Is there a better estimation than $\lambda^d(A) \leq 2^d (\operatorname{diam}A)^d$?

Question

Let $A \subseteq \mathbb{R}^d$ be a Lebesgue-measurable set and
$$ \operatorname{diam} A = \sup\{\|x-y\|_2 \colon x,y \in A\}. $$ For an arbitrary $x=(x_1, \ldots, x_d) \in A$, we can cover $A$ by $$ A \subseteq \prod_{i=1}^{d} (x_i - \operatorname{diam}A , x_i + \operatorname{diam}A) $$ and therefore $\lambda^d(A) \leq 2^d (\operatorname{diam}A)^d$. Is there a better constant than $2^d$?

Answer 1

It's not hard to get from $2^d$ to $1$. Denote by $\pi_j$ the projection of $\mathbb R^d$ onto the $j$th coordinate axis.

$\pi_j(A)$ has diameter at most $\newcommand{\diam}{\operatorname{diam}} \diam A$ and is therefore contained in an interval $I_j$ with length at most $\diam A$.

Moreover $A \subset \pi_j^{-1}(I_j)$ for all $j$, so that $$A \subset \pi_1^{-1}(I_1) \cap \pi_2^{-1}(I_2) \cap \cdots \cap \pi_d^{-1}(I_d).$$

The last set is precisely the $d$-interval $I_1 \times \cdots \times I_d$ so that $$\lambda^d(A) \le \ell(I_1) \cdots \ell(I_d) \le (\diam A)^d.$$

This can be improved, but it is no longer so trivial. Look up "isodiametric inequality".