Why is the mean value theorem used here? I am really struggling to understand the mean value theorem. I have taken calculus up through advanced calculus, and a bunch of analysis courses. I don't "get" it though. Sure, I know the statement, but I don't know how to use it.  Here is an example.

2.47 Theorem. Suppose that $\Omega$ is an open set in $\mathbb{R}^n$ and $G: \Omega \rightarrow \mathbb{R}^n$ 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 $E \subset \Omega$ 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 $x \in \mathbb{R}^n$ and $T=\left(T_{i j}\right) \in G L(n, \mathbb{R})$, 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 $2 h$ centered at $a$.
Let $Q$ be a cube in $\Omega$, say $Q=\{x:\|x-a\| \leq h\}$. By the mean value theorem, $g_j(x)-g_j(a)=\sum_j\left(x_j-a_j\right)\left(\partial g / \partial x_j\right)(y)$ 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 $2.44, m(G(Q)) \leq\left(\sup _{y \in Q}\left\|D_y G\right\|\right)^n m(Q)$. 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
$$
\begin{aligned}
m(G(Q)) &=|\operatorname{det} T| m\left(T^{-1}(G(Q))\right) \\
& \leq|\operatorname{det} T|\left(\sup _{y \in Q}\left\|T^{-1} D_y G\right\|\right)^n m(Q)
\end{aligned}
$$

Could someone help intuitively understand why the mean value theorem is used here? I want to understand the application of the theorem deeply and chew on it until it becomes an obvious clear application. It must be a very fundamental and powerful result if it is the reason Taylor's theorem works. Could someone please share with me their intuition for it?
 A: To me, the Mean Value Theorem is important because it allows you to use facts about the derivative of a function to infer facts about the function itself.
For instance: if a function is constant, its derivative is zero. That follows from the definition. But the converse requires the MVT. The proof is that $f(x) - f(a) = f'(c)(x-a)$ for some $c$ between $x$ and $a$, so if $f'(c) = 0$ for all $c$, $f(x) = f(a)$ for all $x$. This proof also reveals that this fact is only a partial converse: $f$ is constant on every component of the domain. We tacitly assumed that for each $x$, $f$ was continuous and differentiable between $x$ and $a$.
In this proof, the author wants to say that the measure of $G(Q)$ (facts about a function) is controlled by the measure of $Q$ and the supremum over $Q$ of $\Vert D_yG \Vert$ (facts about the derivative). Here some of the details that are skipped: Given $x$ and $a$, define $\gamma(t) = (1-t)a + tx$. Then $\gamma(0) = a$, $\gamma(1) = x$, and $\gamma'(t) = x-a$ for all $t$. Let $h = g_j(\gamma(t))$. Then $h(1) - h(0) = h'(c)$ for some $c$ between $0$ and $1$. Let $y = \gamma(c)$. By the chain rule,
$$
   g_j(x) - g_j(a) = h(1) - h(0) = h'(c) = \sum_{k} \frac{\partial g_j}{\partial x_k}(y)(x_k-a_k)
$$
Therefore,
\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*}
The $y$ in the penultimate line is the specific $y$ satisfying the MVT equation in the $j$th coordinate; the $y$ in the last line is a generic $y$ ranging over all of $Q$. This makes the last bound independent of $j$; therefore
$$
    \left\Vert G(x) - G(a) \right\Vert \leq h \sup_{y \in Q} \left\Vert D_y G \right\Vert 
$$
