Let $S \subset (0, \frac{1}{3}) \times [0, 1]$, be the set such that for each $0 < t < \frac{1}{3}$, $S \cap (\{ t \} \times [0, 1])$ is the standard [Smith-Volterra Cantor set](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set) of parameter $t$.


**Question:** Do there exist measurable subsets $E$ of $(0, \frac{1}{3})$ and $F$ of $[0, 1]$, both of nonzero Lebesgue measure such that $E \times F \subset S$, perhaps up to a set of $2$-dimensional Lebesgue measure $0$?



Does the family of fat Cantor sets contain a measurable rectangle?
No. The intuition here is that sets that are missing a lot of "diagonal strips" cannot contain large product sets (even if the strips are very narrow).


For technical reasons it is convenient to work with the left half $S\_l = S \cap (0,\frac{1}{3}) \times [0,1/2]$ of $S$, so that all the missing intervals in the Smith-Volterra Cantor set move leftwards with $t$ rather than rightwards. By symmetry it suffices to show that $S\_l$ does not contain $E \times F$ up to null sets for some positive measure $E, F$.


By construction, $S\_l$ consists of the trapezoid $T := \{ (t, x): 0 < t < 1/3; 0 \leq x \leq \frac{1-t}{2} \}$ with the sets $I\_{k, \epsilon\_2, \dots, \epsilon\_k}$ deleted for all $k \geq 2$ and $\epsilon\_2,\dots,\epsilon\_k \in \{0,1\}$, where the ``strip'' $I\_{k, \epsilon\_2, \dots, \epsilon\_k}$ consists of all pairs $(t,x)$ with $0 < t < 1/3$ and $x$ obeying the estimate
$$ \left| x - f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t) \right| < \delta\_k$$
where
$$ f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t) := \sum\_{j=2}^{k-1} \epsilon\_j \frac{1-t}{2} \dots \frac{1-t^{j-1}}{2} \frac{1+t^j}{2} + \frac{1-t}{2} \dots \frac{1-t^{k-1}}{2} \frac{1}{2}$$
and
$$ \delta\_k := \frac{1-t}{2} \dots \frac{1-t^{k-1}}{2} \frac{t^k}{2}.$$
The width parameter $\delta\_k$ is quite small, but fortunately our estimates will be uniform in this parameter and so it will not be necessary to control it.


One can think of this strip $I\_{k,\epsilon\_2,\dots,\epsilon\_k}$ as a constant-width interval $[f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t) - \delta\_k, f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t) + \delta\_k]$ moving in time. Crucially, this interval moves leftwards at roughly constant speed (so that the strip is oriented "diagonally"):


**Lemma** For any $k \geq 2$ and $\epsilon\_2,\dots,\epsilon\_k \in \{0,1\}$, the function $f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t)$ is decreasing for $0 < t < 1/3$ with $-f\_{k,\epsilon\_2,\dots,\epsilon\_k}'(t) \sim \sum\_{j=2}^{k-1} \frac{\epsilon\_j}{2^j} + \frac{1}{2^k}$.


**Proof** For $0 < t < 1/3$ and $j \geq 2$, the function $\frac{1-t}{2} \dots \frac{1-t^{j-1}}{2} \frac{1+t^j}{2}$ is comparable to $2^{-j}$, and has a negative log derivative of
$$ \frac{1}{1-t} + \frac{2t}{1-t^2} + \dots + \frac{(j-1)t^{j-2}}{1-t^{j-1}} - \frac{jt^{j-1}}{1+t^j};$$
since $jt^{j-1} < 0.9 (j-1) t^{j-2}$ (say), this quantity can be seen to be comparable to $1$, so this summand in $f\_{k,\epsilon\_2,\dots,\epsilon\_k}$ contributes $\sim \epsilon\_j 2^{-j}$ to the negative derivative. Similarly the final term $\frac{1-t}{2} \dots \frac{1-t^{k-1}}{2} \frac{1}{2}$ contributes $\sim 2^{-k}$, and the claim follows. $\Box$


Now we can establish the claim. Suppose for contradiction that $S\_l$ contained $E \times F$ up to null sets. Then by [Lebesgue's density theorem](https://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem) (or the [Lebesgue differentiation theorem](https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem)), there is a point $(t\_0,x\_0)$ in $S\_l$ such that $t\_0$ is a point of density of $E$ and $x\_0$ is a point of density of $F$. We can take $(t\_0,x\_0)$ to lie in the interior of $T$, in particular $x\_0 > 0$.


Let $k$ be large. Because the intervals in the $k^{th}$ generation of the Smith-Volterra Cantor set have length $\sim 2^{-k}$, there is $\epsilon\_2,\dots,\epsilon\_k \in \{0,1\}$ such that $|x\_0-f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t)| \sim 2^{-k}$. Note that for $k$ large enough we have
$$ x\_0 \sim f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t\_0) \sim \sum\_{j=2}^{k-1} \frac{\epsilon\_j}{2^j} + \frac{1}{2^k}$$
and so for $k$ large enough we see from the lemma that
$$ -f\_{k,\epsilon\_2,\dots,\epsilon\_k}'(t) \sim x\_0$$
for all $0 < t < 1/3$.


Now we advance the time variable $t$ from $t\_0$ to $t\_0 + \frac{1}{2^k x\_0}$, so that the interval $[f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t\_0) - \delta\_k, f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t\_0) + \delta\_k]$ moves left at roughly constant speed $\sim x\_0$, and hence travels a distance $\sim 2^{-k}$. Because of this, we see that for $C$ a sufficiently large absolute constant, the localized strip
$$ I\_{k,\epsilon\_2,\dots,\epsilon\_k} \cap ([t\_0,t\_0+\frac{1}{2^k x\_0}] \times [x\_0-C2^{-k}, x\_0+C2^{-k}])$$
contains rectangles $I\_j \times J\_j$ for $j=1,\dots,J \sim 2^k \delta\_k$, with the $I\_j$ intervals of length $\sim \delta\_k/x\_0$ boundedly overlapping in $[t\_0,t\_0+\frac{1}{2^k x\_0}]$ and the $J\_j$ intervals of length $\sim \delta\_k$ boundedly overlapping in $[x\_0-C2^{-k}, x\_0+C2^{-k}]$. For each such rectangle, either $E$ avoids $I\_j$ up to null sets, or $F$ avoids $J\_j$. From this and the pigeonhole principle, we see that either the density of $E$ in $[t\_0,t\_0+\frac{1}{2^k x\_0}]$ is less than $1-c$ for some absolute constant $c>0$, or the density of $F$ in $[x\_0-C2^{-k}, x\_0+C2^{-k}]$ is less than $1-c$. But, sending $k \to \infty$, this contradicts the fact that $t\_0, x\_0$ are points of density for $E, F$ respectively.


