# Finding an explicit & bijective function that satisfies the following properties?

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 range of $$f$$ is $$\mathbb{R}$$
3. For all real $$x_1,x_2,y_1,y_2$$, where $$-\infty and $$-\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 that no function satisfies that question.

• Crossposted from here Jul 25 at 14:57
• Conway's base 13 function is almost always 0, so it will not have property 4. Also, your requirements are redundant, since property 2 is a consequence of property 4. Jul 25 at 15:07
• Is it known that there is any function $f:\mathbb{R}\to\mathbb{R}$ at all, whose graph has positive outer measure on every rectangle in the plane? In other words, drop the measurability requirement, the bijective requirement, and weaken requirement 4 to just the first part, but with outer measure. Jul 25 at 15:32
• @DavisJohnson It is against manners here on MO to change one's question after a satisfactory answer is given. You should revert your edits to change the question, and instead accept Iosif's answer, which seems to answer your original question. Ask a new question if you want about what I asked in my comment. Jul 25 at 15:45
• New question posted here. Jul 25 at 15:56

$$\newcommand\R{\mathbb R}\newcommand\la{\lambda}$$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 $$-\infty and $$-\infty we have $$\la(([x_1,x_2]\times[y_1,y_2])\cap\{(x,f(x))\colon x\in\R\})=0.$$
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.