Bounds on measure of self-difference and $\epsilon$-neighborhood of convex set I am trying to see how to prove Lemma 10.15 from Kallenberg's Foundations of Modern Probability:

Here $|B|$ denotes the Lebesgue measure of $B$, and $r(B)$ denotes the radius of the largest open ball contained in $B$.
I can see how to reduce both inequalities to the case where $B$ is compact. Then, for (i) I've tried relating $B$ to its Löwner-John covering ellipsoid, but from this I can only get the bound $|B - B| \leq (2d)^d |B|$, which is not good enough. For (ii) I've tried to sandwich $\partial_\epsilon B$ between two dilations of $B$ (based on the center of a ball with radius $r(B)$ contained in $B$), but this hasn't been successful so far.
Does anyone know how to prove these or where to find a reference for these types of results?
 A: (i) The set $B-B$ is called the difference body of $B$. The inequality $|B-B|\le \binom{2n}{n}|B|$ is due to Rogers and Shepard (The difference body of a convex body. Arch. Math., 8, 220–233, 1957). A proof can be found in the book Convex bodies: the Brunn-Minkowski theory by Rolf Schneider, Theorem 7.3.1. (The perusal of this book is highly recommended if you are interested in volume inequalities for convex sets.) The Rogers-Shepard inequality is sharp, with equality attained for any simplex, and its proof is tricky. 
(ii) is a rougher estimate. Translate $B$ so that the largest ball it contains is centered at $0$. The set $(1+\epsilon/r(B)) B = B + (\epsilon/r(B)) B$ contains the $\epsilon$-neighborhood of $B$, since $(\epsilon/r(B)) B$ contains a ball of radius $\epsilon$. The volume of $(1+\epsilon/r(B)) B$ is, of course, $(1+\epsilon/r(B))^d |B|$. Therefore, the "outer $\epsilon$-collar" of $B$, understood as the set of all points that do not belong to $B$ but lie within distance at most $\epsilon$ from it, has volume at most $((1+\epsilon/r(B))^d-1) |B|$. It remains to argue that the "inner $\epsilon$-collar" is smaller than the outer one, which seems obvious, but I don't have a proof right now.
Addendum to complete proof of (ii): Define the inner $\epsilon$-collar as the set of points in $B$ with distance less than $\epsilon$ to $\partial B$. We must show that the volume of the inner $\epsilon$-collar is at most $((1 + \epsilon/r(B))^d - 1)|B|$. If $\epsilon/r(B) \geq 1$, then this is trivially true since in this case $((1 + \epsilon/r(B))^d - 1) \geq 1$ and the inner $\epsilon$-collar has volume at most $|B|$. So assume $\epsilon/r(B)<1$. Since $B$ is convex, we have $(1-\epsilon/r(B))B + (\epsilon/r(B))B \subseteq B$. Since $(\epsilon/r(B))B$ contains the ball of radius $\epsilon$ centered at the origin, this implies that each point of $(1-\epsilon/r(B))B$ has distance at least $\epsilon$ to $\partial B$, so that $(1-\epsilon/r(B))B$ does not intersect the inner $\epsilon$-collar. This implies that the inner collar is contained in the set difference $B \setminus (1-\epsilon/r(B))B$, which has volume $(1 - (1-\epsilon/r(B))^d)|B|$. Noting that $(1 - (1-\epsilon/r(B))^d) \leq ((1 + \epsilon/r(B))^d - 1)$ since the function $f(x)=x^d$ is convex, this completes the proof.
