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Let $I_\alpha\subset[0,1]$ be an $\alpha$-Cantor set of Lebesgue measure $\alpha$ and let $I=I_\alpha+\{1\}=\{1+x:x\in I_\alpha\}$.

Q1. What is the Lebesgue measure of the set $\{\frac{t}{s}:t,s\in I\}$?

Q2. How is $\mu(\{\frac{t}{s}:t,s\in I\}\})$ (where $\mu$ is the Lebesgue measure) compared to $\alpha$?

Context: These two questions have to do with the growth and Følner sequences in $(\mathbb R^*,\times)$?

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    $\begingroup$ Why don't you just apply $\log(1+x)$ to $I$: $J=\log(1+I)$? Then you are asking about $J-J$ (or $\exp(J-J)$. $\endgroup$ – Anthony Quas Jan 13 '17 at 6:35
  • $\begingroup$ @Anthony Quas: The set $J$ has not as much structure as the set $I$. It is not clear to me that is better to work with a simpler operation or a simpler set. $\endgroup$ – Jan-Christoph Schlage-Puchta Jan 17 '17 at 18:34
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Call a set $A\subseteq[a,b]$ thick in the interval $[a,b]$, if for all $t\in(a,b)$ we have $|A\cap[a,t]|>0.6(t-a)$ and $|A\cap[t,b]|>0.6(b-t)$. The pigeon hole principle implies that if $A, B$ are thick in $[a,b]$, $[c,d]$, respectively, $1\leq a,b,c,d\leq 2$, and $b-a, d-c$ are smaller than a certain constant, then $AB=[ac, bd]$. Since a fat Cantor set is the union of finitely many sets which are thick in certain intervals, the set $\{t/s|t, s \in I\}$ is the union of a finite number of intervals, hence for each explicitly given fat Cantor set the measure of this set can be computed.

The measure depends not only on $\alpha$, but also on the exact sequence used to define the length of the removed intervals. For example, if the Cantor set consists of two subsets, which are think in $[0,\epsilon]$, $[1-\epsilon,1]$, respectively, then the product set consists of the intervals $[1, (1+\epsilon)^2]$, $[2-\epsilon, 2+2\epsilon]$, $[(2-\epsilon)^2, 4]$, which is of measure $9\epsilon$, whereas the measure of the Cantor set can be anything between $1.2\epsilon$ and $2\epsilon$. In fact I would expect that if you remove intervals in such a way that the resulting set is rather equally spread out, then the measure of the product set might not even tend to 0 with $\alpha$. On the other hand, the example shows that if you remove most of the complement in a single step, the measure tends to 0 linearly with $\alpha$.

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