Relative volume increase of $\delta$-fattening of a compact set

Question

For a non-empty, compact set $A \subseteq \mathbb{R}^n$, the $\delta$-fattening of $A$, $A_\delta$, is defined to be the set $$ A_\delta = \cup_{a \in A} B_{\delta}(a), $$ where $B_\delta(a)$ denotes the closed ball centered at $a$ with radius $\delta$.

Is it possible to establish an upper bound on $\mu(A_\delta)$ in terms of $\mu(A)$, where $\mu$ is the Lebesgue measure over $\mathbb R^n$?

In a previous post, it was claimed that:

Claim Let $A$ be a nonempty compact subset of $\mathbb R^n$ with $\mu(A)>0$. Then for all $\delta>0$ $$\mu(A_\delta)\le \left(1+\delta\,\frac{\lambda(\partial A)}{n\,\mu(A)}\right)^n\mu(A)\tag1,$$ where $A_\delta$ is the $\delta$-fattening of the set $A$, and $\lambda(\partial A)$ is the Minkowski content $$\lambda(\partial A)=\liminf_{\delta\to 0}\delta^{-1}(\mu(A_\delta)-\mu(A))\tag2.$$ However, the proof utilized the fact that $f(\delta) = \left(\mu(A_\delta)/\mu(A)\right)^{1/n}$ is concave, which per this post is not true. Can the claim above or a similar inequality be established?

Answer 1

The claim is false in general. The example given by George Lowther disproves it. Indeed, in that example, $n=2$ and $A$ is the union of the unit disk and a one-point set at distance $R>1$ from the origin, so that $\mu(A_\delta)=f(\delta):=\pi[(1+\delta)^2+\delta^2]$ for $\delta\in[0,(R-1)/2]$ and $\lambda(\partial A)=f'(0)=2\pi$. So, the ratio of the RHS of your proposed inequality (1) to its LHS is $$\frac{(1+\delta)^2}{(1+\delta)^2+\delta^2}<1$$ for $\delta\in(0,(R-1)/2]$.