数学代写|MATH2400 Mathematical Analysis

(Exercise 3a). The complement of $\mathcal{C}_{\xi}$ in $[0,1]$ is the union of open intervals of total length 1.

Proof. At each stage of the construction, we remove an open interval of length $\xi / l$ from the middle of each remaining interval, where $l$ is the length of each interval. This means that at each stage, the total length of each interval after the removal is $l(1-\xi)$. Since each closed interval from which we remove this interval is the same, the total sum of the remaining intervals after each stage is multiplied by $(1-\xi)$. We begin with $[0,1]$, so after the first stage, the remaining length is $(1-\xi)$, and after the second stage the remaining length is $(1-\xi)^2$. Generally, the remaining length after the $k$ th stage is $(1-\xi)^k$. Thus, the sum of the lengths of all removed intervals up to the $k$ th stage is $1-(1-\xi)^k$. In the limit as $k \rightarrow \infty$, this length of remaining intervals goes to 1 for $\xi \in(0,1)$.

(Exercise 2c). The Cantor-Lebesgue function $F: \mathcal{C} \rightarrow[0,1]$ is surjective.
Proof. Let $y \in[0,1]$. Then $y$ has a binary expansion $y=\sum_{k=1}^{\infty} b_k 2^{-k}$, where each $b_k$ is zero or one. Let $a_k=2 b_k$ for each $k$. Then let $x=\sum_{k=1}^{\infty} a_k 3^{-k}$. We can see from the definition of $F$ that $F(x)=y$. Furthermore, each $a_k$ is zero or two, so $x$ is in the Cantor set by Exercise 2a. Hence $F$ is surjective.

(Exercise 3b). $m_\left(\mathcal{C}{\xi}\right)=0$ Proof. After $k$ iterations of the removal process, there are $2^k$ intervals each with length $$ \frac{1-(1-\xi)^k}{2^k} $$ and the total length of these intervals is $1-(1-\xi)^k$. That is, for each $k \in \mathbb{N}$, there is a covering of $\mathcal{C}{\xi}$ of disjoint cubes (intervals) $Q_j$ with total length $1-(1-\xi)^k$. Thus $\inf \left{\sum_{j=1}^{\infty}\left|Q_j\right|\right}$ is less than or equal to $1-(1-\xi)^k$ for all $k \in \mathbb{N}$. Since $\lim k 1-(1-\xi)^k=0$, this means that this infimum is at most zero. But the infimum cannot be less than zero, since it is an infimum over sums in which all terms are positive. Thus $m\left(\mathcal{C}_{\xi}\right)=0$.

Proposition 0.7 (Exercise 4a). Let $\hat{\mathcal{C}}$ be the Cantor-type set with constants $\left(l_k\right){k=1}^{\infty}$, such that $\sum{k=1}^{\infty} l_k 2^{k-1}<1$. Then
$$
m(\hat{\mathcal{C}})=1-\sum_{k=1}^{\infty} l_k 2^{k-1}
$$
Proof. The complement of $\hat{\mathcal{C}}$ in $[0,1]$ is a countable union of disjoint intervals:
$$
[0,1] \backslash \hat{\mathcal{C}}=B\left(c_{1,1}, l_1 / 2\right) \cup B\left(c_{2,1}, l_2 / 2\right) \cup B\left(c_{2,2}, l_2 / 3\right) \cup \ldots=\bigcup_{k=1}^{\infty} \bigcup_{j=1}^{2^{k-1}} B\left(c_{k, j}, l_k / 2\right)
$$
Where $c_{k, j}$ is the center of the $j$ th interval with length $l_k$. Each of these intervals is measurable, with measure $l_k$ for some $k$. There are $2^{k-1}$ such intervals of length $l_k$ for each $k$. Thus by countable additivity of $m$, we have
$$
[0,1] \backslash \hat{\mathcal{C}}=\sum_{k=1}^{\infty} l_k 2^{k-1}
$$
which converges to less than 1 by hypothesis. By additivity,
$$
m([0,1] \backslash \hat{\mathcal{C}})+m(\hat{\mathcal{C}})=m([0,1])=1
$$
Thus
$$
m(\hat{\mathcal{C}})=1-m([0,1] \backslash \hat{\mathcal{C}})=1-\sum_{k=1}^{\infty} l_k 2^{k-1}
$$