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.

Or you should have! In fact the iterative construction you sketch, removing the middle fifth of each component at each stage, is wrong.

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.)


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