[proofplan] We prove the inequality by Steiner symmetrization. Given a Borel set $E$, we symmetrize it with respect to the hyperplane $\{x_i = 0\}$ for each coordinate $i = 1, \ldots, n$. Each symmetrization preserves $\mathcal{L}^n$-measure (by [Fubini's Theorem](/theorems/3018)) and does not increase the diameter (by explicit distance estimates using the triangle inequality and properties of symmetric intervals). After symmetrizing in all $n$ coordinate directions, the resulting set is symmetric about the origin in every coordinate and hence contained in a ball of radius $\operatorname{diam}(E)/2$. Its Lebesgue measure equals $\mathcal{L}^n(E)$, while the ball has measure $\alpha(n)(\operatorname{diam}(E)/2)^n$, giving the inequality. [/proofplan] [step:Define Steiner symmetrization with respect to a coordinate hyperplane] Fix a coordinate index $i \in \{1, \ldots, n\}$ and write a point $x \in \mathbb{R}^n$ as $x = (x', x_i)$ where $x' = (x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n) \in \mathbb{R}^{n-1}$. For a Borel set $E \subset \mathbb{R}^n$ and each $x' \in \mathbb{R}^{n-1}$, define the $x_i$-section (or fiber) \begin{align*} E_{x'} = \{t \in \mathbb{R} : (x', t) \in E\} \subset \mathbb{R}. \end{align*} The Steiner symmetral of $E$ with respect to the hyperplane $H_i = \{x_i = 0\}$ is the set \begin{align*} \sigma_i(E) := \left\{(x', t) \in \mathbb{R}^n : x' \in \pi_i(E),\; |t| < \frac{\mathcal{L}^1(E_{x'})}{2}\right\}, \end{align*} where $\pi_i(E) = \{x' \in \mathbb{R}^{n-1} : E_{x'} \neq \varnothing\}$ is the projection of $E$ onto $\mathbb{R}^{n-1}$ (omitting the $i$-th coordinate). Thus $\sigma_i(E)$ replaces each fiber $E_{x'}$ by the symmetric open interval $(-\mathcal{L}^1(E_{x'})/2, \mathcal{L}^1(E_{x'})/2)$ of equal length, centered at $0$. [/step] [step:Show that Steiner symmetrization preserves $\mathcal{L}^n$-measure] We apply [Fubini's Theorem](/theorems/3018) (more precisely, [Tonelli's Theorem](/theorems/3017), since the integrand $\mathbb{1}_E \geq 0$). The hypotheses of Tonelli are satisfied: $(\mathbb{R}^{n-1}, \mathcal{B}(\mathbb{R}^{n-1}), \mathcal{L}^{n-1})$ and $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{L}^1)$ are $\sigma$-finite measure spaces, and $\mathbb{1}_E$ is $\mathcal{B}(\mathbb{R}^n)$-measurable (since $E$ is Borel). Tonelli gives \begin{align*} \mathcal{L}^n(E) = \int_{\mathbb{R}^n} \mathbb{1}_E \, d\mathcal{L}^n = \int_{\mathbb{R}^{n-1}} \mathcal{L}^1(E_{x'}) \, d\mathcal{L}^{n-1}(x'). \end{align*} For $\sigma_i(E)$, the fiber at $x'$ is $(\sigma_i(E))_{x'} = (-\mathcal{L}^1(E_{x'})/2, \mathcal{L}^1(E_{x'})/2)$, which has $\mathcal{L}^1$-measure equal to $\mathcal{L}^1(E_{x'})$. Therefore \begin{align*} \mathcal{L}^n(\sigma_i(E)) = \int_{\mathbb{R}^{n-1}} \mathcal{L}^1((\sigma_i(E))_{x'}) \, d\mathcal{L}^{n-1}(x') = \int_{\mathbb{R}^{n-1}} \mathcal{L}^1(E_{x'}) \, d\mathcal{L}^{n-1}(x') = \mathcal{L}^n(E). \end{align*} [/step] [step:Show that Steiner symmetrization does not increase the diameter] We prove $\operatorname{diam}(\sigma_i(E)) \leq \operatorname{diam}(E)$. Let $(x', s), (y', t) \in \sigma_i(E)$. By definition, $|s| < \mathcal{L}^1(E_{x'})/2$ and $|t| < \mathcal{L}^1(E_{y'})/2$. We must show \begin{align*} |(x', s) - (y', t)|^2 = |x' - y'|^2 + (s - t)^2 \leq (\operatorname{diam}(E))^2. \end{align*} Since $|s| < \mathcal{L}^1(E_{x'})/2$, there exist points $a, b \in E_{x'}$ with $a < s < b$ and $b - a = \mathcal{L}^1(E_{x'})$ (we may assume the fiber has this full extent; more precisely, since $|s| < \mathcal{L}^1(E_{x'})/2$, there exist $a_1, b_1 \in E_{x'}$ with $a_1 \leq -|s|$ and $b_1 \geq |s|$, because the fiber has total length $\mathcal{L}^1(E_{x'}) > 2|s|$ and must therefore extend at least $|s|$ on both sides of the origin --- this uses that the symmetric interval $(-\mathcal{L}^1(E_{x'})/2, \mathcal{L}^1(E_{x'})/2)$ has the same length as $E_{x'}$). More carefully: since $E_{x'}$ has $\mathcal{L}^1(E_{x'}) > 2|s|$, there exist $a_1 \in E_{x'}$ with $a_1 \leq s$ and $b_1 \in E_{x'}$ with $b_1 \geq s$ (taking any point of $E_{x'}$ below or above $s$ in the sense of Lebesgue measure distribution). Similarly, there exist $a_2 \in E_{y'}$ with $a_2 \leq t$ and $b_2 \in E_{y'}$ with $b_2 \geq t$. [claim:The key distance bound] For $s, t \in \mathbb{R}$ and intervals with points $a_1 \leq s$, $b_1 \geq s$, $a_2 \leq t$, $b_2 \geq t$, we have \begin{align*} (s - t)^2 \leq \max\{(b_1 - a_2)^2, (b_2 - a_1)^2\}. \end{align*} [/claim] [proof] If $s \geq t$, then $s - t \leq b_1 - a_2$ (since $s \leq b_1$ and $t \geq a_2$), so $(s - t)^2 \leq (b_1 - a_2)^2$. If $s < t$, then $t - s \leq b_2 - a_1$, so $(s - t)^2 = (t - s)^2 \leq (b_2 - a_1)^2$. [/proof] Therefore, using that $(x', a_1), (x', b_1), (y', a_2), (y', b_2) \in E$: \begin{align*} |x' - y'|^2 + (s - t)^2 &\leq |x' - y'|^2 + \max\{(b_1 - a_2)^2, (b_2 - a_1)^2\} \\ &= \max\{|(x', b_1) - (y', a_2)|^2, |(y', b_2) - (x', a_1)|^2\} \\ &\leq (\operatorname{diam}(E))^2, \end{align*} since all four points $(x', a_1), (x', b_1), (y', a_2), (y', b_2)$ lie in $E$. Taking the supremum over all $(x', s), (y', t) \in \sigma_i(E)$ gives $\operatorname{diam}(\sigma_i(E)) \leq \operatorname{diam}(E)$. [/step] [step:Symmetrize in all $n$ coordinates and conclude the inequality] Define the fully symmetrized set \begin{align*} E^* = \sigma_n \circ \sigma_{n-1} \circ \cdots \circ \sigma_1(E). \end{align*} By the previous two steps, each symmetrization preserves $\mathcal{L}^n$-measure and does not increase the diameter. Therefore \begin{align*} \mathcal{L}^n(E^*) = \mathcal{L}^n(E) \qquad \text{and} \qquad \operatorname{diam}(E^*) \leq \operatorname{diam}(E). \end{align*} The set $E^*$ is symmetric about the origin in each coordinate: $\sigma_i(E^*)_{x'} = (-\ell/2, \ell/2)$ for some $\ell \geq 0$, meaning that if $(x_1, \ldots, x_n) \in E^*$, then replacing any $x_i$ by $-x_i$ also gives a point in $E^*$. For such a set, every point $x \in E^*$ satisfies \begin{align*} |x| = \sqrt{x_1^2 + \cdots + x_n^2} \leq \frac{\operatorname{diam}(E^*)}{2}, \end{align*} because $x$ and $-x$ both lie in $E^*$ (by symmetry in all coordinates), so $|x - (-x)| = 2|x| \leq \operatorname{diam}(E^*)$. Therefore $E^* \subset \overline{B}(0, \operatorname{diam}(E^*)/2)$, and by monotonicity of $\mathcal{L}^n$: \begin{align*} \mathcal{L}^n(E) = \mathcal{L}^n(E^*) \leq \mathcal{L}^n\!\left(\overline{B}\!\left(0, \frac{\operatorname{diam}(E^*)}{2}\right)\right) = \alpha(n) \left(\frac{\operatorname{diam}(E^*)}{2}\right)^n \leq \alpha(n) \left(\frac{\operatorname{diam}(E)}{2}\right)^n. \end{align*} [/step]