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Suppose using the lebesgue outer measure $\lambda^{*}$, we restrict $A$ to sets measurable in the Caratheodory sense, 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 graph of $f$, i.e. $\left\{(x,f(x)):x\in\mathbb{R}\right\}$ is dense in $\mathbb{R}\times\mathbb{R}$
  3. The range of $f$ is $\mathbb{R}$
  4. 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 function?

I have also asked a similar question here with an answer suggesting the function doesn’t exist.

Arbuja
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    I could find a function that satisfies the first three conditions, but point 4 seems especially tough to me. Would you happen to know if those two inequalities are even possible if we replaced the graph of $f$ by a general subset of $\mathbb{R}^2$? – Willow Wisp Jul 26 '23 at 15:28
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    @WillowWisp Actually, according to this post such a function is impossible to find. – Arbuja Jul 26 '23 at 16:20

0 Answers0