[proofplan]
The proof establishes both inequalities. For $\mathcal{H}^n(E) \leq \mathcal{L}^n(E)$, we cover $E$ by small balls using the outer regularity of $\mathcal{L}^n$ and observe that each ball contributes $\alpha(n) r^n = \mathcal{L}^n(B(x, r))$ to both the Hausdorff covering sum and the Lebesgue measure. For $\mathcal{H}^n(E) \geq \mathcal{L}^n(E)$, we apply the [Isodiametric Inequality](/theorems/3047) to each covering set: the Lebesgue measure of any set $C$ is at most $\alpha(n)(\operatorname{diam}(C)/2)^n$, so the Lebesgue measure of a union is controlled by the Hausdorff covering sum.
[/proofplan]
[step:Prove $\mathcal{H}^n(E) \leq \mathcal{L}^n(E)$ by covering with small balls]
If $\mathcal{L}^n(E) = +\infty$, the inequality is immediate. Assume $\mathcal{L}^n(E) < +\infty$.
Fix $\delta > 0$ and $\varepsilon > 0$. By outer regularity of Lebesgue measure, there exists an open set $U \supset E$ with $\mathcal{L}^n(U) \leq \mathcal{L}^n(E) + \varepsilon$. Since $U$ is open, for each $x \in U$ there exists $r_x \in (0, \delta/2]$ with $B(x, r_x) \subset U$. The collection $\mathcal{F} = \{\overline{B}(x, r_x) : x \in U\}$ is a Vitali cover of $U$ by closed balls with uniformly bounded diameter $\sup \operatorname{diam}(\overline{B}) \leq \delta < \infty$.
Apply the [Besicovitch Covering Theorem](/theorems/3021): there exists a dimensional constant $N = N(n)$ and at most $N$ countable subfamilies $\mathcal{G}_1, \ldots, \mathcal{G}_N$ of $\mathcal{F}$, each consisting of pairwise disjoint balls, such that $U \subset \bigcup_{i=1}^N \bigcup_{B \in \mathcal{G}_i} B$. Since each $\mathcal{G}_i$ consists of disjoint balls contained in $U$:
\begin{align*}
\sum_{B \in \mathcal{G}_i} \mathcal{L}^n(B) \leq \mathcal{L}^n(U) \leq \mathcal{L}^n(E) + \varepsilon \qquad \text{for each } i = 1, \ldots, N.
\end{align*}
At least one subfamily $\mathcal{G}_{i_0}$ satisfies $\mathcal{L}^n\!\left(E \cap \bigcup_{B \in \mathcal{G}_{i_0}} B\right) \geq \mathcal{L}^n(E)/N$ (since the $N$ subfamilies cover $E$). However, for the upper bound we do not need this selection --- we simply use all balls from all subfamilies as a $\delta$-cover.
Let $\{B_j\}_{j=1}^\infty$ be an enumeration of $\bigcup_{i=1}^N \mathcal{G}_i$. This is a countable $\delta$-cover of $E$ (since $E \subset U$ and each $B_j$ has diameter at most $\delta$). Each ball $B_j = \overline{B}(x_j, r_j)$ satisfies $\alpha(n)(\operatorname{diam}(B_j)/2)^n = \alpha(n) r_j^n = \mathcal{L}^n(B_j)$. Therefore
\begin{align*}
\mathcal{H}^n_\delta(E) \leq \sum_{j=1}^\infty \alpha(n)\left(\frac{\operatorname{diam}(B_j)}{2}\right)^n = \sum_{j=1}^\infty \mathcal{L}^n(B_j) = \sum_{i=1}^N \sum_{B \in \mathcal{G}_i} \mathcal{L}^n(B) \leq N \cdot (\mathcal{L}^n(E) + \varepsilon).
\end{align*}
This gives $\mathcal{H}^n_\delta(E) \leq N(\mathcal{L}^n(E) + \varepsilon)$ with the dimensional constant $N = N(n)$. The constant $N$ is harmless: it does not depend on $\delta$ or $\varepsilon$, but it shows only that $\mathcal{H}^n(E) \leq N \cdot \mathcal{L}^n(E)$. To obtain the sharp constant $1$, we refine the argument as follows.
For $E$ with $\mathcal{L}^n(E) < +\infty$, the set $E$ is Lebesgue measurable. For any $\varepsilon > 0$ and any $\delta > 0$, the [Vitali Covering Lemma for Sets](/theorems/3024) applied to the Vitali cover $\{\overline{B}(x, r) : x \in E, \, 0 < r \leq \delta/2, \, \overline{B}(x,r) \subset U\}$ yields a countable disjoint collection $\{\overline{B}(x_j, r_j)\}$ such that $\mathcal{L}^n(E \setminus \bigcup_j \overline{B}(x_j, r_j)) = 0$. Since the balls are disjoint and contained in $U$:
\begin{align*}
\sum_j \mathcal{L}^n(\overline{B}(x_j, r_j)) \leq \mathcal{L}^n(U) \leq \mathcal{L}^n(E) + \varepsilon.
\end{align*}
The balls $\{\overline{B}(x_j, r_j)\}$ cover $E$ up to a set of $\mathcal{L}^n$-measure zero. By Borel regularity of $\mathcal{L}^n$, the exceptional null set can be covered by additional balls of arbitrarily small total $\mathcal{H}^n_\delta$-measure (less than $\varepsilon$). Combining:
\begin{align*}
\mathcal{H}^n_\delta(E) \leq \sum_j \alpha(n) r_j^n + \varepsilon = \sum_j \mathcal{L}^n(\overline{B}(x_j, r_j)) + \varepsilon \leq \mathcal{L}^n(E) + 2\varepsilon.
\end{align*}
Sending $\varepsilon \to 0$ and then $\delta \to 0$: $\mathcal{H}^n(E) \leq \mathcal{L}^n(E)$.
[/step]
[step:Prove $\mathcal{H}^n(E) \geq \mathcal{L}^n(E)$ using the Isodiametric Inequality]
Fix $\delta > 0$ and let $\{C_j\}_{j=1}^\infty$ be any countable $\delta$-cover of $E$ (so $E \subset \bigcup_j C_j$ and $\operatorname{diam}(C_j) \leq \delta$ for all $j$). By monotonicity and countable subadditivity of $\mathcal{L}^n$:
\begin{align*}
\mathcal{L}^n(E) \leq \mathcal{L}^n\!\left(\bigcup_{j=1}^\infty C_j\right) \leq \sum_{j=1}^\infty \mathcal{L}^n(C_j).
\end{align*}
Applying the [Isodiametric Inequality](/theorems/3047) to each $C_j$:
\begin{align*}
\mathcal{L}^n(C_j) \leq \alpha(n) \left(\frac{\operatorname{diam}(C_j)}{2}\right)^n.
\end{align*}
The Isodiametric Inequality applies since each $C_j$ can be replaced by a Borel set of equal diameter and at least as large Lebesgue measure (by Borel regularity, or directly since $\mathcal{L}^n$ is defined on all subsets of $\mathbb{R}^n$ as an outer measure). Substituting:
\begin{align*}
\mathcal{L}^n(E) \leq \sum_{j=1}^\infty \alpha(n) \left(\frac{\operatorname{diam}(C_j)}{2}\right)^n.
\end{align*}
Taking the infimum over all $\delta$-covers $\{C_j\}$ of $E$:
\begin{align*}
\mathcal{L}^n(E) \leq \mathcal{H}^n_\delta(E).
\end{align*}
This holds for every $\delta > 0$. Taking $\delta \to 0^+$ (equivalently, the supremum over $\delta > 0$):
\begin{align*}
\mathcal{L}^n(E) \leq \mathcal{H}^n(E).
\end{align*}
[/step]
[step:Combine both inequalities to conclude $\mathcal{H}^n(E) = \mathcal{L}^n(E)$]
The previous two steps establish:
\begin{align*}
\mathcal{L}^n(E) \leq \mathcal{H}^n(E) \leq \mathcal{L}^n(E)
\end{align*}
for every Borel set $E \subset \mathbb{R}^n$. Therefore $\mathcal{H}^n(E) = \mathcal{L}^n(E)$.
Since both $\mathcal{H}^n$ and $\mathcal{L}^n$ are Borel regular outer measures (by [$\mathcal{H}^s$ is a Borel Regular Outer Measure](/theorems/3044) for $\mathcal{H}^n$, and by standard properties for $\mathcal{L}^n$), and they agree on all Borel sets, they agree on all subsets of $\mathbb{R}^n$: for any $A \subset \mathbb{R}^n$, Borel regularity gives Borel sets $B_1 \supset A$ and $B_2 \supset A$ with $\mathcal{H}^n(B_1) = \mathcal{H}^n(A)$ and $\mathcal{L}^n(B_2) = \mathcal{L}^n(A)$. Setting $B = B_1 \cap B_2$:
\begin{align*}
\mathcal{H}^n(A) \leq \mathcal{H}^n(B) = \mathcal{L}^n(B) \leq \mathcal{L}^n(A),
\end{align*}
and symmetrically $\mathcal{L}^n(A) \leq \mathcal{H}^n(A)$, so $\mathcal{H}^n(A) = \mathcal{L}^n(A)$ for all $A \subset \mathbb{R}^n$.
[/step]
