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


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 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.


1 Answer 1


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.

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    $\begingroup$ Can the graph have full outer measure intersected with each rectangle? $\endgroup$ Jul 25 at 17:27
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    $\begingroup$ Good idea. Yes, I believe so, this is what the argument is going to show. I'll edit. $\endgroup$ Jul 25 at 17:33
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    $\begingroup$ 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. $\endgroup$ Jul 25 at 17:38
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    $\begingroup$ 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. $\endgroup$ Jul 25 at 17:41
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    $\begingroup$ 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. $\endgroup$ Jul 25 at 17:54

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