This is a 3 part problem:

Consider a set $E\subset [0,1]$, and let $(\xi_n)$ to be a sequence of constants in $(0,1/2)$ which are defined as $(\xi_n)=(\frac{1}{2}exp({\frac{-a}{n^2}}))$ for $a>0$. Construct sets $U_{k,j}$ which will be the open middle intervals deleted at various stages in the construction. Consider the examples: $U_{1,1}=(\xi_1, 1-\xi_1)$, $U_{2,1}=(\xi_{1}\xi_{2}, (1-\xi_2)\xi_{1})$

Define $E_n=[0,1]\setminus \bigl(\bigcup_{k=1}^{n} \bigcup_{j=1}^{2^{k-1}}U_{k,j} \bigr)$

for each $n\geq 1$ we have $2^{n-1}$ open intervals $U_{n,j}$ which are middle segments of length $1-2\xi_n$. Let $E=\bigcap_{n\geq 1}E_n$

Note if $(\xi_n)=(1/3)$ when we have the classic ternary cantor set.


$(1)$ find Lebesgue measure $m(E_n)$ and $lim_{n\rightarrow \infty} m(E_n)$.

I know for the ordinary cantor set, when we take the limit, we get $lim_{n\rightarrow \infty}m(E_n)=(\frac{2}{3})^n=0$, because at the $nth$ stage of construction, we have closed sets of length $(\frac{2}{3})^n$ remaining. I know this set is supposed to have positive measure but I am not finding the infinite sum to show this

$(2)$ prove this is a Cantor set (perfect set that's nowhere dense)

I think this one more or less follows the proof that the ordinary cantor set is nowhere dense, but I wouldn't be surprised if I made a mistake

$(3)$ Let $s\in (0,1)$. Find a sequence $(\xi_n)$ where $lim_{n\rightarrow \infty}m(E_n)=s$

I tried a few series, but nothing worked out.


I am not sure if this is the construction you are trying to to do. Fix a real number $\epsilon\in (0,1)$ such that $0<\epsilon <1/3.$ Let $\delta = ((1/3)-\epsilon)$. Then $0 < \delta < 1/3$. Now, repeat the construction of the usual middle-thirds Cantor set as follows. Remove an open segment of width ${\delta}/3$ centered at $1/2$. In the next step, remove two segments of width ${\delta}/9$ each centered at the midpoints of $[0,1/3]$, and $[2/3,1]$. Continue the process, with each of the four closed intervals remaining. To work out the measure of what was removed we sum the series $$ \sum_{k=1}^{\infty }\frac{2^{k-1}\left( \delta \right) }{3^{k}}=\left( \delta \right) \frac{1}{3}\sum_{k=1}^{\infty }\left( \frac{2}{3}\right) ^{k-1}=\left( \delta \right) \frac{1}{3}\sum_{k=0}^{\infty }\left( \frac{2}{3% }\right) ^{k}=\delta .$$ So the measure of the set that remains must be $\epsilon$. The set that remains is actually homeomorphic to the middle-thirds Cantor set. (Consider a straight line of slope $\delta$ passing through the origin which maps $[0,1]$ homeomorphically onto $[0,\delta]$, and restrict this map to the usual middle-thirds Cantor set.) This argument can easily be modified by enlarging the middle third interval, or shortening to to accommodate any $\delta\in (0,1)$, and thereby constructing sets homeomorphic to the usual middle-thirds Cantor set which have arbitrary positive measure in $(0,1)$.


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