Suppose $\lambda^{\*}$ is the Lebesgue outer measure.


**Question:**


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


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


**Attempt:** I'm not sure how to answer this question but 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 [the question](https://math.stackexchange.com/q/4710339/1176963).


**Edit:** I changed the question since there is a simpler question that has to be answered.



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?
The answer is yes.


We construct $f$ by transfinite recursion using a well ordering of the reals. (So this may not be explicit enough for you.)


In fact, we can make the function $f$ bijective, with the graph of $f$ having full outer measure in every rectangle.


**Theorem.** There is a bijective function $f:\mathbb{R}\to\mathbb{R}$ with full outer measure in every rectangle in the plane.


**Proof.** There are only continuum many Borel sets, and so we may enumerate a list $(N\_\alpha,R\_\alpha)$, for $\alpha<\frak{c}$ in type continuum, where $N\_\alpha\subset R\_\alpha$ is a set with less than full measure in rectangle $R\_\alpha$ in the plane, and all such combinations arise in the list.


We now define the function $f$ in stages. At any stage $\alpha<\frak{c}$, we will have defined $f$ on fewer than continuum many points. At stage $\alpha$, we consider the set $N\_\alpha\subset R\_\alpha$ sitting inside that rectangle. Because it has has less than full measure, the complement $R\_\alpha\setminus N\_\alpha$ has positive measure, and so there must be continuum many $x$ on whose section in $R\_\alpha$ there are points in $R\_\alpha\setminus N\_\alpha$. So there must some such $x$ on which $f$ is not yet defined, and we may define $f(x)$ so that $(x,f(x))$ is one of those points in $R\_\alpha\setminus N\_\alpha$. Continue this process for continuum many stages, and then extend $f$ on the remaining points arbitrarily. Since on any rectangle, the graph of $f$ is not contained in any measurable set of less than full measure, the graph must have full outer measure on all such rectangles.


We can arrange that $f$ is onto by changing its values on a size continuum measure zero set, which will not affect the outer measure property of its graph.


We can actually make $f$ bijective, since at each stage in the construction, there will be a not-yet-defined section $x$ on which $N\_\alpha$ omits continuum many points in $R\_\alpha$, and so at that stage we can let $f(x)$ be a totally new value, while still avoiding $N\_\alpha$. And then at the end, we can change $f$ on measure zero set so as to hit all the other values, so $f$ will be bijective. $\Box$


Regarding the question of "explicit", we have seen on the other question that there can be no measurable function with a postive outer measure graph. In particular, there can be no Borel function, and this is a common sense of explicitness. Meanwhile, it is consistent with ZFC that there is a projectively definable well-ordering of the reals, and in this model the function $f$ I provide can be projectively definable, at a fairly low level of complexity. Meanwhile, it is also relatively consistent with ZFC that every projective set is measurable. So the question of whether there is an explicit function $f$ as desired, if this is taken to mean projectively definable, is independent of ZFC.


