Outer measure allowing only finite covers Let $\lambda$ be the usual length function of intervals of $\mathbb R$ and set
$\overline \lambda(A) = \inf \{\sum^m_{n=1} \lambda(I_n): A_n \text{ interval and } A \subset \bigcup^m_{n=1} I_n\}$.
Consider the set $A = \mathbb Q \cap [0,1]$. The usual definition of outer measure which allows countable covers in the above infimum would give a measure of $0$ to this set $A$. Which is showed by taking a countable sequence of intervals $I_n$ which contain all points in $A$ (if $(r_n)_n$ is an enumeration of $A$ then $r_n \in I_n  \forall n$) such that $\lambda(I_n) < \varepsilon \cdot 2^{-n}$.
But if we only allow finite covers its intuively clear that the length of a cover is at least $1$. I try to write down a formal argument: If a cover of $A$ were smaller than $1$, we assume wlog the cover were disjoint. But now I am stuck. How do I show that the complement of $A$ and the cover contains at least one rational number?

HINT: Use induction on the number of open intervals in your cover . . .

Specifically, prove the following statement by induction on $n$:

If $\mathcal{C}$ is a cover of $\mathbb{Q}\cap (a, b)$ by $n$ intervals, then the sum of the diameters of the elements of $\mathcal{C}$ is $>b-a$.

