数学代写|MATH2400 Mathematical Analysis

Let $B_1 \subset \mathbb{R}^d$ be the unit ball, $B_1=\left{x \in \mathbb{R}^d:|x|<1\right}$. Let $B=B(x, r)$ be any ball in $\mathbb{R}^d$ with center $x$ and radius $r>0$. Then $m(B)=r^d m\left(B_1\right)$.
Proof. We observe that $B=x+r B_1$, that is, $B$ is equal to a translation of a dilation of the unit ball. By translation and dilation properties of $m$, we have
$$
m(B)=m\left(x+r B_1\right)=m\left(r B_1\right)=r^d m\left(B_1\right)
$$

Let $E \subset \mathbb{R}^d$ be measurable, and let $\delta=\left(\delta_1, \delta_2, \ldots \delta_d\right)$ with each $\delta_i>0$. We define
$$
\delta E=\left{\left(\delta_1 x_1, \delta_2 x_2, \ldots \delta_d x_d\right):\left(x_1, x_2, \ldots x_d\right) \in E\right}
$$
Then $\delta E$ is measurable, and more specifically,
$$
m(\delta E)=\left(\prod_{i=1}^d \delta_i\right) m(E)
$$
Proof. First we observe that if $Q=\prod_{i=1}^d\left[a_i, b_i\right]$ is a closed rectangle, then $\delta Q=\prod_{i=1}^d\left[\delta_i a_i, \delta_i b_i\right]$. Then
$$
|\delta Q|=\prod_{i=1}^d\left(\delta_i b_i-\delta_i a_i\right)=\prod_{i=1}^d \delta_i\left(b_i-a_i\right)=\left(\prod_{i=1}^d \delta_i\right)\left(\prod_{i=1}^d\left(b_i-a_i\right)\right)=\left(\prod_{i=1}^d \delta_i\right)|Q|
$$
Let $\epsilon>0$. Since $E$ is measurable, there exists an open set $\mathcal{O}$ such that $m_(E \backslash \mathcal{O}) \leq \epsilon$. Let $\left{Q_j\right}_{j=1}^{\infty}$ be a covering of $E \backslash \mathcal{O}$ by closed cubes. Then $\left{\delta Q_j\right}$ is a covering of $\delta E \backslash \delta \mathcal{O}$ by closed cubes. (To see this, let $y \in \delta E \backslash \delta \mathcal{O}$. Then $y=\left(\delta_1 x_1, \ldots \delta_d x_d\right)$ where $x=\left(x_1, \ldots x_d\right) \in E \backslash \mathcal{O}$. Then $x \in Q_j$ for some $j$, so then $y \in \delta Q_j$.) We note that $\delta \mathcal{O}$ must be open since $\mathcal{O}$ is open. Thus $$ m_(\delta E \backslash \delta \mathcal{O}) \leq \sum_{j=1}^{\infty}\left|\delta Q_j\right|=\left(\prod_{i=1}^d \delta_i\right) \sum_{j=1}^{\infty}\left|Q_j\right| \leq\left(\prod_{i=1}^d \delta_i\right) \epsilon
$$
Since $\prod_i \delta_i$ is some constant, we can make $m_*(\delta E \backslash \delta \mathcal{O})$ arbitrarily small, hence $\delta E$ is measurable.
Now let $\left{Q_j\right}_{j=1}^{\infty}$ be a covering of $E$ by closed cubes. Then $\left{\delta Q_j\right}$ is a cover of $\delta E$ by closed cubes. (For analogous reasons to the above parenthetical note.) We also define
$$
\delta^{-1} E=\left{\left(\delta_1^{-1} x_1, \ldots \delta_d^{-1} x_d\right):\left(x_1, \ldots x_d\right) \in E\right}
$$
Using this notation, we can say that if $\left{R_j\right}_{j=1}^{\infty}$ is a covering of $\delta E$ by closed cubes, then $\left{\delta^{-1} R_j\right}$ is a cover of $E$ by closed cubes. With this identification, we can say that for any covering $\left{Q_j\right}$ of $E$ by closed cubes,
$$
\begin{aligned}
m(\delta E) & =\inf \left{\sum_{j=1}^{\infty}\left|\delta Q_j\right|\right}=\inf \left{\prod_{i=1}^d \delta_i \sum_{j=1}^{\infty}\left|Q_j\right|\right} \
& =\left(\prod_{i=1}^d \delta_i\right) \inf \left{\sum_{j=1}^{\infty}\left|Q_j\right|\right}=\left(\prod_{i=1}^d \delta_i\right) m(E)
\end{aligned}
$$

(#3 in Rudin) Prove Proposition 1.15: The axioms of multiplication imply the following statements for all $x, y$, and $z$ in a field $F$.
(a) If $x \neq 0$ and $x y=x z$ then $y=z$.
(b) If $x \neq 0$ and $x y=x$ then $y=1$.
(c) If $x \neq 0$ and $x y=1$ then $y=1 / x$.
(d) If $x \neq 0$ then $1 /(1 / x)=x$.
Note that (b) and (c) assert the uniqueness of the multiplicative identity and the multiplicative inverse respectively.

(#5 in Rudin) Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that
$$
\inf A=-\sup (-A) .
$$
Where the $\inf A$ is defined to be the greatest lower bound of $A$.