### Begin a new journey !!

Today, I went back to blogging and I decided to recreate my personal website to begin a new journey. My upcoming posts will be fascinating topics in Math, Physics and some interesting stories of daily life around us. Thank for following and supporting me !!

### Change of Variable Formula

We will start the series of my posts with pleasant topics. In our earlier study of the integral in the one-dimensional case, we have already used an important formula for change of variable to compute some complicated integral. Such formula has the form

$\displaystyle \int_{\alpha(a)}^{\alpha(b)}f \mathrm{d}\alpha =\int_{a}^{b}f(\alpha(x))\alpha'\mathrm{d}x$
In this post, I would like to provide readers with a similar formula in the general case and its proof. The main result we will obtain is

Theorem 1. If $\varphi : D_t \rightarrow D_x$ is a diffeomorphism of a bounded open set $D_t \subset \mathbb{R}^n$ onto a set $D_x = \varphi(D_t) \subset \mathbb{R}^n$, $f$ is a function integrable over $D_x$ and is supported in $D_x$ , then $f \circ \varphi |\mathrm{det} \varphi'|$ is integrable over $D_t$. The following formula holds

$\displaystyle \int_{D_x=\varphi(D_t)} f \mathrm{d}x= \int_{D_t}f \circ \varphi |\mathrm{det}\varphi'| \mathrm{d}t$ (1)

We now divide the proof of this result into a few steps

Step 1: We now prove this result in the one-dimensional case. Firstly, we proceed with the proof for the case of $D_x, D_t$ are closed intervals in $\mathbb{R}^1$.

Indeed, let $P_x=\left\{x_0 of $D_t$ through the mapping $\varphi^{-1}$. We have $\mathrm{diam}(P_x) \to 0 \Leftrightarrow \mathrm{diam}(P_t) \to 0$ from the uniform continuity of $\varphi, \varphi^{-1}$. Since Lagrange ‘s Theorem, $|x_i-x_{i-1}|=|\varphi(t_i)-\varphi(t_{i-1}|=|\varphi'(\tau_i)|.|(t_i-t_{i-1}|$.
We now write the Riemann sums for partitions $P_x,P_t$ with chosen points $\xi_i=\varphi(\tau_i)$

$\displaystyle \sum_{i}f(\xi_i)|x_i-x_{i-1}|= \sum_{i} f(\varphi(\tau_i))|\varphi'(\tau_i)||t_i-t_{i-1}|$

Send $\mathrm{diam}(P_x), \mathrm{diam}(P_t) \to 0$ , we obtain the formula (1).

Next, We point out that (1) holds for $D_x, D_t$ are arbitrary bounded open sets in $\mathbb{R}^1$. Let $K= \mathrm{supp}f \in D_x$, It follows the boundedness and closedness of $K$ that $K$ is compact. Then, $K$ can be covered by a finite collection of closed intervals $I_i$ in $D_x$ such that two of which have no common interior points. Therefore,

$\displaystyle \int_{D_x} f \mathrm{d}x= \int_{K} f \mathrm{d}x = \sum_{i} \int_{I_i} f \mathrm{d}x = \sum_{i} \int_{\varphi^{-1}(I_i)} f \circ \varphi |det \varphi'| \mathrm{d}t =\int_{\varphi^{-1}(K)} f \circ \varphi |det \varphi'| \mathrm{d}t=\int_{D_t} f \circ \varphi |det \varphi'| \mathrm{d}t$

Step 2: Now we prove (1) for the class of diffeomorphisms have the relatively “simple” form, that is element diffeomorphism. We recall the following definition.

Definition. The diffeomorphism $\varphi: D_t \rightarrow D_x$ is elementary if its coordinate representation has the form

$x^1=\varphi^1(t^1,t^2,\dots , t^n)=t^1$
$x^2=\varphi^2(t^1,t^2,\dots , t^n)=t^2$
$\cdots$
$x^{k-1}=\varphi^{k-1}(t^1,t^2,\dots , t^n)=t^{k-1}$
$x^{k+1}=\varphi^{k+1}(t^1,t^2,\dots , t^n)=t^{k+1}$
$\cdots$
$x^{n}=\varphi^n(t^1,t^2,\dots , t^n)=t^{n}$

We need to prove (1) hold for $\varphi$ is elementary.

Indeed, assume without loss of generality that $\varphi$ changes only the nth coordinate. Denote

$D_{x^n}(\tilde{x_0}):= \left\{(\tilde{x},x^n) \in D_x | \tilde{x}=\tilde{x_0} \in \mathbb{R}^{n-1} \right\}$
$D_{t^n}(\tilde{t_0}):= \left\{(\tilde{t},t^n) \in D_t | \tilde{t}=\tilde{t_0} \in \mathbb{R}^{n-1} \right\}$
Let $D_x \subset I_{\tilde{x}} \times I_{x^n}$ with $I_{\tilde{x}}$ is a closed (n-1)-dimensional interval and $I_{x^n}$ is a closed interval of nth coordinate.
Respectively, we can choose an interval $I_{\tilde{t}} \times I_{t^n}$ contains $D_t$, taking account of the coincidence of first (n-1) coordinates of $D_t,D_x$ we can choose $I_{\tilde{t}} \equiv I_{\tilde{x}}$. By Fubini ‘s theorem,

$\displaystyle \int_{D_x}f \mathrm{d}x=\int_{ I_{\tilde{x}} \times I_{x^n}}f \chi_{D_x}\mathrm{d}x=\int_{I_{\tilde{x}}}\mathrm{d}\tilde{x} \int_{I_{x^n}} f\chi_{D_{x^n}(\tilde{x})}\mathrm{d}x^n=\int_{I_{\tilde{x}}}\mathrm{d}\tilde{x} \int_{D_{x^n}(\tilde{x})} f\mathrm{d}x^n$

$\displaystyle = \int_{I_{\tilde{t}}}\mathrm{d}\tilde{t} \int_{D_{t^n}(\tilde{t})} f(\tilde{t},\varphi^n(\tilde{t},t^n))|\partial_{t^n}\varphi^n|\mathrm{d}t^n=\int_{I_t}(f \circ \varphi|\mathrm{det} \varphi'| \chi_{D_t})\mathrm{d}t$

$\displaystyle =\int_{D_t}(f \circ \varphi|\mathrm{det} \varphi'|)\mathrm{d}t$

Where $\chi_A$ is indicator function of the set $A$ . In this computation, we have already used the fact $\partial_{t^n} \varphi^n = \det( \varphi')$.

Step 3: Now we admit without proof the following result about the local resolution of the diffeomorphism. The proof will be provided in my later post.

Lemma 1. If $g: G_{\mbox{open}} \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a diffeomorphism, then for any $x_0 \in G$ there is a neighborhood of the point in which the representation $g=f_1 \circ f_2 \cdots \circ f_m$ holds, where $f_1,f_2, \dots, f_m$ are elementary diffeomorphisms.

We need another result as a bridge to use the lemma 1, the result states as follows:
Lemma 2. If $\psi: D_{\tau} \rightarrow D_t, \varphi: D_t \rightarrow D_x$ are diffeomorphisms for each of which formula (1) holds, then it holds also for the composition $\psi \circ \varphi$

Because the proof of Lemma 2 is quite simple, so it is left as an exercise for readers.

Complete the proof. In the final step, using results that we got above, we now provide a complete proof for the general case.

For each point of $p$ of the compact set $K_t = \mbox{supp}((f \circ \varphi) |\det( \varphi')|) \subset D_t$, there exist a ball $B(p,\delta_p) \subset D_t$ in which $\varphi$ decomposes into a composition of elementary diffeomorphisms. Since the compactness of $K_t$ we can choose a finite covering $B(p_1, \delta_1), \dots , B(p_k,\delta_k)$ of $K_t$. Let $\delta = \frac{1}{2} \min\{\delta_1, \dots, \delta_2 \}$. Then any set whose diameter is smaller than $\delta$ and its closure intersects $K_t$ must be contained in at least one of the neighborhoods $B(p_1,\delta_1), B(p_2,\delta_2), \dots, B(p_k,\delta_k)$.

Let $\{I_i\}$ is a collection of disjoint closed interval cover $D_t$ such that $\mbox{diam}(I_i) < \delta$. Then intervals $\{I_j\}$ that have nonempty intersection with $K_t$ are contained in some $U_i=B(p_i,\delta_i) \subset D_t$ and the integral over $I_i$ is equal to 0 iff it has an empty intersection with $K_t$ . Then,

$\displaystyle \int_{D_t} (f \circ \varphi) |\det(\varphi')| \mathrm{d}t = \sum_{j} \int_{I_j}(f \circ \varphi) |\det(\varphi')| \mathrm{d}t$ (2)

Similarly,

$\displaystyle \int_{D_x} f(x) \mathrm{d}x = \sum_{j} \int_{\varphi(I_j)} f(x) \mathrm{d}x$ (3)

Finally, Since $I_j \subset U_i$, by Lemma 1, 2 and Step 2, we obtain that

$\int_{\varphi(I_j)} f(x) \mathrm{d}x = \int_{I_j} (f \circ \varphi)|\det(\varphi')| \mathrm{d}t$ (4)

Combining (2),(3), (4) , we get a complete proof for Theorem 1.

A straightforward corollary that we have often used is

Corollary . Let $g: A \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$ an injective, continuously differentiable such that the set $\{ x \in A| \det g'(x) = 0 \}$ has the measure zero. Then formula (1) holds for all integrable function $f: g(A) \rightarrow \mathbb{R}$.