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)$ ?

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,$ $ $ 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?