# Show that this set, defined similarly to the Cantor set, also has measure 0

The standard, middle-thirds Cantor set can be thought of as the set of all numbers on the interval $[0, 1]$ whose ternary expansions contain no 1s, that is, numbers of the form $$\sum_{n=1}^{\infty} a_n 3^{-n} \text{ where } (a_n) \in \{0, 2\}^\mathbb{N}$$ It is well known that this Cantor set has Lebesgue measure $0$.

I am trying to show that a very similar-looking set, the set of points of the form $$\sum_{n=1}^{\infty} a_n e^{-n} \text{ where } (a_n) \in \{-1, 1\}^\mathbb{N}$$ also has Lebesgue measure $0$. I initially tried to take a bijection between these sets and show this bijection is a homeomorphism, but then realized homeomorphisms don't necessarily preserve measure.

Is there any way to use properties of the Cantor set to show that the latter set has measure $0$?

• You probably can't directly use properties of the Cantor set here. Instead, you should repeat the definition of the Cantor set in terms of an intersection of unions of intervals for this set. Dec 15, 2014 at 20:18

First note that the set $S$ is a subset of the interval $[-a, a]$, where $$a = \sum_{n=1}^\infty e^{-n} = \frac{1}{e - 1} \approx 0.582$$ The set is self-similar via two affine transformations, each scaling down by a factor of $e$ then shifting either right $(+)$ or left $(-)$ by $e^{-1}$. In symbols, the functions $f_\pm(x) = e^{-1}x \pm e^{-1}$ cover the set: $$S = f_+(S) \cup f_-(S).$$ Here's an image of the first six partial partial sums to guide your intuition.

Now construct a decreasing sequence of sets who each cover $S$ and whose total measure decreases to $0$ in the limit. Let $K_0 = [-a, a]$. By similarity, the intervals $[f_\pm(-a), f_\pm(a)]$ covers $f_\pm(S)$, so set $$K_1 = [f_-(-a), f_-(a)] \cup [f_+(-a), f_+(a)],$$ and continuing with the pattern, \begin{align} K_2 = &[f_-f_-(-a), f_-f_-(a)] \cup [f_-f_+(-a), f_-f_+(a)] \\ & \quad\cup [f_+f_-(-a), f_+f_-(a)] \cup [f_+f_+(-a), f_+f_+(a)]. \end{align} The set $K_n$ consists of $2^n$ disjoint intervals, each of length $e^{-n}a$, so the total measure of $K_n$ is $$2^n \cdot \frac{a}{e^n} = \Bigl( \frac{2}{e} \Bigr)^{\!n} \! \frac{1}{e - 1} \longrightarrow 0 \qquad \text{as } n \to \infty.$$