Union of closed subsets with empty interior has Lebesgue measure $1$

Question

I would like to find a sequence of closed subsets of $\left[0,1\right]$ with empty interior whose union has Lebesgue measure equal to $1$.
I tried combining the Cantor set, the fat Cantor set and $\mathbb{Q}$ but nothing worked.
Any ideas?

Answer 1

There are actually many different so-called "fat Cantor sets." This Wikipedia page on them describes how one can construct Cantor-type subsets of $[0, 1]$ of measure $\frac{1 - 3a}{1 - 2a}$ for $a \in (0, 1/3)$ (see the "other fat Cantor sets" section), and notes that $\lim_{a \to 0^+} \frac{1 - 3a}{1 - 2a}$. Let $C_n$ be the fat Cantor set made with parameter $\frac{1}{3n + 1}$. Then $\bigcup_{n = 1}^\infty C_n$ contains $C_k$ for each $k \in \mathbb{N}$, and thus $\bigcup_{n = 1}^\infty C_n$ has measure $\geq \frac{1 - 3 \frac{1}{3k + 1}}{1 - 2 \frac{1}{3k + 1}}$, and as $k \to \infty$, this will approach $1$. Thus $\bigcup_{n = 1}^\infty C_n$ has measure $\geq 1$. On the other hand, we know that $\bigcup_{n = 1}^\infty C_n \subseteq [0, 1]$, so the measure is also $\leq 1$.

Each $C_n$ is a compact subset of $[0, 1]$ with empty interior.