# Math4121 Lecture 20

## Continue on Chapter 4

### Properties of the Cantor Set

Monotonicity: If $S\subseteq T$, then $c_e(S)\leq c_e(T)$.

Sub-additivity: $c_e(S\cup T)\leq c_e(S)+c_e(T)$.

Example: $S=\mathbb{Q}\cap[0,1]$, $T=[0,1]\setminus\mathbb{Q}$.

Then $c_e(S)=1$, $c_e(T)=1$, even though $S\cap T=\emptyset$.

$S\cup T=[0,1]$, $c_e(S\cup T)=1\leq 1+1=c_e(S)+c_e(T)$

The above example shows that:

> The following is **not true**: $c_e(S\cup T)=c_e(S)+c_e(T)$ if $S\cap T=\emptyset$.

However, the following is true:

(In $\mathbb{R}$)

If $S=\bigcup_{n=1}^{\infty} I_n$, $T=\bigcup_{n=1}^{\infty} J_n$, where $I_n$ and $J_n$ are intervals, and $S\cap T=\emptyset$, then $c_e(S\cup T)=c_e(S)+c_e(T)$.

### Back to Osgood's Lemma

#### Osgood's Lemma

Let $S$ be a closed, bounded set in $\mathbb{R}$, and $S_1\subseteq S_2\subseteq \ldots$, and $S=\bigcup_{n=1}^{\infty} S_n$. Then $\lim_{k\to\infty} c_e(S_k)=c_e(S)$.

<details>
<summary>Proof of Osgood's Lemma</summary>

Trivial that $c_e(S_k)\leq c_e(S)$.

We need to show that $\forall \epsilon>0, \exists K$ such that $c_e(S_k)>c_e(S)-\epsilon$ for all $k\geq K$.

Let $U_k$ be finite union of open intervals containing $S_k$ such that $c_e(U_k)<c_e(S_k)+\frac{\epsilon}{2^k}$.

So $\{U_k\}_{k=1}^{\infty}$ are an open cover of $S$.

Since $S$ is closed and bounded in $\mathbb{R}$, it is compact.

So, $\exists N$ such that $S\subseteq \bigcup_{k=1}^{N} U_k$.

Then we split the $U_k$ into two parts:

$U_k=(U_k\cap U_N)\cup (U_k\setminus U_N)$, we denote $U_k^{(1)}=U_k\cap U_N$, $U_k^{(2)}=U_k\setminus U_N$, for $k\leq N$.

So, since $U_k^{(1)}, U_k^{(2)}$ disjoint intervals, and $S_k\subseteq U_k^{(1)}$, we have

$$
\begin{aligned}
c_e(U_k^{(1)})+c_e(U_k^{(2)})&=c_e(U_k)\\
c_e(S_k)+c_e(U_k^{(2)})&<c_e(S_k)+\frac{\epsilon}{2^k}\\
c_e(U_k^{(2)})&<\frac{\epsilon}{2^k}\\
\end{aligned}
$$

So, 

$$
\begin{aligned}
c_e(S)&\leq c_e(U)\\
&\leq c_e(U_N)+\sum_{k=1}^{N-1} c_e(U_k^{(2)})\\
&<c_e(S_N)+\frac{\epsilon}{2^{N}}+\sum_{k=1}^{N-1}\frac{\epsilon}{2^k}\\
&\leq c_e(S_N)+\epsilon\\
&<c_e(S_N)
\end{aligned}
$$

</details>

### Convergence Theorems for sequences of functions

Is

$$
\lim_{n\to\infty}\int f_n(x)\ dx=\int \lim_{n\to\infty} f_n(x)\ dx
$$

?

Yes when $f_n\to f$ uniformly.

Uniform convergence also means $\lim_{n\to\infty} \sup_{x\in [a,b]}|f_n(x)-f(x)|=0$.

But there exists some cases that does not converge to the limit but still satisfies the above condition.

#### Theorem 4.5 (Arzela-Osgood Theorem)

If $\{f_n\}_{n=1}^{\infty}$ is a sequence of continuous, uniformly bounded function and $f(x)=\lim_{n\to\infty} f_n(x)$ exists for all $x\in [a,b]$ (pointwise convergence), then

$$
\lim_{n\to\infty}\int_a^b f_n(x)\ dx=\int_a^b f(x)\ dx
$$

<details>
<summary>Proof of Arzela-Osgood Theorem (incomplete)</summary>

Define $\Gamma_{\alpha}=\{x:\forall m\in \mathbb{N} \textup{ and }\forall \delta>0, \exists n\geq m \textup{ s.t. } |y-x|<\delta \textup{ and } |f_n(y)-f_m(y)|>\alpha\}$.

_$\Gamma_{\alpha}$ is the negation of $(\alpha,\delta)$ definition of limit._

$\Gamma_{\alpha}$ is closed and nowhere dense.

Continue on next lecture.
</details>