TITLE: A proof regarding Lebesgue measure and a differentiable function
QUESTION [3 upvotes]: Could anyone kindly provide a hint on the following problem? My guess is to do some change of variables? Thank you!

Let $f:\Bbb{R}\rightarrow\Bbb{R}$ be a continuously differentiable function. Define $\phi:\Bbb{R}^2\rightarrow\Bbb{R}^2$ by $\phi(x,y)=(x+f(x+y),y-f(x+y))$. Prove that $|\phi(E)|=|E|$ for every measurable set $E\subset \Bbb{R}^2$, where $|E|$ denotes the Lebesgue measure of $E$.

REPLY [3 votes]: Clearly, the function $\phi$ has continuous partial derivatives. Its Jacobian is given as follows: $$\mathbf J_{\phi}(x,y)=\left[\begin{array}{rr}1+f'(x+y)&f'(x+y)\\-f'(x+y)&1-f'(x+y)\end{array}\right]$$ and $\det\mathbf J_{\phi}(x,y)=1$ for all $(x,y)\in\mathbb R^2$.
It can be shown also that $\phi$ is injective. To see this, let $(x_1,y_1)\in\mathbb R^2$ and $(x_2,y_2)\in\mathbb R^2$ be such that $\phi(x_1,y_1)=\phi(x_2,y_2)$. That is,
\begin{align*}
x_1+f(x_1+y_1)=&\,x_2+f(x_2+y_2),\\
y_1-f(x_1+y_1)=&\,y_2-f(x_2+y_2).
\end{align*}
Adding the two equations yields $x_1+y_1=x_2+y_2$, from which it follows that $f(x_1+y_1)=f(x_2+y_2)$. Taking another look at the two equations above, one can conclude that $x_1=x_2$ and $y_1=y_2$. Hence, $\phi$ is injective.
Now one can use the usual change-of-variables formula (see, for example, Folland, 1999, Theorem 2.47, p. 74):
\begin{align*}
\mu(\phi(E))=&\,\int\mathbb 1_{\phi(E)}(x,y)\,\mathrm d(x,y)=\int_{\phi\left(\mathbb R^2\right)}\mathbb 1_{\phi(E)}(x,y)\,\mathrm d(x,y)\\
=&\,
\int_{\mathbb R^2}\mathbb 1_{\phi(E)}(\phi(x,y))\times\underbrace{|\det\mathbf J_{\phi}(x,y)|}_{=1}\,\mathrm d(x,y)=\int_{\mathbb R^2}\mathbb 1_{E}(x,y)\,\mathrm d(x,y)=\mu(E)
\end{align*}
whenever $E\subseteq\mathbb R^2$ is Lebesgue-measurable. Here, $\mathbb 1_X$ denotes the indicator function of the set $X\subseteq\mathbb R^2$. That $$\mathbb 1_{\phi(E)}(\phi(x,y))=\mathbb 1_E(x,y),$$
which is the same thing as $$\phi(x,y)\in\phi(E)\Longleftrightarrow (x,y)\in E$$ for all $(x,y)\in\mathbb R^2$, follows from the fact that $\phi$ is injective.

REPLY [2 votes]: Consider the change of variables
$$\begin{align}
\xi&=x+f(x+y,)\\ \eta &= y-f(x+y).
\end{align}$$
It is bijective, the inverse being
$$\begin{align}
x&=\xi-f(\xi+\eta),\\ y&= \eta+f(\xi+\eta).
\end{align}$$
Its jacobian is
$$
\frac{\partial(\xi,\eta)}{\partial(x,y)}=1.
$$ change of variables in multiple integrals gives the result.
The formula for
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