[Lebesgue integral] Change of variable

hbghlyj

Integration 2017
Theorem 7.5. Suppose $\varphi:(a, b) \rightarrow(c, d)$ is one to one, onto, and differentiable, $\varphi' \geq 0$. If $f:(c, d) \rightarrow[-\infty, \infty]$ is measurable, then $f \in L^1(c, d)$ if and only if $\varphi' f \circ \varphi \in L^1(a, b)$. In this case
$$
\int_c^d f(x)\rmd x=\int_a^b f(\varphi(t)) \varphi'(t)\rmd x .
$$
Proof. By construction of Lebesgue integrals, we only need to consider the case that $f$ is non-negative and Lebesgue measurable. However, any non-negative and measurable function is the limit of an increasing sequence of non-negative simple measurable functions, by MCT and the linearity of the Lebesgue integration, we only need to show the statement is valid for $f=1_E$ where $E \subset(c, d)$ is a measurable subset. In this case $\int_c^d 1_E\rmd m=m(E)<\infty$. By definition, for every $\varepsilon>0$ there is a sequence of intervals $\left(a_n, b_n\right)$ such that $\bigcup_{n=1}^{\infty}\left(a_n, b_n\right) \supset E$ and
$$\tag{7.3}\label1
\sum_{n=1}^{\infty}\left(b_n-a_n\right)-\varepsilon \leq m(E) \leq \sum_{n=1}^{\infty}\left(b_n-a_n\right) .
$$
On the other hand,
$$
1_E(\varphi(t)) \varphi'(t) \leq \sum_{n=1}^{\infty} 1_{\left(a_n, b_n\right)}(\varphi(t)) \varphi'(t)
$$
on $(a, b)$, so that$$
\begin{aligned}
\int_a^b 1_E(\varphi(t)) \varphi'(t)\rmd t & \leq \sum_{n=1}^{\infty} \int_a^b 1_{\left(a_n, b_n\right)}(\varphi(t)) \varphi'(t) \\
& \leq \sum_{n=1}^{\infty}\left(b_n-a_n\right)
\end{aligned}
$$
which follows that, together with \eqref{1}
$$
\int_a^b 1_E(\varphi(t)) \varphi'(t)\rmd t \leq m(E)+\varepsilon=\int_c^d 1_E(x)\rmd x+\varepsilon .
$$
Since $\varepsilon>0$ is arbitrary, we have
$$
\int_a^b 1_E(\varphi(t)) \varphi'(t)\rmd t \leq \int_c^d 1_E(x)\rmd x
$$
so that
$$
\int_a^b f(\varphi(t)) \varphi'(t)\rmd t \leq \int_c^d f(x)\rmd x
$$
for non-negative measurable function $f$. Applying the inequality to the inverse of $\varphi$ we also have
$$
\int_c^d f(x)\rmd x \leq \int_a^b f(\varphi(t)) \varphi'(t)\rmd t
$$
so that we must have$$
\int_c^d f(x)\rmd x=\int_a^b f(\varphi(t)) \varphi'(t)\rmd x .
$$
which completes the proof. $\small\blacksquare$

hbghlyj

$\Bbb R$上的积分换元 是否可以看作 “1维曲线的reparameterization后, 线积分不变”

(1的线积分是曲线的长度)
The length of a differentiable curve in $\Bbb R^n$ is independent of its parameterization
\begin{aligned} \text { length of curve } & =\int_{\alpha}^{\beta}\left\|\mathbf{d}^{\prime}(s)\right\| d s \\ & =\int_{\alpha}^{\beta}\left\|(\mathbf{c} \circ p)^{\prime}(s)\right\| d s \\ & =\int_{\alpha}^{\beta}\left\|\mathbf{c}^{\prime}(p(s)) p^{\prime}(s)\right\| d s \\ & =\int_{\alpha}^{\beta}\left\|\mathbf{c}^{\prime}(s)\right\| p^{\prime}(s) d s, \quad \text { since } p^{\prime}(s) \text { is a positive scalar } \\ & =\int_{a=p(\alpha)}^{b=p(\beta)}\left\|\mathbf{c}^{\prime}(t)\right\| d t, \quad \text { where } t=p(s) \text { and } d t=p^{\prime}(s) d s\end{aligned}

hbghlyj

on $(a, b)$, so that\begin{aligned}
\int_a^b 1_E(\varphi(t)) \varphi'(t)\rmd t & \leq \sum_{n=1}^{\infty} \int_a^b 1_{\left(a_n, b_n\right)}(\varphi(t)) \varphi'(t) \\
& \color{red}{\leq \sum_{n=1}^{\infty}\left(b_n-a_n\right)}
\end{aligned}

I am a little confused:
I know the first $\leq$ interchanging $\sum$ and $\int$ follows from Theorem 6.9(MCT for series of non-negative measurable functions)
But why is the second $\leq$ ?