b_{n+m}\leq \cdots \leq b_n$. $\leadsto \forall n,m \in \mathbb{N}, \ a_n \leq b_m$, which eliminates dependency of index. Since $(a_n)$ is monotone increasing and bounded above, by MCT, $\exists \lim a_n = \sup a_n \leq b_m \forall m\in\mathbb{N}$. Similarly, since $(b_n)$ is monotone decreasing, by MCT, $\lim b_n = \inf b_n \leadsto \sup a_n \leq \inf b_n$. Therefore, $\forall n \in \mathbb{N}, a_n \leq \sup a_n \leq \inf b_n \leq b_n$ $\leadsto \emptyset \ne \cap^\infty_{n=1} I_n \ni [\sup a_n, \inf b_n]$

## Remark

$LUP \Leftrightarrow MCT \Leftrightarrow NIP \nonumber$

# Limsup and Liminf

Def. Given a sequence $(S_n) \subseteq \mathbb{R}, \ \limsup S_n := \inf_{n\in \mathbb{N}}\sup_{k\geq n} S_k$ and $\liminf S_n := \sup_{n\in \mathbb{N}}\inf_{k\geq n} S_k$

Given a sequence $(S_n) \subseteq \mathbb{R}$, define $a_1 := \sup_{k \geq 1}S_k =\sup \{S_k \mid k \geq 1\}, \cdots, a_n := \sup_{k\geq n}S_k = \{S_k \mid k \geq n\}$. Then, $(a_n) \subseteq \mathbb{R} \cup \{ \infty \}$ and $a_1 \geq a_2 \geq \cdots \geq a_n$ for any $n\in \mathbb{N}$, i.e., $(a_n)$ is monotone decreasing. Similarly, define $b_1 = \inf_{k \geq 1} S_k, \cdots, b_n = \inf_{k \geq n} S_k$. Then, $(b_n) \subseteq \mathbb{R} \cup \{ -\infty \}$ and $b_1 \leq b_2 \leq \cdots \leq b_n$ for any $n \in \mathbb{N}$, i.e., $(b_n)$ is monotone increasing. For each $n\in\mathbb{N}, a_n \geq b_n$. So, $a_1 \geq a_2 \geq \cdots \geq a_n \geq b_n$ and $b_1 \leq b_2 \leq \cdots \leq b_n \leq a_n$. By MCT, $\exists \lim a_n = \inf a_n = \inf_{n\in\mathbb{N}}\sup_{k\geq n}S_k$, and $\exists \lim b_n = \inf b_n = \sup_{n\in\mathbb{N}}\inf_{k\geq n}S_k$.

NB; $\liminf S_n \leq \limsup S_n$.

## Property

$\text{Given a sequence } (S_n) \subseteq \mathbb{R}. \text{ Suppose } \limsup S_n \in \mathbb{R}. \text{ Then, } \limsup S_n = \beta \text{ iff } \forall \epsilon > 0: \nonumber$
1. $\exists n_0 \in \mathbb{N}$ s.t. $\forall n \geq n_0, \ S_n < \beta + \epsilon$; all but finitely many elements of $S_n$ should be less than $\beta + \epsilon$.
2. $\forall n \in \mathbb{N}, \exists k \in \mathbb{N}$ s.t. $k \geq n, \ S_k \geq \beta - \epsilon$; infinitely many elements exist in $[\beta - \epsilon, \beta]$.

### Proof

1. $(\Rightarrow)$ Let $\limsup S_n = \inf_{n\in\mathbb{N}}\sup_{k\geq n}S_k = \beta, \ \epsilon >0$ be fixed. Observe that $\beta + \epsilon$ is not a lower bound of $\{ \sup_{k\geq n} S_k \}$. Thus, $\exists n_0 \in \mathbb{N}$ s.t. 1. $\sup_{k \geq n_0} S_k < \beta + \epsilon$. This implies that $\forall k \geq n_0, S_k < \beta + \epsilon$. Since $\forall n \in \mathbb{N} \ \beta - \epsilon < \sup S_k,\ \beta -\epsilon$ is not an upper bound of $\{S_k \mid k \geq n \} \leadsto 2. \ \exists k \in \mathbb{N}$ s.t. $k \geq n$ and $\beta -\epsilon < S_k$.
2. $(\Leftarrow)$ By 1., $\forall \epsilon >0, \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \ S_n-\beta < \epsilon$ $\Leftrightarrow \sup S_n - \beta < \epsilon \leadsto \sup_{k \geq n}S_k - \beta < \epsilon.$
By 2., $\forall \epsilon >0, \forall n \in \mathbb{N}, \ \exists k \in \mathbb{N}$ s.t. $k \geq n, \ S_k \geq \beta - \epsilon \Leftrightarrow -\epsilon \leq \sup_{k \geq n} S_k - \beta$. Therefore, $\forall \epsilon >0, \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \ | \sup_{k \geq n} S_k - \beta | < \epsilon$ which concludes proof.

## Corollary

$\text{Let } (S_n) \subseteq \mathbb{R}. \text{ Then, } \limsup S_n = \liminf S_n = \beta \text{ iff } \exists \lim S_n = \beta. \nonumber$

### Proof

1. $(\Rightarrow)$ Since $\inf_{k \geq n} S_k \leq S_n \leq \sup_{k \geq n} S_k$, by Sandwich theorem, $\liminf S_k \leq \lim S_n \leq \limsup S_k$ implies $\lim S_n = \beta$.
2. $(\Leftarrow)$ may be proved by two ways;
• Since sequences $(\inf_{k \geq n} S_k),\ (\sup_{k \geq n} S_k) \subseteq (S_n) \subseteq \mathbb{R}$, i.e., subseqeunce of $(S_n)$, and by Asm. $\lim S_n = \beta$, $\limsup S_n = \liminf S_n = \beta$.
• By Asm., $\forall \epsilon >0, \exists N_\epsilon$ s.t. $\forall n \geq N_\epsilon, \ \beta -\epsilon < S_n < \beta + \epsilon$. $\leadsto \beta -\epsilon < \inf_{k \geq n} S_k \leq S_n \leq \sup_{k \geq n} S_k < \beta + \epsilon$.

## Remark

1. Although $(S_n)$ is not convergent, $\limsup S_n$ and $\liminf S_n$ exist in $\bar{\mathbb{R}}$.
2. If $(S_n)$ is bounded above by $M\in\mathbb{R}$, $| \limsup S_n | \leq M, | \liminf S_n | \leq M$, i.e., $\limsup S_n, \liminf S_n \in \mathbb{R}$.
3. Bolzano–Weierstrass theorem: for any bounded sequence in $\mathbb{R}$, there exists a convergent subsequence.
4. For bounded sequences $(a_n), (b_n) \subseteq \mathbb{R}$, (i.e., do not assume convergence) $\liminf a_n + \liminf b_n \leq \liminf (a_n+b_n) \leq \liminf a_n + \limsup b_n \leq \limsup (a_n + b_n) \leq \limsup a_n + \limsup b_n$