Let $e(r)$ be a monotone decreasing function in $r\in [0, \infty)$. Suppose that $\int_{\mathbb{R}} |e(r)|dr< \infty$. Let $x_i$,$x_i'$ in $\mathbb{R}^n$, $i=1,\ldots,k$ and $k\leq n+1$ such that $||x_i-x_i||\geq ||x_i'-x_j'||$. Let $e_{x_i}(y) = e(||x_i - y||)$.
I want to prove that $\int_{\mathbb{R}^n} \Pi e_{x_i} \leq \int_{\mathbb{R}^n} \Pi e_{x'_i}$. This is found in page 3 of this article (at the bottom) https://www.ihes.fr/~gromov/wp-content/uploads/2018/08/1658.pdf.
Here, Gromov claims that it follows inmediately from the fact that $\text{vol}(\cap_{i=1}^{k} B(x_i,r_i) ) \leq \text{vol}(\cap_{i=1}^{k} B(x'_i,r_i) )$ for all $r_i \geq 0$.
But that is not obvious to me. Let me show you my attempts to prove that theorem:
Let $A(r_i)= \cap_{i=1}^{k} B(x_i,r_i)$ and $A'(r_i)= \cap_{i=1}^{k} B(x'_i,r_i)$.
I think that is enough to prove that $\int_{A(r_i)} \Pi e_{x_i} \leq \int_{A'(r_i)} \Pi e_{x'_i}$ for every $(r_i)$, and then note that $\mathbb{R}^n = \cup_{r_i} A(r_i) $ and $\mathbb{R}^n = \cup_{r_i} A'(r_i) $, so the inequality follows for $\mathbb{R}^n$. However, I can't prove this statement. But I have some ideas: I think that if, for every $A = A(r_i)$, I can find an $U(A) \subset A'$ such that $\int_{A(r_i)} \Pi e_{x_i} = \int_{U(A)} \Pi e_{x'_i}$ then, as volume is an increasing function, then $\int_{A(r_i)} \Pi e_{x_i} = \int_{U(A)} \Pi e_{x'_i} \leq \int_{A(r_i)} \Pi e_{x'_i}$ and I am done. I think that I may want an argument be like of: as $\text{vol}(A) \leq \text{vol}(A')$, then exist a set $B\subset A'$ such that $\text{vol}(A)=\text{vol}(B)$ so blablabla.
So, as you can see, I'm very stuck.
Do you know how the claim follows? Is it very obvious as Gromov says? I am forgetting something? Thanks in advance.