# 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?

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 $$-\infty and $$-\infty: $$\lambda^{*}(\left([x_1,x_2]\times[y_1,y_2]\right)\cap\left\{(x,f(x)):x\in\mathbb{R}\right\})>0$$

Attempt: I'm not sure how to answer this question but heard of Conway’s Base-13 function.

I have also asked a similar question here with an answer suggesting that no function satisfies the question.

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

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.

• Can the graph have full outer measure intersected with each rectangle? Jul 25 at 17:27
• Good idea. Yes, I believe so, this is what the argument is going to show. I'll edit. Jul 25 at 17:33
• This answers the question in the title, about "any function ... at all". The question in the text below asks for "an explicit $f$", and that's independent of ZFC. Jul 25 at 17:38
• I have edited to get full measure. And @AndreasBlass, yes, I had stated in my answer that because the construction uses a well-ordering of the reals, it may not be explicit enough for the OP. But if projective counts as explicit, then it is independent of ZFC. Jul 25 at 17:41
• I think a further refinement will make $f$ have positive outer measure in every rectangle, but never full outer measure. One should fix a set with positive but not full measure in every rectangle, and then make the choices for $f$ avoid that set. Jul 25 at 17:54