Let $C$ be the Cantor middle set, show that $C+\frac{1}{2}C$ is measurable.

Let $C$ be the Cantor middle set, show that $C+\frac{1}{2}C$ is measurable.

We know $C+C=[0,2], C-C=[-1,1]$. But how to show $C+\frac{1}{2}C$ is measurable. It seems that this set can no algorithmically constructed.

Solution 1:

Note that for compact sets $A,B \subset \mathbb R$, set $A+B$ is again compact (can be easily proven either by taking subsequences or since $A+B$ is an image of compact $A \times B$ via continuous function $f(a,b) = a + b$). Note that Cantor set $C$ is compact. Hence (by the result above) $C + \frac{1}{2}C$ is compact, so measurable

Solution 2:

The best argument is that by compactness. Nevertheless, we can describe $C+\frac12C$ explicitly, using the fact that elements of $C$ allow a ternary representation using only zeroes and twos as digits:

Clearly, $C+\frac12C\subseteq [0,1]+[0,\frac12]=[0,\frac32]$. To see that equality holds, let $x\in[0,\frac32]$. If $x=\frac32$, then clearly $x\in C+\frac12C$, so we may assume that either $x<1$ or $1\le x<\frac32$.

If $x<1$, pick a ternary representation $$ x=\sum_{n=1}^\infty c_n3^{-n}$$ with $c_n\in\{0,1,2\}$. Then with $$ a_n=\begin{cases}2&c_n=2\\0&\text{otherwise}\end{cases},\qquad b_n=\begin{cases}2&c_n=1\\0&\text{otherwise}\end{cases},$$ we have $$ x = \sum_{n=1}^\infty a_n3^{-n}\quad +\quad \frac12\sum_{n=1}^\infty b_n3^{-n},$$ as desired.

On the other hand, if $\frac32>x\ge1$, the ternary representation looks like $$ x=1+\sum_{n=1}^\infty c_n3^{-n}$$ and as $x<\frac32$, this representation does not consist solely of ones and non-one is a zero. Let $m$ be minimal with $c_m=0$. Then with $$ a_n=\begin{cases}2&c_n=2\lor n\le m\\0&\text{otherwise}\end{cases},\qquad b_n=\begin{cases}2&c_n=1\lor n\le m\\0&\text{otherwise}\end{cases},$$ we again have (check this!) $$ x = \sum_{n=1}^\infty a_n3^{-n}\quad +\quad \frac12\sum_{n=1}^\infty b_n3^{-n},$$ as desired.