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Math4121Introduction to Lebesgue Integration (Lecture 18)

Math4121 Lecture 18

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Small sets

A set that is nowhere dense, has zero outer content yet is uncountable.

By modifying this example, we can find similar with any outer content between 0 and 1.

Definition: Perfect Set

S[0,1]S\subseteq[0,1] is perfect if S=SS=S'.

Examples of perfect set

  • [0,1][0,1] is perfect
    • perfect sets are closed
  • Finite collection of points is not perfect because they do not have limit points.
    • perfect sets are uncountable (no countable sets can be perfect)

Middle third Cantor set

We construct the set by removing the middle third of the interval.

Let C0=[0,1]C_0=[0,1], C1=[0,13][23]C_1=[0,\frac{1}{3}]\cup[\frac{2}{3}]

Continuing this process indefinitely, we define the Cantor set as

C=n=0CnC=\bigcap_{n=0}^{\infty}C_n
  1. CnCn1C_n\subseteq C_{n-1}
  2. (Cn)=(Cn1)\ell(C_n)=\ell(C_{n-1})
  3. Each CnC_n is closed.

The algebraic expression for CnC_n, where a[0,1]a\in[0,1], we write as a decimal expansion in base 33.

a=n=1an3n a=\sum_{n=1}^{\infty} \frac{a_n}{3^n}, where an{0,1,2}a_n\in\{0,1,2\}.

In this case, C0C1C_0\to C_1 means deleting all numbers with a1=1a_1=1. (the same as deleting the interval [13,23][\frac{1}{3},\frac{2}{3}])

C1C2C_1\to C_2 means deleting all the numbers with a2=1a_2=1.$

So we can write the set as C={n=1an3n,an{0,2}}C=\left\{\sum_{n=1}^{\infty}\frac{a_n}{3^n},a_n\in\{0,2\}\right\}

Proposition 4.1

CC is perfect and nowhere dense, and outer content is 0.

Proof

(i) ce(C)=0c_e(C)=0

Let ϵ>0\epsilon>0, then n\exists n such that (23)<ϵ\left(\frac{2}{3}\right)<\epsilon. Then CnC_n is a cover of CC, and (Cn)<ϵ\ell(C_n)<\epsilon.

(ii) CC is perfect

Since CnC_n is closed, CC is closed (any intersection of closed set is closed) so CCC'\subseteq C.

Let aCa\in C, and we need to show aa is a limit point. Let ϵ>0\epsilon>0, and we need to find aC{a}a^*\in C\setminus\{a\} and aa<ϵ|a^* - a| < \epsilon. Suppose a=n=1an3n,an{0,2}a=\sum_{n=1}^{\infty} \frac{a_n}{3^n}, a_n \in \{0, 2\}, Notive that if aCa^*\in C has the expansion as aa except the k-th term.

So aa=23k|a^*-a|=\frac{2}{3^k}, which can be made arbitrarily small by choosing a sufficiently large kk. Thus, aa is a limit point of CC, proving that CC is perfect.

(iii) CC is nowhere dense

It is sufficient to show CC contains no intervals.

Any open intervals has a real number with 1 in it’s base 3 decimal expansion (proof in homework)

take some interval in (a,b)(a,b) we can change the digits that is small enough and keep the element still in the set

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