Folland Real Analysis: Theorem 2.47 I am trying to understand this theorem but I have problems with some parts of the proof:

2.47 Theorem. Suppose that $\Omega$ is an open set in $\mathbb{R}^{n}$ and $G:\Omega\to\mathbb{R}^{n}$ is a $C^1$ diffeomorphism.
a. If $f$ is a Lebesgue measurable function on $G(\Omega)$, then $f\circ G$ is Lebesgue measurable on $\Omega$. If $f\geq 0$ or $f\in L^1(G(\Omega),m)$, then
$$\int_{G(\Omega)} f(x) d x=\int_{\Omega} f \circ G(x)\left|\operatorname{det} D_x G\right| d x .$$
b. If $E\subsetΩ$ and $E \in \mathcal{L}^n$, then $G(E) \in \mathcal{L}^n$ and $m(G(E))=\int_E\left|\operatorname{det} D_x G\right| d x$.
Proof. It suffices to consider Borel measurable functions and sets. Since $G$ and $G^{−1}$ are both continuous, there are no measurability problems in this case, and the general case follows as in the proof of Theorem 2.42. A bit of notation: For $x∈\mathbb{R^n}$ and $T=(T_{ij})∈GL(n,\mathbb{R})$, we set
$$\|x\|=\max _{1 \leq j \leq n}\left|x_j\right|, \quad\|T\|=\max _{1 \leq i \leq n} \sum_{j=1}^n\left|T_{i j}\right|$$
We then have $\|T x\| \leq\|T\|\|x\|$, and $\{x:\|x-a\| \leq h\}$ is the cube of side length 2h centered at a.
Let $Q$ be a cube in $Ω$, say $Q=\{x:\|x-a\| \leq h\}$. By the mean value theorem, $g_j(x)-g_j(a)=\sum_j\left(x_j-a_j\right)\left(\partial g_j / \partial x_j\right)(y)$ for some $y$ on the line segment joning $x$ and $a$, so that for $x \in Q,\|G(x)-G(a)\| \leq h\left(\sup _{y \in Q}\left\|D_y G\right\|\right)$. In other words, $G(Q)$ is contained in a cube of side length $\sup _{y \in Q}\left\|D_y G\right\|$ times that of $Q$, so that by Theorem 2.44, $m(G(Q)) \leq\color{red}{\left(\sup _{y \in Q}\left\|D_y G\right\|\right)^n} m(Q)$. If $T \in G L(n, \mathbb{R})$, we can apply this formula with $G$ replaced by $T^{−1}\circ G$ together with Theorem 2.44 to obtain
\begin{aligned}
m(G(Q)) &=|\operatorname{det} T| m\left(T^{-1}(G(Q))\right) \\
& \leq|\operatorname{det} T|\left(\sup _{y \in Q}\left\|T^{-1} D_y G\right\|\right)^n m(Q)\hspace{1cm}(2.48)
\end{aligned}
Since $D_yG$ is continuous in $y$, for any $\varepsilon>0$ we can choose $\delta>0$ so that $\color{red}{\|(D_zG)^{-1}D_yG\|^n\leq 1+\varepsilon}$ if $y,z\in Q$ and $\|y-z\|\leq\delta$. Let us now subdivide
$Q$ into subcubes $Q_1, ... ,Q_N$ whose interiors are disjoint, whose side lengths are at
most $\delta$, and whose centers are $x_1, . . . ,x_N$ . Applying (2.48) with $Q$ replaced by $Q_j$
and with $T = D_{x_j}G$, we obtain
\begin{aligned}
m(G(Q)) &=\sum_1^N m(G(Q_j))\\
&\leq \sum_1^N |\operatorname{det} D_{x_j}G| \left(\sup _{y \in Q}\left\|(D_{x_j}G)^{-1} D_y G\right\|\right)^n m(Q_j) \\
& \leq(1+\varepsilon)\sum_1^N |\operatorname{det} D_{x_j}G| m(Q_j)
\end{aligned}

This are my questions:

*

*Where did that supreme come from and why is it raised to $n$?

*I know that by $D_yG$ continuity, for any $\varepsilon>0$ I can choose $\delta>0$ such that $\|(D_zG)^{-1}D_yG\|\leq 1+\varepsilon$, and I also know that in the proof $\|(D_zG)^{-1}D_yG\|$ it is raised to $n$ so that in the last inequality $1+\varepsilon$ is not raised to $n$, but why can I claim that $\|(D_zG)^{-1}D_yG\|^n$ is bounded by $1+\varepsilon$?

 A: For the first question, \begin{align*}
    |g_j(x) - g_j(a)| &= \left|  \sum_{k} \frac{\partial g_j}{\partial x_k}(y)(x_k-a_k) \right|
    \\&\leq \sum_{k} \left|  \frac{\partial g_j}{\partial x_k}(y)(x_k-a_k) \right|
      = \sum_{k} \left|  \frac{\partial g_j}{\partial x_k}(y)\right| \left|x_k-a_k \right|
    \\&\leq \sum_{k}\left|  \frac{\partial g_j}{\partial x_k}(y)\right| \, \left\Vert x-a \right\Vert
     \leq h \sum_{k}\left|  \frac{\partial g_j}{\partial x_k}(y)\right|
    \\&\leq h \left\Vert D_y G \right\Vert 
    \\&\leq h \sup_{y \in Q} \left\Vert D_y G \right\Vert ,
\end{align*}therefore we have $$
    \left\Vert G(x) - G(a) \right\Vert \leq h \sup_{y \in Q} \left\Vert D_y G \right\Vert .
$$To see why it is raised to $n$, define $T’$ by$$T’(x_1,…,x_j,…,x_n)=(x_1,…,\sup_{y\in Q}||D_yG||x_j,…,x_n).$$Then$$m(G(Q))\leq m(T’(Q))=|\det T’|m(Q)=(\sup_{y\in Q}||D_yG||)^nm(Q).$$
For the second question, since $n$ is a constant, we can replace $(1+\epsilon)^n$ by $1+\epsilon’$ for a smaller $\epsilon’$.
