For a non-empty, compact set $A \subseteq \mathbb{R}^n$, the [$\delta$-fattening](https://www.wikiwand.com/en/Hausdorff_distance#/Definition) 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](https://math.stackexchange.com/a/338941/525562), 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](http://en.wikipedia.org/wiki/Minkowski%E2%80%93Steiner_formula)
$$\lambda(\partial A)=\liminf\_{\delta\to 0}\delta^{-1}(\mu(A\_\delta)-\mu(A))\tag2.$$
However, the proof utilized the fact that $f(\delta) = \left(\mu(A\_\delta)/\mu(A)\right)^{1/n}$ is concave, which per [this post](https://math.stackexchange.com/questions/1641264/concavity-of-the-nth-root-of-the-volume-of-r-neighborhoods-of-a-set) is not true. Can the claim above or a similar inequality be established?



Relative volume increase of $\delta$-fattening of a compact set
The claim is false in general. The example given by [George Lowther](https://math.stackexchange.com/questions/1641264/concavity-of-the-nth-root-of-the-volume-of-r-neighborhoods-of-a-set) 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 $\delta\in(0,(R-1)/2]$. 


