if $[a,b]\subset \bigcup\limits_{i=1}^n(a_i,b_i)$ then $b-a\leq \sum_\limits{i=1}^n(b_i-a_i)$ I need to prove: if $[a,b]\subset \bigcup\limits_{i=1}^n(a_i,b_i)$ then $b-a\leq \sum_\limits{i=1}^n(b_i-a_i)$ .
If I apply the monotony of the Lebesgue Measure:$\mu([a,b])\leq\mu( \bigcup\limits_{i=1}^n(a_i,b_i) )=\sum_\limits{i=1}^n \mu((a_i,b_i))$
and the result is immediate. But how can we show this without using properties of the Lebesgue Measure? I think there is a more basic proof.
 A: For a positive integer $N$, and a bounded set $A\subset\Bbb R$ let $\mu_N(A)$
be the number of integers $k$ such that $k/N\in A$ (finitely
many, since $A$ is bounded).
It is plain that $\mu_N(A_1\cup A_2)\le\mu_N(A_1)+\mu_N(A_2)$ and that $B\subseteq
A$ implies $\mu_N(B)\le \mu_N(A)$. Consequently if $B\subseteq\bigcup_{i=1}^n A_i$ then $\mu_N(B)\le\sum_{i=1}^n\mu_N(A_i)$. Therefore, in your set-up
$$\mu_N([a,b])\le\sum_{i=1}^n\mu_N((a_i,b_i)).$$
Divide by $N$:
$$\frac1N\mu_N([a,b])\le\sum_{i=1}^n\frac1N\mu_N((a_i,b_i)).\tag{*}$$
But as $N\to\infty$,
$$\frac1N\mu_N([a,b])\to b-a$$
and 
$$\frac1N\mu_N((a_i,b_i))\to b_i-a_i.$$
Letting $N\to\infty$ in (*) then gives
$$b-a\le\sum_{i=1}^n(b_i-a_i).$$
A: Proof is by induction on $n$.  
base case: $n=1$.  If $[a,b] \subset (a_1,b_1)$, then $b-a \le b_1-a_1$.  This is because $a_1 < a$ and $b_1 > b$.
induction step: Suppose this result is known for some value of $n$.  Now suppose
$$
[a,b] \subseteq \bigcup_{j=1}^{n+1} (a_j,b_j)
$$
Point $a$ is in $[a,b]$, so it is in $\bigcup_{j=1}^{n+1} (a_j,b_j)$.  Renumbering these, we may assume $a \in (a_{n+1},b_{n+1})$.  So $a_{n+1} < a < b_{n+1}$. In particular
$$
b_{n+1}-a < b_{n+1}-a_{n+1}
\tag1$$
Now
$$
[b_{n+1},b] \subseteq [a,b] \subseteq \bigcup_{j=1}^{n+1} (a_j,b_j)
$$
and interval $[b_{n+1},b]$ is disjoint from $(a_{n+1},b_{n+1})$, so
$$
[b_{n+1},b] \subseteq \bigcup_{j=1}^{n} (a_j,b_j).
$$
From the induction hypothesis,
$$
b-b_{n+1} \le \sum_{j=1}^n (b_j-a_j) .
\tag2$$
Now add (1) and (2) to get
$$
b-a \le \sum_{j=1}^{n+1} (b_j-a_j),
$$
as required.
A: You can do this with Riemann integrals. The function
$$f(x) = \chi_{[a,b]}(x) = \left\{ \begin{array}{cc} 1 & x \in [a,b] \\ 0 & x \notin [a,b] \end{array}\right.$$ is Riemann integrable on any interval $[c,d]$ containing $[a,b]$ and
$$\int_c^d f(x) \, dx = b-a.$$
Define $f_i(x) = \chi_{[a_i,b_i]}(x)$ for each $i$ and observe that the inclusion $[a,b] \subset \bigcup_{i=1}^n [a_i,b_i]$ implies the numerical inequality
$$f(x) \le \sum_{i=1}^n f_i(x).$$
Let $[c,d]$ be an interval containing $[a,b]$ and all the $[a_i,b_i]$. Then
$$b-a = \int_c^d f(x) \, dx \le \sum_{i=1}^n \int_c^d f_i(x) \, dx = \sum_{i=1}^n (b_i - a_i).$$
