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.
On
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 $b-a = \int 1_{[a,b]}(x)dx \le \int \sum_k 1_{(a_k,b_k)}(x) dx = \sum_k \int 1_{(a_k,b_k)}(x) dx = \sum_k (b_k-a_k)$.
On
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.
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.