Suppose using the lebesgue outer measure $\lambda^{\*}$, we restrict $A$ to sets [measurable in the Caratheodory sense](https://en.m.wikipedia.org/wiki/Carath%C3%A9odory%27s_criterion), defining the Lebesgue measure $\lambda$.


**Question:**


Does there exist an explicit and *bijective* $f:\mathbb{R}\to\mathbb{R}$, where:


1. The function $f$ is measurable
2. The range of $f$ is $\mathbb{R}$
3. For all real $x\_1,x\_2,y\_1,y\_2$, where $-\infty<x\_1<x\_2<\infty$ and $-\infty<y\_1<y\_2<\infty$:
$$\lambda(\left([x\_1,x\_2]\times[y\_1,y\_2]\right)\cap\left\{(x,f(x)):x\in\mathbb{R}\right\})>0$$


$\quad\quad\!$ and
$$\lambda(\left([x\_1,x\_2]\times[y\_1,y\_2]\right)\cap\left\{(x,f(x)):x\in\mathbb{R}\right\})\neq(x\_2-x\_1)(y\_2-y\_1)$$


**Attempt:** I'm not sure how to answer this question but I heard of [Conway’s Base-13](https://en.m.wikipedia.org/wiki/Conway_base_13_function) function?


I have also asked a similar question [here](https://math.stackexchange.com/q/4710339/1176963) with an [answer](https://math.stackexchange.com/a/4714521/1176963) suggesting that no function satisfies [that question](https://math.stackexchange.com/q/4710339/1176963).



Finding an explicit & bijective function that satisfies the following properties?
$\newcommand\R{\mathbb R}\newcommand\la{\lambda}$No. Indeed, let $G:=\{(x,f(x))\colon x\in\R\}$. Then, by the Tonelli theorem,
$$\la(G)=\int\_\R dx\,\int\_{[f(x),f(x)]}dy=\int\_\R dx\,0=0.$$
So, for all real $x\_1,x\_2,y\_1,y\_2$ such that $-\infty<x\_1<x\_2<\infty$ and $-\infty<y\_1<y\_2<\infty$ we have
$$\la(([x\_1,x\_2]\times[y\_1,y\_2])\cap\{(x,f(x))\colon x\in\R\})=0.$$




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Here we used the fact that the graph $G$ of $f$ is measurable, which follows because $f$ can be uniformly approximated from below and from above by measurable functions taking at most countable many values. Modifying this latter argument slightly, we can do even without the Tonelli theorem.


