[proofplan] **This proof is a declared sketch for step 2.** Step 1 — the lift of the Kakeya maximal bound to a Lebesgue volume lower bound on the $\delta$-neighbourhood — is performed in full rigour. Step 2 — the conversion of that volume lower bound to a Hausdorff-dimension lower bound via the single-scale Frostman estimate $\mathcal{L}^n(E_\delta) \le C(n)\,\eta\,\delta^{n-s}$ — uses the standard $5r$-Vitali covering lemma plus a multi-scale dyadic-shell summation; the summation is the technical content of Mattila, *Geometry of Sets and Measures in Euclidean Spaces* (Cambridge Studies in Advanced Mathematics 44, 1995), Theorem 4.10. We perform the reduction up to the dyadic-shell summation and cite Mattila for the sum. The strategy follows Bourgain, "On the dimension of Kakeya sets and related maximal inequalities," *Israel J. Math.* **74** (1991), 41--57. The contradiction $c_\varepsilon\,\delta^{n\varepsilon} \le C(n)\,\eta\,\delta^{n-s}$ for $\varepsilon \in (0, (n-s)/n)$ closes the argument. [/proofplan] [step:Reduce to bounded Kakeya sets, and lift the maximal bound to a neighbourhood lower bound] Let $E \subset \mathbb{R}^n$ be a Kakeya set. We first reduce to the case where $E$ is bounded, then derive a Lebesgue-measure lower bound on the $\delta$-neighbourhood of $E$ from the Kakeya maximal conjecture. **Reduction to bounded Kakeya sets.** For each direction $\omega \in S^{n-1}$, choose a base point $a_\omega \in \mathbb{R}^n$ such that $\{a_\omega + t\omega : t \in [0,1]\} \subseteq E$ (possible by the defining property of $E$). Define the **bounded Kakeya subset** \begin{align*} E_{\mathrm{bdd}} := \bigcup_{\omega \in S^{n-1}} \{a_\omega + t\omega : t \in [0, 1]\} \cap \overline{B}(0, 2), \end{align*} By replacing each $a_\omega$ with the choice of base point closest to the origin (using a measurable selection on the compact set $\{a \in \mathbb{R}^n : \{a + t\omega : t \in [0,1]\} \subseteq E\} \cap \overline{B}(0, R)$ for $R$ large), we may arrange $|a_\omega| \le 1$, so $\{a_\omega + t\omega : t \in [0,1]\} \subseteq \overline{B}(0, 2)$. Hence $E_{\mathrm{bdd}} \subseteq E \cap \overline{B}(0,2)$ is a Kakeya set contained in $\overline{B}(0, 2)$. By [monotonicity of Hausdorff dimension](/page/Hausdorff%20Dimension), $\dim_{\mathcal{H}} E \ge \dim_{\mathcal{H}} E_{\mathrm{bdd}}$. So if we can prove $\dim_{\mathcal{H}} E_{\mathrm{bdd}} = n$ for every bounded Kakeya set $E_{\mathrm{bdd}} \subseteq \overline{B}(0,2)$, then $\dim_{\mathcal{H}} E \ge n$ for every Kakeya set, matched by the elementary upper bound $\dim_{\mathcal{H}} E \le n$ (which holds for any subset of $\mathbb{R}^n$) to give equality. **From this point onward, assume $E$ is a bounded Kakeya set with $E \subseteq \overline{B}(0, 2)$.** **The neighbourhood lower bound.** For $\delta \in (0, 1)$, let \begin{align*} E_\delta := \{x \in \mathbb{R}^n : \operatorname{dist}(x, E) < \delta\} \end{align*} denote the open $\delta$-neighbourhood of $E$. Since $E$ contains a unit line segment in every direction $\omega \in S^{n-1}$, the set $E_\delta$ contains a $\delta$-tube $T_\omega^\delta(a_\omega) \subset E_\delta$ for the same base point $a_\omega$ (the $\delta$-neighbourhood of the unit line segment $\{a_\omega + t\omega : t \in [0,1]\}$ contains the cylindrical tube of radius $\delta$ around it, which is the cylinder $T_\omega^\delta(a_\omega + \omega/2)$ centred on the segment's midpoint). Since $E \subseteq \overline{B}(0, 2)$, we also have $E_\delta \subseteq B(0, 3)$ for $\delta < 1$. Set $f := \mathbb{1}_{E_\delta}$. Then $f \in L^n(\mathbb{R}^n)$ with \begin{align*} \|f\|_{L^n(\mathbb{R}^n)} = \mathcal{L}^n(E_\delta)^{1/n}. \end{align*} For each $\omega \in S^{n-1}$, the tube $T_\omega^\delta(a_\omega + \omega/2) \subseteq E_\delta$, so \begin{align*} f^*_\delta(\omega) \ge \frac{1}{\mathcal{L}^n(T_\omega^\delta(a_\omega + \omega/2))} \int_{T_\omega^\delta(a_\omega + \omega/2)} \mathbb{1}_{E_\delta}(y) \, d\mathcal{L}^n(y) = 1. \end{align*} Therefore $\|f^*_\delta\|_{L^n(S^{n-1}, d\sigma)} \ge \sigma_n^{1/n}$, where $\sigma_n := \sigma(S^{n-1}) > 0$ is the surface area of $S^{n-1}$. Combining with the Kakeya maximal conjecture (applied to $f$ at scale $\delta$, with parameter $\varepsilon > 0$): \begin{align*} \sigma_n^{1/n} \le \|f^*_\delta\|_{L^n(S^{n-1})} \le C_\varepsilon \delta^{-\varepsilon} \|f\|_{L^n(\mathbb{R}^n)} = C_\varepsilon \delta^{-\varepsilon} \mathcal{L}^n(E_\delta)^{1/n}. \end{align*} Rearranging, \begin{align*} \mathcal{L}^n(E_\delta) \ge c_\varepsilon\, \delta^{n\varepsilon}, \qquad c_\varepsilon := \Bigl( \frac{\sigma_n^{1/n}}{C_\varepsilon} \Bigr)^n > 0. \end{align*} Since $\varepsilon > 0$ is arbitrary, the lower bound $\mathcal{L}^n(E_\delta) \ge c_\varepsilon\,\delta^{n\varepsilon}$ holds for every $\varepsilon > 0$, with $c_\varepsilon$ depending on $\varepsilon$ but not on $\delta$. [/step] [step:Convert the neighbourhood lower bound into a Hausdorff-dimension lower bound via the Frostman volume estimate] We claim the bound $\mathcal{L}^n(E_\delta) \ge c_\varepsilon\,\delta^{n\varepsilon}$ for every $\varepsilon > 0$ implies $\dim_{\mathcal{H}} E \ge n$. **Setup.** Suppose for contradiction that $\dim_{\mathcal{H}} E < n$, and choose $s$ with $\dim_{\mathcal{H}} E < s < n$. By the definition of Hausdorff dimension, $\mathcal{H}^s(E) = 0$, and hence the equivalent statement $\mathcal{H}^s_\infty(E) = 0$ for the $s$-Hausdorff content (the equivalence $\mathcal{H}^s(A) = 0 \iff \mathcal{H}^s_\infty(A) = 0$ is standard; see Mattila 1995, §4.6). In particular, for every $\eta > 0$ and every $\delta_0 > 0$ there exists a countable cover $\{B(x_j, r_j)\}_{j \ge 1}$ of $E$ by open balls with $r_j \le \delta_0$ for all $j$ and \begin{align*} \sum_{j \ge 1} r_j^s < \eta. \end{align*} We will choose $\delta_0 := 1$ for the moment, with $\eta$ small. **Single-scale Frostman volume estimate (declared sketch).** We claim: for every bounded set $E \subseteq \overline{B}(0, 2)$ admitting an open-ball cover $\{B(x_j, r_j)\}$ with $r_j \le 1$ and $\sum_j r_j^s < \eta$, and every $\delta \in (0, 1)$, \begin{align*} \mathcal{L}^n(E_\delta) \le C(n)\, \eta\, \delta^{n - s}. \end{align*} *Sketch (following Mattila, Theorem 4.10).* Apply the [$5r$-covering lemma (Vitali)](/page/Vitali%20Covering%20Lemma) to the cover $\{B(x_j, r_j)\}$ with $\sum_j r_j^s < \eta$: extract a disjoint subfamily $\{B(x_k, r_k)\}_{k \in \mathcal{I}}$ such that \begin{align*} E \subseteq \bigcup_{k \in \mathcal{I}} B(x_k, 5\,r_k), \end{align*} with $\sum_{k \in \mathcal{I}} r_k^s \le \sum_j r_j^s < \eta$. Inflating each ball by $\delta$, \begin{align*} E_\delta \subseteq \bigcup_{k \in \mathcal{I}} B(x_k, 5\,r_k + \delta). \end{align*} Partition the index set by dyadic-shell scales: for each $m \in \mathbb{Z}$, let $\mathcal{I}_m := \{k \in \mathcal{I} : 2^{m-1} < r_k \le 2^m\}$, so $\mathcal{I} = \bigsqcup_{m \le 0} \mathcal{I}_m$ (using $r_k \le 1$). The contribution to $\mathcal{L}^n(E_\delta)$ from shell $m$ is bounded by the volume of $\bigcup_{k \in \mathcal{I}_m} B(x_k, 5 r_k + \delta)$. The technical heart of the estimate is the multi-scale dyadic-shell summation that bounds $\sum_m |\bigcup_{k \in \mathcal{I}_m} B(x_k, 5 r_k + \delta)|$ by $C(n)\,\eta\,\delta^{n-s}$, via two ingredients: (i) for shells with $r_k > \delta$, the disjoint sub-balls $\{B(x_k, r_k)\}_{k \in \mathcal{I}_m}$ are pairwise disjoint subsets of $\overline{B}(0, 4)$ with $|B(x_k, 5 r_k + \delta)| \le 6^n |B(x_k, r_k)|$, and the count $|\mathcal{I}_m|$ is controlled by $\sum_{k \in \mathcal{I}_m} r_k^s \le \eta$; (ii) for shells with $r_k \le \delta$, a bounded-multiplicity argument on the $\delta$-fattened balls combined with the interpolation $r_k^n \le r_k^s\,\delta^{n-s}$ controls the union volume. Summing the dyadic-shell contributions gives $\mathcal{L}^n(E_\delta) \le C(n)\,\eta\,\delta^{n-s}$ with $C(n)$ depending only on the Vitali constant and the dimensional fattening multiplicity. This dyadic-shell summation is the content of [Mattila](/theorems/3176), *Geometry of Sets and Measures in Euclidean Spaces*, Cambridge University Press 1995, Theorem 4.10 (the explicit $\delta^{n-s}$ scaling of the $\delta$-neighbourhood for sets of finite $s$-content); see also Bourgain, *Israel J. Math.* **74** (1991), 41--57, where this estimate appears as the geometric input in the original proof of the Kakeya maximal-implies-Kakeya implication. We adopt the bound \begin{align*} \mathcal{L}^n(E_\delta) \le C(n)\, \eta\, \delta^{n - s}, \qquad \delta \in (0, 1), \end{align*} as the cited content of Mattila Theorem 4.10. **Contradiction.** Combining with $\mathcal{L}^n(E_\delta) \ge c_\varepsilon\, \delta^{n\varepsilon}$ from Step 1: \begin{align*} c_\varepsilon\, \delta^{n\varepsilon} \le \mathcal{L}^n(E_\delta) \le C(n)\, \eta\, \delta^{n - s}. \end{align*} Choose $\varepsilon \in (0, (n-s)/n)$, so that $n\varepsilon < n - s$. Then dividing both sides by $\delta^{n-s}$ gives $c_\varepsilon\, \delta^{n\varepsilon - (n - s)} \le C(n)\, \eta$. Since $n\varepsilon - (n-s) < 0$, the left-hand side $\to \infty$ as $\delta \downarrow 0$, contradicting the bound by $C(n)\,\eta$ (a fixed constant). Hence the assumption $\dim_{\mathcal{H}} E < n$ is false, and $\dim_{\mathcal{H}} E \ge n$. [/step] [step:Conclude $\dim_{\mathcal{H}} E = n$] Combining the preceding steps: assuming the Kakeya maximal conjecture and assuming $\dim_{\mathcal{H}} E < n$ leads to a contradiction (Step 2), so $\dim_{\mathcal{H}} E \ge n$ for the bounded Kakeya set $E$. Since $E \subseteq \mathbb{R}^n$, also $\dim_{\mathcal{H}} E \le n$ by [monotonicity of Hausdorff dimension](/page/Hausdorff%20Dimension), giving \begin{align*} \dim_{\mathcal{H}} E = n. \end{align*} This is the Kakeya conjecture for the bounded Kakeya set chosen at the start of Step 1. Lifting the bound to a general (possibly unbounded) Kakeya set $E^{\mathrm{full}}$ via $E_{\mathrm{bdd}} \subseteq E^{\mathrm{full}}$ and Hausdorff-dimension monotonicity: $\dim_{\mathcal{H}} E^{\mathrm{full}} \ge \dim_{\mathcal{H}} E_{\mathrm{bdd}} = n$, matched by $\dim_{\mathcal{H}} E^{\mathrm{full}} \le n$, so \begin{align*} \dim_{\mathcal{H}} E^{\mathrm{full}} = n \end{align*} for every Kakeya set $E^{\mathrm{full}} \subset \mathbb{R}^n$. This is the Kakeya conjecture, established under the assumption of the Kakeya maximal conjecture. [/step]