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I need to prove the following statement:

Let $B_s(z)$ be a ball centered at $z$ of radius $s$ s.t. $0\not\in B_s(z)$. Moreover let $K_s(z)$ the convex hull of $\{0\}\cup B_s(z)$. Then $$ \left|K_{\frac{s}{r}}\left(\frac{z}{r}\right)\cap B\right|\geq\alpha>0, \text{for $r$ large enough}, $$ where $|\cdot|$ is the Lebesgue measure, $B=B_1(0)$.

I have no clue on how to prove it...
By picturing $K_s(z)$ it is a cone with vertex in $0$ and the ball as the "top".

As $r$ goes to infinity, the ball $B_{\frac{s}{r}}\!\left(\frac{z}{r}\right)$ goes inside the unit ball centered in the origin, so for $r$ large enough $\big|K_{\frac{s}{r}}\!\left(\frac{z}{r}\right)\cap B\big|=\big|K_{\frac{s}{r}}\left(\frac{z}{r}\right)\big|$. By intuition $|K_r(z)|=C_nr^n$ so to me seems false.. Is my intuition false?

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    $\begingroup$ Yes, this statement is obviously false, Where did you get it from? $\endgroup$
    – Iosif Pinelis
    Commented Mar 26, 2020 at 14:35
  • $\begingroup$ "Sets of finite perimeter and geometric measure theory" F.Maggi, page 264 in the middle of the page ^^' Maybe i formulated it wrong.. It is strange that the book made an error like this $\endgroup$
    – A. Ninno
    Commented Mar 26, 2020 at 17:14

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