Is it possible to cover line segment [0, 1] with a set of segments having total size less than 1? It is well known (although its highly counterintuitive) that it is possible to cover all the rational points of segment [0, 1] with a set of segments having arbitrary small total length.
But is it possible to cover ALL the points of segment [0, 1] with a set of segments with total length less than 1?
 A: The function which assigns to an interval $(a,b) \subset \mathbb{R}$ its length $\ell((a,b)) = b-a$ has a property named countable additivity. This means that if $A_{1}, A_{2}, \ldots$ are a sequence of pairwise disjoint intervals (which union to an interval), then
$$\ell\left(\bigcup_{n=1}^{\infty} A_{n}\right) = \sum_{n=1}^{\infty} \ell(A_{n}).$$
Therefore, if you use countably many intervals which union up to $[0,1]$, then their lengths must add up to 1.
To make sense of the answer below, I think it is necessary to extend the notion of length to finite unions of intervals, in the obvious way. The most general notion of a set which accepts a notion of length is called measurability, and I am trying to keep it beyond the scope of this answer.
Proof: The hard part is showing
$$\ell\left(\bigcup_{n=1}^{\infty} A_{n}\right) \le \sum_{n=1}^{\infty} \ell(A_{n}).$$
Let $(A_n)$ be a sequence of disjoint intervals which union to $[0,1]$. By removing at most countably many points (some of the endpoints) we may assume that the $A_n$ are all open intervals $(a_n, b_n)$. We would like to construct an open cover of $[0,1]$. Note that both $0$ and $1$ are excluded.
Let $\varepsilon > 0$, and let $c_1, c_2, \ldots$ be the excluded points (in any order). Let $B_n$ be the open interval
$$B_n = (c_n-\varepsilon/2^n, c_n+\varepsilon/2^n).$$
Note that $\sum \ell(B_n) = \varepsilon$.
Therefore the set, $\mathcal{C} = \{A_1, B_1, A_2, B_2, \ldots\}$ is an open cover of $[0,1]$, and its union is the interval $(-\varepsilon/2^N, 1+\varepsilon/2^M)$ for some $N, M$. Since $[0,1]$ is compact, we can take a finite subcover, $\tilde{\mathcal{C}}$ (depending on $\varepsilon$), which will consist of some $A$s and some $B$s. Then
$$\sum_{n=1}^\infty \ell(A_n) + \varepsilon \ge \sum_{A \in \tilde{\mathcal{C}}}\ell(A) + \varepsilon \ge \sum_{A \in \tilde{\mathcal{C}}}\ell(A) + \sum_{B \in \tilde{\mathcal{C}}}\ell(B) > 1.$$
Taking $\varepsilon \to 0$ gives
$$\sum_{n=1}^\infty A_n \ge 1$$
as required.
