Math Problems Vault

Problem 4.6.1

If $x \in [0,1]$ has a finite ternary expansion that ends in two or more $1$ 's, then $x \notin C$ .

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Remark.

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The set of dyadic rationals of $(0,1)$ is $D=\left\{ d \in (0,1): d=m / 2^{n}, n \in \mathbb {N}, m=1, \cdots , 2^{n}-1\right\}$ , written in lowest terms. The Cantor Set is constructed in stages: at Stage $1$ , remove the open middle third $J_{1/2}=(1/3,2/3)$ from $[0,1]$ , leaving the two closed intervals $[0,1/3]$ and $[2/3,1]$ . At Stage $n$ , there are $2^{n-1}$ disjoint closed intervals remaining, each of length $(1/3)^{n-1}$ ; from each, remove the open middle third, denoting these $2^{n-1}$ removed open intervals by $J_{1/2^{n}}, J_{3/2^{n}}, \ldots , J_{(2^{n}-1)/2^{n}}$ , leaving $2^{n}$ disjoint closed intervals, each of Lebesgue measure $(1/3)^{n}$ . Carrying this out for every $n \in \mathbb {N}$ removes the intervals $\left\{ J_{d}: d \in D\right\}$ from $[0,1]$ ; the Cantor Set is $C=[0,1]-\bigcup_{d \in D} J_{d}$ . Claim 1(b): $x \in C$ iff $x$ has a unique ternary expansion [terminating or nonterminating] consisting exclusively of $0$ 's and $2$ 's.

Problem 4.6.2

(a)(b)

Let $x \in C$ .

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(a)

$x$ has exactly two different ternary expansions iff there exists $d \in D$ such that $x$ is an endpoint of the removed interval $J_{d}$ .

(b)

For each $d \in D$ , either $J_{d}=(0.1,0.2)$ [base 3], or for some $n \in \mathbb {N}$ the interval $J_{d}$ has the form $(0.a_{1} \cdots a_{n} 1, 0.a_{1} \cdots a_{n} 2)$ , where $a_{1}, \cdots , a_{n} \in \left\{ 0,2\right\}$ .

Problem 4.6.3

$1 / 4 \in C$ , even though $1 / 4$ fails to be an endpoint of any removed interval $J_{d}$ . Also, $0,1 \in C$ .

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Problem 4.6.4

(a)(b)

There is a certain symmetry concerning $C$ .

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(a)

$C$ is symmetric about $\frac{1}{2}$ : $x \in C$ iff $1-x \in C$ .

(b)

Let $d \in D$ , and write $J_{d}=(\alpha , \beta )$ . Then $1-d \in D$ and $J_{1-d}=(1-\beta , 1-\alpha )$ .

Problem 4.6.5

$\left\{ x-y: x, y \in C\right\} =[-1,1]$ and $\left\{ x+y: x, y \in C\right\} =[0,2]$ , so that

\lambda (\left\{ x-y: x, y \in C\right\} )=\lambda (\left\{ x+y: x, y \in C\right\} )=2.

That is, while $\lambda (C)=0$ , both $C \oplus C$ and $C \ominus C$ have positive Lebesgue measure.

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Problem 4.6.6

$C$ is perfect: each $x \in C$ is the limit of a sequence of distinct points of $C$ .

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Problem 4.6.7

Let $I_{C}$ be the indicator function for $C$ : $I_{C}(x)$ is $1$ or $0$ as $x \in C$ or $x \notin C$ . Then $I_{C}$ is continuous on $[0,1]-C$ , and $I_{C}$ is discontinuous at each $x \in C$ .

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Problem 4.6.8

(a)(b)(c)

This problem reviews some basic analysis facts that will be used in the next part of this section.

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(a)

Every collection of disjoint nonempty open intervals is amc.

(b)

Let $f:(a, b) \rightarrow \mathbb {R}$ be monotone. Then the set of points of discontinuity of $f$ is amc, and the only discontinuities are jump discontinuities.

(c)

If $f:[a, b] \rightarrow [c, d]$ is monotone and onto, then $f$ is continuous.

Remark.

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For (a): each such interval contains a distinct rational, and there are only countably many rationals. For (b): recall that $f$ has a jump discontinuity at $x$ if and only if both $f(x-)$ and $f(x+)$ exist, where at least one of $f(x-)$ or $f(x+)$ differs from $f(x)$ . For (c): use (b).

Problem 4.6.9

We actually have $\left\{ x \in [0,1]: F'(x) \text{ exists and equals zero}\right\} \in \mathscr {B}$ .

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Remark.

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Show that this set actually equals the Borel set $\bigcup_{d \in D} J_{d}$ . The Lebesgue [Singular] Function $F: [0,1] \rightarrow \mathbb {R}$ is defined by writing $F(x)=\sup \left\{ f(y): y \in C, y \leqslant x\right\}$ for every $x \in [0,1]$ , where $f: C \rightarrow \mathbb {R}$ is defined by: if $x=0.a_{1} a_{2} \cdots$ is the unique ternary expansion of $x \in C$ consisting exclusively of $0$ 's and $2$ 's, then $f(x)=\sum_{n=1}^{\infty } \frac{a_{n}/2}{2^{n}}=0.\frac{a_{1}}{2} \frac{a_{2}}{2} \cdots$ [base 2]. Claim 3: $F=f$ on $C$ , with $F(0)=0$ and $F(1)=1$ ; $F$ is onto $[0,1]$ , nondecreasing, and continuous [in fact uniformly continuous]; and for each $d \in D$ , $F(x)=d$ for each $x \in J_{d}$ [so $F'(x)$ exists and equals $0$ on each $J_{d}$ ].

Problem 4.6.10

(a)(b)

Let $n \in \mathbb {N}$ .

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(a)

If $x, y \in C$ and $\left|x-y\right|<3^{-n}$ , then $\left|f(x)-f(y)\right| \leqslant 2^{-n}$ .

(b)

If $x, y \in [0,1]$ and $\left|x-y\right|<3^{-n}$ , then $\left|F(x)-F(y)\right| \leqslant 2^{-n}$ . This constitutes another proof that $F$ is uniformly continuous.

Problem 4.6.11

There is an alternative way of constructing the Lebesgue function. Define $g: \bigcup_{d \in D} J_{d} \rightarrow [0,1]$ by writing $g(x)=d$ for all $x \in J_{d}$ , $d \in D$ .

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(a)

$g$ is nondecreasing.

(b)

$g$ is uniformly continuous.

(c)

So far $g$ is defined only on $\bigcup_{d \in D} J_{d}$ . Define a new function $G$ with domain $[0,1]$ by writing $G(x)=\lim_{n} g(x_{n})$ , where $\left\{ x_{n}\right\}_{n=1}^{\infty }$ is a sequence of points in $\bigcup_{d \in D} J_{d}$ such that $x_{n} \rightarrow x$ . For each $x \in [0,1]$ the value of $G(x)$ does not depend on the sequence $\left\{ x_{n}\right\}_{n=1}^{\infty }$ of points in $\bigcup_{d \in D} J_{d}$ that converges to $x$ , so $G$ is well-defined.

(d)

$G=g$ on $\bigcup_{d \in D} J_{d}$ , and part (b) forces $G$ to be continuous on $[0,1]$ .

(e)

$G=F$ , where $F$ is the Lebesgue Singular Function as we have constructed it.

Problem 4.6.12

Let $D_{\alpha }$ denote a subset of $[0,1]$ constructed in the same fashion as the Cantor Set $C$ , except for the fact that [for each $n \in \mathbb {N}$ ] at the $n$ th step, instead of removing the middle third of the $2^{n-1}$ remaining closed intervals, we instead remove the middle portion of length $\alpha 3^{-n}$ , where $\alpha \in [0,1]$ . [The set $C$ corresponds to $D_{1}$ , and $[0,1]$ corresponds to $D_{0}$ .] Then $D_{\alpha }$ is closed, hence in $\mathscr {B}$ . Also, $D_{\alpha }^{c}$ is a dense subset of $[0,1]$ , and $\lambda (D_{\alpha })=1-\alpha$ .

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Remark.

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Such sets $D_{\alpha }$ for $0<\alpha <1$ are called fat Cantor Sets because they are constructed in a fashion similar to that by which $C$ was constructed, but unlike $C$ , they have positive Lebesgue measure.