数学代写|MAST20026 Real Analysis

Proposition 0.1 (Exercise 8). Let $L: \mathbb{R}^d \rightarrow \mathbb{R}^d$ be a linear transformation and let $E \subset \mathbb{R}^d$ be measurable. Then $L(E) \subset \mathbb{R}^d$ is measurable.

Proof. First, note that if $A$ is compact, then $L(A)$ is compact, since $L$ is continuous. Thus if $F$ is an $F_\sigma$ set, then $L(F)$ is an $F_\sigma$ set. Let $F=\bigcup_{k=1}^{\infty} F_k$ where $F_k$ are all closed. Then
$$
L(F)=L\left(\bigcup_{k=1}^{\infty} F_k\right)=\bigcup_{k=1}^{\infty} L\left(F_k\right)
$$
so $L(F)$ can also be written as a union of closed sets, so $L(F)$ is also an $F_\sigma$ set.
Since $E$ is measurable, there exists an $F_\sigma$ set $F$ with $m(F \backslash E)=0$ by Corollary 3.5. Fix $\epsilon>0$. Then there exists a covering of $F \backslash E$ by closed cubes $\left{Q_j\right}_{j=1}^{\infty}$ with $F \backslash E \subset \bigcup_j Q_j$ and $\sum_{j=1}^{\infty}\left|Q_j\right|<\epsilon$. Then
$$
L(F \backslash E) \subset L\left(\bigcup_{j=1}^{\infty} Q_j\right)=\bigcup_{j=1}^{\infty} L\left(Q_j\right)
$$
As noted in part (b), if $Q_j$ is a closed cube of side length $l$, then $L\left(Q_j\right)$ is a closed cube of side length $c_d M l$. Thus
$$
m_\left(L\left(Q_j\right)\right)=\left|L\left(Q_j\right)\right|=\left(c_d M\right)^d\left|Q_j\right| $$ Now we can compute $m_(L(F \backslash E))$.
$$
m_(L(F \backslash E)) \leq m_\left(\cup_j L\left(Q_j\right)\right) \leq \sum_{j=1}^{\infty} m_\left(L\left(Q_j\right)\right)=\sum_{j=1}^{\infty}\left(c_d M\right)^d\left|Q_j\right|=\left(c_d M\right)^d \sum_{j=1}^{\infty}\left|Q_j\right|<\left(c_d M\right)^d \epsilon $$ Thus $m_(L(F \backslash E))=0$. Since $L(F) \backslash L(E) \subset L(F \backslash E)$, by monotonicity we get $m_*(L(F) \backslash$ $L(E))=0$. As noted at the beginning, $L(F)$ is an $F_\sigma$ set. Thus $L(E)$ differs from an $F_\sigma$ set by a set of measure zero, so $L(E)$ is measurable by Corollary 3.5 .

Proposition 0.2 (Exercise 9). There exists an open set $\mathcal{O}$ such that the boundary of the closure of $\mathcal{O}$ has positive Lebesgue measure.

Proof. Let $\hat{\mathcal{C}}$ be a Cantor-like set with constants $\left{l_k\right}_{k=1}^{\infty}$ as in Exercise 4, chosen so that $\sum_{k=1}^{\infty} l_k 2^{k-1}<1$, and so then by 4 (a), $\hat{\mathcal{C}}$ has positive measure. Let $\mathcal{O}k$ be the union of open intervals removed at the $k$ th stage of constructing $\hat{\mathcal{C}}$. Then $\mathcal{O}_k$ is a disjoint union of $2^{k-1}$ open balls of radius $l_k / 2$. Denote the centers by $a_k^j$ for $j=1, \ldots 2^{k-1}$, so we can write $$ \mathcal{O}_k=\bigcup{j=1}^{2^{k-1}} B\left(a_k^j, l_k / 2\right)
$$
Then let
$$
\mathcal{O}=\bigcup_{n=1}^{\infty} \mathcal{O}_{2 n-1}=\mathcal{O}_1 \cup \mathcal{O}_3 \cup \mathcal{O}_5 \cup \ldots
$$
We claim that the boundary of the closure of $\mathcal{O}$ contains $\hat{\mathcal{C}}$. Suppose that $x \in \hat{\mathcal{C}}$. In 4(b), we constructed a sequence $x_n$ where each $x_n \in \hat{\mathcal{C}}^c$ and $x_n \rightarrow x$ by choosing $x_k$ to be the center the interval of $\mathcal{O}_k$ nearest to $x$. We choose the subsequence $\left(x_1, x_3, x_5, \ldots\right)$. By the same reasons as in $4(\mathrm{~b})$, we have $x_n \rightarrow x$, and now we have $x_k \in \mathcal{O}$. Thus $x$ is a limit point of $\mathcal{O}$. Since $\mathcal{O}$ and $\mathcal{\mathcal { C }}$ are disjoint by construction, $x \notin \mathcal{O}$, so $x$ is in the boundary of the closure of $\mathcal{O}$. Thus $\hat{\mathcal{C}}$ is a subset of the boundary of the closure of $\mathcal{O}$.

Since $\sum_{k=1}^{\infty} l_k 2^{k-1}$ by hypothesis, $\hat{\mathcal{C}}$ has positive measure. Then by monotonicity, since the boundary of the closure of $\mathcal{O}$ contains $\hat{\mathcal{C}}$, this boundary has positive measure.

Proposition 0.3 (Exercise 10a). Let $F_n, f_n$ be as described. The for $n \geq 1, x \in[0,1]$, we have
$$
\begin{array}{r}
0 \leq f_n(x) \leq 1 \
f_{n+1}(x) \leq f_n(x)
\end{array}
$$
Thus $f_n(x)$ converges to a limit as $n \rightarrow \infty$ for each $x$.
Proof. As defined, each $F_k$ is a function from $[0,1]$ into $[0,1]$. Thus $f_n(x)=\prod_{k=1}^n F_k(x)$ is a finite product of values in $[0,1]$, so $0 \leq f_n(x) \leq 1$ for all $n \in \mathbb{N}, x \in[0,1]$. Additionally,
$$
f_{n+1}(x)=\prod_{k=1}^{n+1} F_k(x)=F_{n+1}(x) \prod_{k=1}^n F_k(x)=F_{n+1}(x) f_n(x)
$$
Since $F_{n+1}(x) \in[0,1]$, this implies that $f_{n+1}(x) \leq f_n(x)$. Hence for each fixed $x, f_n(x)$ is a bounded and nonincreasing sequence, by the Monotone Sequence Theorem, $f_n(x)$ is convergent.

Proposition 0.4 (Exercise 10b). Let $f=\lim {n \rightarrow \infty} f_n$ for $f_n$ described above. Then $f$ is discontinuous at $x \in \hat{\mathcal{C}}$. Proof. First, note that for $x \in \hat{\mathcal{C}}$, we have $$ \begin{aligned} F_k(x) & =1 \ \Longrightarrow f_n(x) & =\prod{k=1}^n F_k(x)=\prod_{k=1}^n 1=1 \
\Longrightarrow f(x) & =\lim {n \rightarrow \infty} f_n(x)=\lim {n \rightarrow \infty} 1=1
\end{aligned}
$$

Fix $x \in \hat{\mathcal{C}}$. For $n \in \mathbb{N}$, let $x_n$ be the center of the removed interval nearest to $x$. As shown in 4(b), this sequence $x_n$ converges to $x$. However, by definition of $F_n$, we have $F_n\left(x_n\right)=0$. Thus for $k \geq n$, we have
$$
f_k\left(x_n\right)=\prod_{j=1}^k F_j\left(x_n\right)=0
$$
And thus we have
$$
f\left(x_n\right)=\lim _{k \rightarrow \infty} f_k\left(x_n\right)=0
$$
for all $n$. Thus $\lim _n f\left(x_n\right)=0$ and $x_n \rightarrow x$ but $f(x)=1$. Thus $f$ is discontinuous at $x$.