What is an example of a covering of the set of irrational numbers in the interval $[0,1]$, with a total length of $1$?

Question

Let $E$ be the set of rationals in the interval $A=[0,1]$, and let $F$ be the set of irrationals in $A$.
Then, $m^*(E)=0$ since $E$ is countable, and so, by the definition of Lebesgue measurable sets, $m^*(F)=1$, since $m^*(A)=1$.
That is,
$m^*(A)=m^*(A \cap E)+m^*(A \cap E^C) \implies 1=m^*(E)+m^*(F)$.

However, I find the result $m^*(F)=1$ rather surprising.

What is an example of a covering of open sets of $F$ with total length $1$? The obvious answer would be $\{(0,1)\}$, but how do I show that there does not exist a covering with a smaller length, e.g. of length $0$?

In other words, I'm looking for an alternative derivation of the result $m^*(F)=1$. One that is done more directly from the definition of $m^*$, i.e. involving a covering of open sets of the set $F$.

Answer 1

Take as an example the sets $I_n=(\frac{1}{n+1},\frac{1}{n})$

Let $\epsilon>0$

By definition of the outer measure there is a covering $\{I_n:n \in \Bbb{N}\}$ of $F$ such that $$\sum_nl(I_n)-\epsilon \leq m^*(F) \leq \sum_nl(I_n)$$

If $\sum_nl(I_n)<1$, then exists an interval $J$ such that $J \cap \bigcup_nI_n=\emptyset$

But density of irrationals contradicts the fact that $\{I_n:n \in \Bbb{N}\}$ is a covering of $F$

Answer 2

Suppose that $\mathcal C$ is a cover of $F$ of length $0 \le L < 1$.

We know that $E$ is countable. Let's say that $E = \{r_n \mid n \in \mathbb N\}$. Denote $\epsilon = \frac{1-L}{2}$. Then $$\mathcal U = \{(r_n - \epsilon/2^n, r_n + \epsilon/2^n) \cap [0,1] \mid n \in \mathbb N\}$$ is an open cover of $E$ of length less or equal to $\epsilon$. Therefore $\mathcal C \cup \mathcal U$ is an open cover of $[0,1]$ of length less than one. A contradiction.