# Middle Fifths Cantor Set is Borel and Has Measure =?

I've been working on the following problem; I think I have the answer, I would just like to confirm that there are no gaps in my logic.

Consider the "middle fifths" Cantor set $$\mathscr{C}=\left\lbrace\sum_{j=1}^{\infty}\frac{a_{j}}{5^{j}}~:~a_{j}=0,1,3,\mathrm{or}~4~\mathrm{for~each}~j\right\rbrace.$$ Prove that $$\mathscr{C}$$ is a Borel set. What is the Lebesgue measure of $$\mathscr{C}$$?

My Solution: We consider the construction of the above Cantor set as follows: let $$C_{0}=[0,1]$$, let $$C_{1}=[0,2/5]\cup[3/5,1]$$, and in general, let $$C_{n}$$ be the set remaining after removing an open interval of length $$1/5$$ from the middle of each connected component of $$C_{n-1}$$. Then we put $$\mathscr{C}=\bigcap_{n=0}^{\infty}C_{n}$$.

Let $$\mathscr{B}$$ be the Borel $$\sigma$$-algebra on $$\mathbb{R}$$. Then, by definition, $$\mathscr{B}$$ contains all the open subsets of $$\mathbb{R}$$. Since $$\sigma$$-algebras are closed under set complements, it follows that $$\mathscr{B}$$ also contains all the closed subsets of $$\mathbb{R}$$. Hence, $$C_{n}\in\mathscr{B}$$ for each $$n$$. Moreover, since $$\sigma$$-algebras are closed under countable intersections and $$C_{n}\in\mathscr{B}$$ for each $$n$$, it follows that $$\mathscr{C}=\bigcap_{n=0}^{\infty}C_{n}$$ is Borel.

We claim that $$m(\mathscr{C})=0$$, where $$m$$ is the Lebesgue measure on $$\mathbb{R}$$. Indeed, each set $$C_{n}$$ is the disjoint union of precisely $$2^{n}$$ intervals, each of length $$1/5^{n}$$. By the finite additivity of the Lebesgue measure, we have $$m(C_{n})=2^{n}(1/5^{n})=(2/5)^{n}$$. Also, by the monotonicity of the Lebesgue measure, we have $$m(\mathscr{C})\leq m(C_{n})=(2/5)^{n}.$$ Since the inequality above holds for all $$n$$, we conclude that $$m(\mathscr{C})=0$$.

Does all of this look okay? I wanted to make sure, since I didn't end up using the original description of $$\mathscr{C}$$ that was given in the problem statement. Thanks in advance for any help!

Edit: After reading up on Fat Cantor Sets, I think my work above is incorrect. Should this Cantor set have positive measure?

$$\newcommand\C{\mathscr C}$$
"I didn't end up using the original description of $$\mathscr C$$ that was given in the problem statement": You used that in showing that $$\C$$ is the same as the $$\C$$ you get by that middle-fifths construction.
You start by removing the middle fifth of $$[0,1]$$, leaving $$[0,2/5]\cup[3/5,1]$$. That's great - that removes the numbers $$\sum a_j/5^j$$ with $$a_1=2$$. Now at the next stage you want to remove the numbers with $$a_2=2$$. To do that you need to remove the middle fifth from each of the four intervals $$[0,1/5],[1/5,2/5],[3/5,4/5],[4/5,1]$$, which is not the same as removing the middle fifth of each component.
(Yes, the measure is $$0$$. The simplest way to see that is to note that at each stage the measure gets multiplied by $$4/5$$. A "fat Cantor set" requires a different ratio at each step.)