MHB Problem of the Week #77 - November 18th, 2013

[Chris L T521]

Here's this week's problem.

-----

Problem: Show that for each $\alpha\in\left[0,1\right)$, there is a perfect symmetric set with Lebesgue measure $\alpha$.

-----

Hint: [sp]Construct this set in a way that is similar to how the Cantor set is constructed.[/sp]

 

[Chris L T521]

No one answered this week's problem. You can find my solution below.

[sp]Proof: We construct this perfect symmetric set in a similar fashion to the Cantor set. Let $C_0=[0,1]$ and let $\alpha\in[0,1)$. Let $R_1:= \left(\dfrac{1+\alpha}{4},\dfrac{3-\alpha}{4}\right)$ be the portion removed from $C_0$. It should be clear that $m(R_1)=\dfrac{1-\alpha}{2}$. Now, define $C_1=C_0\backslash R_1$. We continue this process and define $$R_2 := \left(\frac{1+\alpha}{16}, \frac{3-\alpha}{16}\right)\cup\left(\frac{13+\alpha}{16}, \frac{15-\alpha}{16}\right).$$
It should be clear that $m(R_2)=\dfrac{1-\alpha}{4}$. Now define $C_2=C_1\backslash R_2$. We continue this process ad infinitum. Observe that at the $k$-th step, $m(R_k)=\dfrac{1-\alpha}{2^k}$. In this construction, each $R_k$ is being removed from $C_0$ and $R_i\cap R_j=\emptyset$ for $i\neq j$. Thus, the measure of this set is
$$\begin{aligned} m\left(C_0\setminus \bigcup\limits_{k=1}^{\infty}R_k\right) &= m(C_0)-\sum\limits_{k=1}^{\infty}m(R_k)\\ &= 1 -(1-\alpha) \sum\limits_{k=1}^{\infty}\frac{1}{2^k}\\ &= 1-(1-\alpha)\\ &= \alpha.\end{aligned}$$
This completes the construction.$\hspace{.25in}\blacksquare$[/sp]