Problem. Let $h\in C(\mathbb{R})$ be a continuous function, and let $\Phi:\Omega:=[0,1]^{2}\rightarrow\mathbb{R}^{2}$ be the map defined by \begin{align*} \Phi(x_{1},x_{2}):=\left(x_{1}+h(x_{1}+x_{2}),x_{2}-h(x_{1}+x_{2})\right) \tag{1} \end{align*} What is the (Lebesgue) measure (denoted by $\left|\cdot\right|$) of the set $\Phi(\Omega)$?

Implicit in the problem statement is that $\Phi(\Omega)$ is Lebesgue measurable, but this is obvious as $\Phi$ is the composition of continuous functions, so $\Phi(\Omega)$ is a compact subset of $\mathbb{R}^{2}$.

If we knew that $h$ was $C^{1}$, then the Jacobian of $\Phi$ is \begin{align*} J\Phi(x_{1},x_{2})=\begin{bmatrix} {1+h'(x_{1}+x_{2})} & {h'(x_{1}+x_{2})} \\ {-h'(x_{1}+x_{2})} & {1-h'(x_{1}+x_{2})} \end{bmatrix} \end{align*} which has determinant $1$ everywhere. Whence, \begin{align*} \left|\Phi(\Omega)\right|=\int_{\mathbb{R}^{2}}\chi_{\Phi(\Omega)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2}=\int_{\mathbb{R}^{2}}\chi_{\Omega}(x_{1},x_{2})\left|J\Phi(x_{1},x_{2})\right|\mathrm{d}x_{1}\mathrm{d}x_{2}=1 \end{align*}

My thought was to approximate $h$ uniformly in a neighborhood of $[-2,2]$ by a smooth function $g$, so that $\left|h(x_{1}+x_{2)}-g(x_{1}+x_{2})\right|<\epsilon$ for all $(x_{1},x_{2})\in\Omega$, given $\epsilon>0$. Denote the analogue of $\Phi$ with $g$ instead of $h$ by $\Phi_{\epsilon}$. One can verify that

$$\forall x=(x_{1},x_{2})\in\Omega,\quad \left|\Phi(x)-\Phi_{\epsilon}(x)\right|=\sqrt{2}\left|h(x_{1}+x_{2})-g(x_{1}+x_{2})\right|<\sqrt{2}\epsilon$$

So $\Phi(\Omega)\subset U_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ and $\Phi_{\epsilon}(\Omega)\subset V_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi(\Omega))<\sqrt{2}\epsilon\right\}$.

We obtain that \begin{align*} 1=\left|\Phi_{\epsilon}(\Omega)\right|\leq\int_{\mathbb{R}^{2}}\chi_{V_{\epsilon}}(x)\mathrm{d}x \end{align*} Since $\Phi(\Omega)$ is compact, in particular closed, $\chi_{V_{\epsilon}}\rightarrow\chi_{\Phi(\Omega)}$ a.e. as $\epsilon\downarrow 0$. From monotone convergence, we obtain that \begin{align*} 1\leq\left|\Phi(\Omega)\right| \tag{2} \end{align*}

My issue is with establishing the reverse inequality. I know that \begin{align*} \left|\Phi(\Omega)\right|\leq\left|U_{\epsilon}\right|=1+\left|\left\{x\in\mathbb{R}^{2}:0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}\right| \end{align*} But I do not know how to control the measure of set $\left\{x\in\mathbb{R}^{2}: 0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ as $\epsilon\downarrow 0$, as $\Phi(\Omega)\setminus\Phi_{\epsilon}(\Omega)$ is contained in this set for all $\epsilon>0$.

  • 1
    $\begingroup$ great idea to approximate with a smooth function! $\endgroup$
    – orangeskid
    Commented Jul 10, 2015 at 23:19

2 Answers 2



$$\Phi^{-1}(y_1, y_2) = (y_1- h(y_1, y_2), y_2 + h(y_1, y_2)\,) $$

Adding some details:

If $\Phi_n \overset{\text{u}}{\to} \Phi$ and $K$ compact then $\mu( \Phi(K)) \ge \limsup_n \mu(\Phi_n(K))$ like you showed with $\epsilon$-neighborhoods. Now take $\Phi_n$ smooth, measure preserving like you showed, and get $\mu(\Phi(K)) \ge \mu(K)$.

Conversely, take $h_n$ smooth converging to $h$. Then $\Phi^{-1}_n \to \Phi^{-1}$. Therefore, we have

$$\mu(K) = \mu (\Phi^{-1}(\Phi(K))) \ge \limsup_n \mu (\Phi_n^{-1}(\Phi(K))$$

However, the smooth $\Phi_n$ preserve the measure so $\mu (\Phi_n^{-1}(\Phi(K)))= \mu( \phi(K))$.

$\bf{Added:}$ What OP used was to approximate $\Phi$ with smooth maps that preserve the area, and then show that $\mu(\Phi(\Omega))\ge \limsup_n \mu(\Phi_n(\Omega))$. This follows from the fact that $\Phi_n(\Omega)\to \Phi(\Omega)$ in the Hausdorff metric and then using the fact that the volume is upper continuous for that metric. However, we do have in general $$\mu(\Phi(\Omega))=\lim_{n\to \infty} \mu(\Phi_n(\Omega))$$

whenever $\Phi_n$ converges to $\Phi$ uniformly on the square $\Omega$. Indeed, for any continous map $\Psi$ defined on the square $\Omega$ the image $\Psi(\Omega)$ contains all the points $y \in \mathbb{R}^2$ whose index relative to the image of the boundary $\Psi(\partial\Omega)$ is not zero.

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    $\begingroup$ Perhaps, I'm just being dense, but I don't see how this observation gets me unstuck in my line of argument. $\endgroup$ Commented Jul 10, 2015 at 23:41
  • $\begingroup$ @Matt Rosenzweig: You are right, the argument was not that straightforward, in fact it uses your proof but with $\Phi^{-1}$ $\endgroup$
    – orangeskid
    Commented Jul 11, 2015 at 0:21
  • $\begingroup$ This looks good. One slight point is that in my original question statement, I should have chosen my approximations so that I get a sequence which converges locally uniformly on $\mathbb{R}$, not just uniformly on a fixed neighborhood. Otherwise, I don't know that $\Phi_{n}^{-1}\stackrel{u}\rightarrow\Phi^{-1}$ on $\Phi(K)$. $\endgroup$ Commented Jul 11, 2015 at 0:33
  • $\begingroup$ @Matt Rosenzweig: Yes, however you can get some bounds for the range, in fact $\Phi^{-1}$ uses again only $h$, so you are safe. $\endgroup$
    – orangeskid
    Commented Jul 11, 2015 at 0:39

Here is another proof which is not so much in the spirit of the OP (approximating $h$ by smooth functions), but which I nevertheless find interesting, in particular since we can now allow $h$ to be an arbitrary (Borel) measurable function. Furthermore, we show that $\Phi$ preserves the measure of arbitrary Borel measurable sets.

Let us first factor the map $\Phi$. In fact, if we set $$ \Psi:\mathbb{R}^{2}\to\mathbb{R}^{2},\left(x,y\right)\mapsto\left(\begin{matrix}x+h\left(y\right)\\ y \end{matrix}\right), $$ then we have \begin{eqnarray*} \left(\begin{matrix}1 & 0\\ -1 & 1 \end{matrix}\right)\Psi\left(\left(\begin{matrix}1 & 0\\ 1 & 1 \end{matrix}\right)\left(\begin{matrix}x\\ y \end{matrix}\right)\right) & = & \left(\begin{matrix}1 & 0\\ -1 & 1 \end{matrix}\right)\Psi\left(\begin{matrix}x\\ x+y \end{matrix}\right)\\ & = & \left(\begin{matrix}1 & 0\\ -1 & 1 \end{matrix}\right)\left(\begin{matrix}x+h\left(x+y\right)\\ x+y \end{matrix}\right)\\ & = & \left(\begin{matrix}x+h\left(x+y\right)\\ x+y-\left[x+h\left(x+y\right)\right] \end{matrix}\right)\\ & = & \left(\begin{matrix}x+h\left(x+y\right)\\ y-h\left(x+y\right) \end{matrix}\right)=\Phi\left(\begin{matrix}x\\ y \end{matrix}\right). \end{eqnarray*} Since the linear maps induced by the matrices $\left(\begin{smallmatrix}1 & 0\\ -1 & 1 \end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 0\\ 1 & 1 \end{smallmatrix}\right)$ are measure preserving, it suffices to show that $\Psi$ is measure preserving. Note that $\Psi$ is Borel measurable with Borel measurable inverse $$ \Psi^{-1}:\mathbb{R}^{2}\to\mathbb{R}^{2},\left(x,y\right)\mapsto\left(\begin{matrix}x-h\left(y\right)\\ y \end{matrix}\right). $$

Now, for a Borel measurable set $M\subset\mathbb{R}^{2}$ and $y\in\mathbb{R}$, let $$ M_{y}:=\left\{ x\in\mathbb{R}\,\mid\,\left(x,y\right)\in M\right\} $$ be the section of $M$ at height $y$. By Cavallieri's principle (which just means "by applying Fubini's theorem to the indicator function $\chi_{M}$"), we then have $$ \lambda_{2}\left(M\right)=\int_{\mathbb{R}}\lambda_{1}\left(M_{y}\right)\,{\rm d}y, $$ where $\lambda_{1},\lambda_{2}$ denote the $1$- and $2$-dimensional Lebesgue measure.

Now, note \begin{eqnarray*} x\in\left[\Psi\left(M\right)\right]_{y} & \Longleftrightarrow & \left(x,y\right)\in\Psi\left(M\right)\\ & \Longleftrightarrow & \left(\begin{matrix}x-h\left(y\right)\\ y \end{matrix}\right)=\Psi^{-1}\left(x,y\right)\in M\\ & \Longleftrightarrow & x-h\left(y\right)\in M_{y}\\ & \Longleftrightarrow & x\in M_{y}+h\left(y\right), \end{eqnarray*} so that translation invariance of the Lebesgue measure yields $$ \lambda_{1}\left(\left[\Psi\left(M\right)\right]_{y}\right)=\lambda_{1}\left(M_{y}+h\left(y\right)\right)=\lambda_{1}\left(M_{y}\right). $$ By Cavallieri's principle as above, we conclude $$ \lambda_{2}\left(\Psi\left(M\right)\right)=\int\lambda_{1}\left(\left[\Psi\left(M\right)\right]_{y}\right)\,{\rm d}y=\int\lambda_{1}\left(M_{y}\right)\,{\rm d}y=\lambda_{2}\left(M\right). $$

So essentially, the idea was that since all $\Psi$ does is a measurable shift (depending on $y$) in $x$-direction, all sections of $\Psi\left(M\right)$ are just translates of the sections of $M$ and hence $\lambda_{2}\left(\Psi\left(M\right)\right)=\lambda_{2}\left(M\right)$ by Cavallieri.

Concluding remark: Since $\Psi$ is measure preserving on Borel sets, it is not hard to show that $\Psi$ indeed maps Lebesgue measurable sets to Lebesgue measurable sets and is also measure preserving on these sets.

  • $\begingroup$ I really like this! I had tried an argument using the translation invariance of the Lebesgue measure but got nowhere because the idea to factor $\Phi$ didn't occur to me. One small thing: I don't think it's necessary for the problem to show that $\Psi$ is measure-preserving on Lebesgue measurable sets, as $\Omega$ and $\Phi(\Omega)$ are compact. Or am I missing something? $\endgroup$ Commented Jul 11, 2015 at 15:04
  • $\begingroup$ @MattRosenzweig: Yes, for the claim you at interested in, it is not necessary. I just wanted to mention that one can show it. $\endgroup$
    – PhoemueX
    Commented Jul 11, 2015 at 15:54
  • $\begingroup$ Sorry I thought about some more, and I'm not convinced that $\Psi$ maps Lebesgue measurable sets to Lebesgue measurable sets. Why don't we run into problems if $h$ is a Lebesgue measurable function which is not Borel measurable? $\endgroup$ Commented Jul 11, 2015 at 16:18
  • $\begingroup$ @MattRosenzweig: In my answer, I am assuming that $h$ is Borel measurable. Second, you can write each Lebesgue measurable set $L$ as $L = B \cup N$ with $B$ a Borel set (you can even take it to be $\sigma$-compact) and $N$ a Lebesgue null-set, which means that there is a Borel null-set $N'$ with $N \subset N'$. Now $\Psi(L) = \Psi(B) \cup \Psi(N)$, where $\Psi(B)$ is Borel and $\Psi(N) \subset \Psi(N')$ with $\Psi(N')$ of measure zero. This shows that $\Psi(L)$ is Lebesgue measurable. $\endgroup$
    – PhoemueX
    Commented Jul 16, 2015 at 22:04

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