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).

Proof of Osgood's Lemma

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}

Convergence Theorems for sequences of functions


\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

Proof of Arzela-Osgood Theorem (incomplete)

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.

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