In Rudin's Real and Complex analysis, there is the change-of-variables theorem, which is Thm 7.26.
There are some conditions. I will write some of them relating to my question:
(i) $X \subset V \subset \mathbb{R}^k$, $V$ is open, $T: V \to \mathbb{R}^k$ is continuous.
(ii) $X$ is Lebesgue measurable, $T$ is one-to-one on $X$, and $T$ is differentiable at every point of $X$
Also, $Y = T(X)$ and $E = T^{-1}(A) = \{x \in V: T(x) \in A\}$. Here, $A$ is a Borel set in $\mathbb{R}^k$.
Then, why is $T(E \cap X) = A \cap Y$. Isn't it $T(E \cap X) \subset A \cap Y$? If $T(E \cap X)$ are to be equal to $A \cap Y$, shouldn't $T$ be one-to-one on $V$, not $X$? Please help me.
\begin{align*} &y\in T(E\cap X) \\ \iff &\exists x\in E\cap X : y = T(x) \\ \iff &y\in T(E)\cap T(X) \\ \iff &y\in T(T^{-1}(A)) \cap Y \\ \iff &y\in A\cap Y \end{align*}
Correction
I simply thought that $T$ is one-to-one on $X$ and $m(T(V-X)) = 0$ implies $T$ is one-to-one on $V$ and $T(E\cap X) = T(E)\cap T(X)$. However, I could check that it is wrong from $f(A\cap B)=f(A)\cap f(B)$ $\iff$ $f$ is injective. and Is a continuous function one-to-one if it is one-to-one almost everywhere on its range?. Maybe I believed in Rudin too much. Nevertheless, Theorem 7.26 holds as it is. To see this, observe that actually $m(T(E\cap X)) = m(A\cap Y)$ is used in (10). We can check $m(T(E\cap X)) = m(A\cap Y)$ from $T(E\cap X) \subset T(E)\cap T(X)$, $m(T(E-X)) \le m(T(V-X)) = 0$, and \begin{equation*} m(A\cap Y) = m(T(E)\cap T(X)) \le m(T(E)) \le m(T(E\cap X)) + m(T(E-X)) = m(T(E\cap X)). \end{equation*}