Sum of lengths of intervals

Question

I was reading Ivan Niven's Maxima and Minima without calculus - more precisely the section regarding the Jeep crossing the Desert but that's not the point.

In that section is given an "almost self-evident lemma" which I understand but I can't see how I could prove it.

Consider a finite number of closed intervals. possibly overlapping on a straight line segment AB of length r. If each point of AB belongs to at least s of the intervals. then the sum of the lengths of the intervals is at least rs.

I think the length of AB is of no use (we can scale AB however I want) so the lemma can be rewritten as this :

Consider a finite number of closed intervals. possibly overlapping on a straight line segment AB of length 1. If each point of AB belongs to at least s of the intervals. then the sum of the lengths of the intervals is at least s.

There's a figure illustrating the lemma - and I see why it is true, but I wonder how I could prove it (which may be harder than understanding the lemma)

Answer 1

Here is a proof using, I beg your pardon, (elementary) properties of integration:

Let $I_k$ be the $k$th interval.

Let $\chi_k$ be the characteristic function of $I_k$.

($\chi(x)=1 \iff x \in I_k$, $ \ \chi_k(x)=0$ otherwise)

Let

$$f(x):=\sum \chi_k(x)$$

As, for any $x$, $f(x)\ge s$, we have:

$$\int_0^r f(x)dx \ge s \times length([0,r])=sr$$

But

$$\int_0^r f(x)dx=\sum \int_0^r \chi_k(x)dx=\sum length(I_k)$$

proving the result.

Answer 2

Let $L$ be the line segment of length $r$.

Suppose $L$ is covered by finite number of closed intervals be such that each point of $L$ contained in at least $s$ closed intervals . Then $L\subset \bigcup_{i=1}^{m}\bigcap_{j=1}^{n\ge s} C_{ij}$ where $C_{ij}$ 's are closed intervals.

$\begin{align}r=\ell(L) &\le \ell(\bigcup_{i=1}^{m}\bigcap_{j=1}^{n\ge s} C_{ij}) \\&\le \sum_{i=1}^{m} \ell( \bigcap_{j=1}^{n\ge s} C_{ij}) \space\quad\quad ...1\\&\le \frac{1}{n}\sum_{i=1}^{m} \sum_{j=1}^{n\ge s} \ell(C_{ij})\quad ... 2\end{align}$

Hence $\sum_{i=1}^{m} \sum_{j=1}^{n\ge s} \ell(C_{ij})\ge rn\ge rs$

---

$... 1$

$\text{finite sub-aditivity of Length function}$

---

$...2$

$\bigcap_{j=1}^{n\ge s} C_{ij}\subset C_{ij}$ forall $j=1, 2,\ldots , n\ge s$

Implies $n\cdot\ell(\bigcap_{j=1}^{n\ge s} C_{ij}) \le \sum_{j=1}^{n\ge s} \ell(C_{ij})$

---