Prove $b-a \le \sum^n_{i=1}(b_i-a_i)$ by induction Show that if the closed interval $[a,b]$ is covered by finitely many open intervals $(a_1,b_1), ...,(a_n,b_n)$, then $$b-a \le \sum^n_{i=1}(b_i-a_i)$$. 
I know that $(a_1,b_1), ...,(a_n,b_n)$ form an open covering of $[a,b]$, and my thought is to show the inequality by mathematical induction, but not sure how to prove this. Could someone provide a complete proof please? Thanks a lot. 
 A: The base case is clear. Let $\{(a_i,b_i)\}_{i=1}^n$ be an open cover of $[a,b]$. Suppose the sets are not nested. Then there are two sets $(a_i,b_i)$ and $(a_j,b_j)$ with $a_i \leq a_j \leq b_i \leq b_j$. Without loss of generality we may assume $i=1, j=2$. Taking their union, i.e. forming $(a_1,b_2)$ we get a smaller open cover and the induction step tells us that
$$
b - a \leq b_2 - a_1 + \sum_{i = 3}^n b_i - a_i.
$$
However we have that $b_2 - a_1 \leq b_2 - a_2 + b_1 - a_1$. This proves the inequality. If the sets are all nested, a similar method will do it.
A: Here is a direct proof:
Since the $(a_k,b_k)$ form a cover of $[a,b]$ then for all $x \in [a,b]$
there is some $k$ such that $x \in (a_k,b_k)$.
Hence $1_{[a,b]}(x) \le \sum_k 1_{(a_k,b_k)}(x)$ for all $x$.
It is clear
that both sides are integrable, hence
$b-a = \int 1_{[a,b]}(x)dx \le \int \sum_k 1_{(a_k,b_k)}(x) dx = \sum_k \int 
1_{(a_k,b_k)}(x) dx = \sum_k (b_k-a_k)$.
A: Take any $n$ element cover. 
The case $n=1$ is clear. Assume $n >1$.
If no two intervals intersect each other, then notice that no $a_i$ or $b_i$ is covered, therefore $a_i ,b_i \in \mathbb R \setminus [a,b]$ for all $i$. But $[a,b]$ is covered so there has to be an $i$ such that $a_i < a < b < b_i$ so the RHS above is at least $b-a$.
If there are at least two intersecting sets, then choose two and take their union, this way you have reduced the original cover to an $n-1$ element cover and thus you can apply induction. Finally noting that the union's length is at most the sum of the lengths, you should be done.
