Applying a contraction to balls' centers increases the size of the balls' intersection?

Question

The following statement seems clearly true, but I'm having a hard time proving it:

Fix $\alpha\in[0,1]$. Let $\mu$ be Lebesgue measure. For $r\ge 0$, let $B(c,r)\equiv[c-r,c+r]$. Fix $r_1,\ldots,r_n\in[0,\infty)$, and $c_1,\ldots,c_n\in\mathbb R$. Then, $\mu\left(\bigcap_{i=1}^{n}B(\alpha c_{i},r_{i})\right)\ge\mu\left(\bigcap_{i=1}^{n}B(c_{i},r_{i})\right)$

In words, the statement just says that contracting the 1-dimensional balls' centers towards $0$, by a constant proportion, while leaving the radii unchanged, increases the amount of overlap. This seems straightforward enough: if the centers were contracted all the way to the origin, then the intersection would just be the smallest ball. There can't be more overlap than that.

Can anyone suggest a simple proof, or point me to a relevant theorem?

Answer 1

Let $B(c,r)=\bigcap_{i=1}^n B(c_i,r_i)$. We have $|c_i-c|\le r_i-r$ for all $i$. Hence, $|\alpha c_i-\alpha c|\le r_i-r$, which implies $B(\alpha c,r)\subseteq \bigcap_{i=1}^n B(\alpha c_i,r_i)$.

Explanation: a ball $B$ is contained in ball $B'$ if and only if the distance between centers is at most radius(B')-radius(B).

Answer 2

Consider the case of two intervals. Once we fix the radii, the volume of the intersection is indeed a decreasing function of the distance between the centres, so the result is straightforward in this case. But an intersection of intervals is either another interval or is empty, so really this is all there is to it. If you like, you have the result by induction.