[proofplan] This proof is a sketch following Davies' 1971 argument; the core mechanism is reproduced rigorously, but a few combinatorial estimates are recorded as named [claim] sub-results whose proofs are themselves substantial. The strategy is by contraposition on Hausdorff dimension. Suppose $E$ has dimension strictly less than $2$. Then there exists $\beta < 2$ with $\mathcal{H}^\beta(E) < \infty$; we choose covers by sets of diameter $\le \delta$ summing to a controlled $\beta$-content and pass to a $\delta$-thickening of $E$, obtaining a planar set of Lebesgue area $\lesssim \delta^{2 - \beta}$ that nonetheless contains a $\delta$-tube in every direction. We then exploit the bound on direction-separated tubes: in the plane, any family of $\sim \delta^{-1}$ tubes of dimensions $\delta \times 1$ pointing in distinct $\delta$-separated directions has total area $\gtrsim |\log \delta|^{-1}$, far larger than $\delta^{2-\beta}$ as $\delta \downarrow 0$ for any $\beta < 2$. The contradiction forces $\dim_{\mathcal{H}} E \ge 2$, and the elementary upper bound from $E \subseteq \mathbb{R}^2$ gives equality. The proof has four steps: pass to $\delta$-thickenings, formalise the tube family, apply the planar tube-area bound, derive the contradiction. [/proofplan]

[step:Pass from a hypothetical $\beta$-dimensional cover to a $\delta$-neighbourhood with area $\lesssim \delta^{2-\beta}$] Suppose for contradiction that \begin{align*} \dim_{\mathcal{H}} E < 2. \end{align*} Then there exists $\beta \in [1, 2)$ with $\mathcal{H}^\beta(E) < \infty$ (the lower endpoint $\beta \ge 1$ comes from [Theorem 3207](/theorems/3207), which gives $\dim_{\mathcal{H}} E \ge 1$). By the [definition of Hausdorff measure](/page/Hausdorff%20Measure), for every $\delta > 0$ there exists a countable cover $\{B_j\}_{j \in \mathbb{N}}$ of $E$ with $\operatorname{diam}(B_j) \le \delta$ and \begin{align*} \sum_j (\operatorname{diam}(B_j))^\beta \le \mathcal{H}^\beta(E) + 1 =: M < \infty. \end{align*} Without loss of generality (by replacing each $B_j$ with the closed Euclidean ball of the same diameter centred at a point of $B_j$), we may assume each $B_j$ is a closed Euclidean ball $B_j = \overline{B}(x_j, r_j)$ with $r_j = \operatorname{diam}(B_j)/2 \le \delta/2$. Define the $\delta$-neighbourhood \begin{align*} E_\delta := \bigcup_{j \in \mathbb{N}} B_j \subseteq \mathbb{R}^2, \end{align*} which is an open superset of $E$ (after a slight enlargement, replace $B_j$ by an open ball of the same centre and slightly larger radius; the area estimate below absorbs the constants). The Lebesgue area of $E_\delta$ is bounded via subadditivity by \begin{align*} \mathcal{L}^2(E_\delta) \le \sum_j \mathcal{L}^2(B_j) = \sum_j \pi r_j^2 \le \pi \sum_j r_j^\beta r_j^{2 - \beta} \le \pi (\delta/2)^{2-\beta} \sum_j r_j^\beta \le \pi (\delta/2)^{2-\beta} M, \end{align*} where we used $r_j^{2-\beta} \le (\delta/2)^{2-\beta}$ (since $r_j \le \delta/2$ and $2 - \beta > 0$) and $r_j^\beta \le (\operatorname{diam} B_j)^\beta \cdot 2^{-\beta} \le (\operatorname{diam} B_j)^\beta$ (after collecting the $2^{-\beta}$ factor into the constant). Therefore \begin{align*} \mathcal{L}^2(E_\delta) \le C_1 \delta^{2 - \beta} \end{align*} for a constant $C_1 = C_1(\beta, M) > 0$ independent of $\delta$. [/step]

[step:Construct a family of $\sim \delta^{-1}$ direction-separated $\delta$-tubes inside $E_{C\delta}$] Recall that for every direction $e \in S^1$ there exists $a_e \in \mathbb{R}^2$ with the unit segment $S_e := \{a_e + t e : t \in [0, 1]\} \subseteq E$. Define the **$\delta$-tube** in direction $e$ as the closed Euclidean $\delta$-neighbourhood of $S_e$: \begin{align*} T_e^\delta := \{x \in \mathbb{R}^2 : \operatorname{dist}(x, S_e) \le \delta\}, \end{align*} which is a closed convex set congruent to a $1 \times 2\delta$ rectangle with two semicircular caps of radius $\delta$, of Lebesgue area \begin{align*} \mathcal{L}^2(T_e^\delta) = 2\delta + \pi \delta^2. \end{align*} For $\delta \le 1$, $\mathcal{L}^2(T_e^\delta) \in [2\delta, (2 + \pi)\delta]$, so $\mathcal{L}^2(T_e^\delta) \asymp \delta$. Choose a maximal $\delta$-separated subset $\{e_1, \ldots, e_N\} \subset S^1$, where "$\delta$-separated" means $|e_i - e_j| \ge \delta$ for $i \neq j$. The cardinality satisfies \begin{align*} c_0 \delta^{-1} \le N \le C_2 \delta^{-1} \end{align*} for absolute constants $c_0, C_2 > 0$ (by a standard volume-packing estimate on $S^1$, whose $1$-dimensional Hausdorff measure is $2\pi$). For each $i$, since $S_{e_i} \subseteq E$ and the cover from Step 1 covers $E$, the $\delta$-thickening $T_{e_i}^\delta$ is contained in the $\delta$-thickening $E_\delta$ enlarged by an additional $\delta$, that is, in $E_{2\delta}$. The bound from Step 1 applies (with $\delta$ replaced by $2\delta$): \begin{align*} \mathcal{L}^2(E_{2\delta}) \le C_1 (2\delta)^{2 - \beta} = C_1 \cdot 2^{2-\beta} \cdot \delta^{2-\beta} \le C_3 \delta^{2-\beta}. \end{align*} [/step]

[step:Apply the planar tube area bound: direction-separated $\delta$-tubes have total area $\gtrsim |\log \delta|^{-1}$] We invoke the central combinatorial estimate of the proof, the planar bushiness lemma: [claim:Planar bushiness lemma] There exists a constant $c_4 > 0$ such that for every $\delta \in (0, 1/2)$ and every family $\{T_{e_i}^\delta\}_{i=1}^N$ of $\delta$-tubes pointing in $\delta$-separated directions $\{e_1, \ldots, e_N\} \subset S^1$ with $N \ge c_0 \delta^{-1}$, \begin{align*} \mathcal{L}^2 \!\left( \bigcup_{i=1}^N T_{e_i}^\delta \right) \ge \frac{c_4}{|\log \delta|}. \end{align*} [/claim] [proof] We give Davies' planar incidence argument explicitly. The strategy is a Cauchy--Schwarz $L^2$ estimate on the sum of indicator functions of the tubes, exploiting the bilinear "two tubes intersect in a small parallelogram" geometry. Let $T_i := T_{e_i}^\delta$ denote the $i$-th tube, $i = 1, \ldots, N$ with $N \asymp \delta^{-1}$, and write \begin{align*} F : \mathbb{R}^2 &\to \mathbb{N} \\ x &\mapsto \sum_{i=1}^N \mathbb{1}_{T_i}(x). \end{align*} The integral and the support of $F$ are \begin{align*} \int_{\mathbb{R}^2} F\,d\mathcal{L}^2 = \sum_{i=1}^N \mathcal{L}^2(T_i) \asymp N \cdot \delta \asymp 1, \qquad \operatorname{supp}(F) = \bigcup_{i=1}^N T_i. \end{align*} By the Cauchy--Schwarz inequality applied to $F = F \cdot \mathbb{1}_{\bigcup T_i}$, \begin{align*} \Bigl(\int F\,d\mathcal{L}^2\Bigr)^2 \le \mathcal{L}^2\!\Bigl(\bigcup_i T_i\Bigr) \cdot \int F^2\,d\mathcal{L}^2. \end{align*} Hence \begin{align*} \mathcal{L}^2\!\Bigl(\bigcup_i T_i\Bigr) \ge \frac{(\int F\,d\mathcal{L}^2)^2}{\int F^2\,d\mathcal{L}^2} \asymp \frac{1}{\int F^2\,d\mathcal{L}^2}. \end{align*} It remains to estimate $\int F^2\,d\mathcal{L}^2 = \sum_{i,j=1}^N \mathcal{L}^2(T_i \cap T_j)$ from above by $C\,|\log \delta|$. **Pairwise tube-intersection estimate.** Fix $i \neq j$ and let $\theta_{ij} := \angle(e_i, e_j) \in (0, \pi)$ be the angle between the two directions. The intersection $T_i \cap T_j$ is contained in a parallelogram with sides parallel to $e_i, e_j$ of widths $\le 2\delta$ each in the orthogonal direction; its area is \begin{align*} \mathcal{L}^2(T_i \cap T_j) \le \min\Bigl(\mathcal{L}^2(T_i), \frac{(2\delta)^2}{\sin \theta_{ij}}\Bigr) \le \min\Bigl(C\delta, \frac{4\delta^2}{\sin\theta_{ij}}\Bigr). \end{align*} For small $\theta$, $\sin \theta \asymp \theta$, so for $\theta_{ij} \ge \delta$ (the regime in which the directions are $\delta$-separated), \begin{align*} \mathcal{L}^2(T_i \cap T_j) \le \frac{C\delta^2}{\theta_{ij}}. \end{align*} For $\theta_{ij} < \delta$ — which does not occur because the directions are $\delta$-separated, so $|e_i - e_j| \ge \delta$, giving $\sin \theta_{ij} \asymp \theta_{ij} \ge c\delta$ for an absolute constant $c > 0$ — the elementary bound $\mathcal{L}^2(T_i \cap T_j) \le C\delta$ from the inclusion $T_i \cap T_j \subseteq T_i$ would suffice; however this case is empty by the $\delta$-separation hypothesis. **Sum the pairwise estimates.** Since the directions $\{e_1, \ldots, e_N\}$ are $\delta$-separated on $S^1$ — a circle of circumference $2\pi$ — for each fixed $i$ the angles $\{\theta_{ij} : j \ne i\}$ form a $\delta$-separated subset of $(0, \pi)$. Group $j \ne i$ by the dyadic angle scale $\theta_{ij} \in [2^k \delta, 2^{k+1}\delta)$ for $k = 0, 1, 2, \ldots, K$ where $K \le \log_2(\pi/\delta) \asymp |\log \delta|$. The number of indices $j$ in each dyadic group is $\le C$ (an absolute constant, by $\delta$-separation: each group is an angular arc of length $\asymp 2^k \delta$, containing at most $C \cdot 2^k$ angles at $\delta$-separation — but only those within distance $2^{k+1}\delta$ of $e_i$, so at most $\le C\cdot 2^{k+1}$ indices). More carefully, by a one-dimensional ball-packing argument on $S^1$, \begin{align*} \#\{j : \theta_{ij} \in [2^k\delta, 2^{k+1}\delta)\} \le C\cdot 2^{k+1}, \qquad k = 0, 1, \ldots, K. \end{align*} Hence for each fixed $i$, \begin{align*} \sum_{j \neq i} \mathcal{L}^2(T_i \cap T_j) &\le \sum_{k=0}^K \#\{j : \theta_{ij} \in [2^k\delta, 2^{k+1}\delta)\} \cdot \frac{C\delta^2}{2^k\delta} \\ &\le \sum_{k=0}^K C \cdot 2^{k+1} \cdot \frac{C\delta}{2^k} = 2C^2 \delta\,(K+1) \le C'\delta\,|\log\delta|. \end{align*} The diagonal $j = i$ contributes $\mathcal{L}^2(T_i) \le C\delta$ to the sum. Summing over $i$ from $1$ to $N$ with $N \asymp \delta^{-1}$, \begin{align*} \int F^2\,d\mathcal{L}^2 = \sum_{i, j} \mathcal{L}^2(T_i \cap T_j) \le N \cdot (C\delta + C'\delta\,|\log\delta|) \le C''\,|\log\delta|. \end{align*} **Conclude the area lower bound.** Substituting back into Cauchy--Schwarz, \begin{align*} \mathcal{L}^2\!\Bigl(\bigcup_i T_i\Bigr) \ge \frac{c}{\int F^2\,d\mathcal{L}^2} \ge \frac{c_4}{|\log\delta|}, \end{align*} for an absolute constant $c_4 > 0$. This is the asserted bound. **Remark on the logarithmic loss.** The factor $|\log\delta|^{-1}$ comes precisely from the harmonic-series sum over dyadic angle scales in the tube-intersection estimate: the contribution of pairs at angular separation $\theta \asymp 2^k\delta$ is $\delta\cdot N$ (independent of $k$), and there are $\asymp |\log\delta|$ scales. The $L^2$ method cannot avoid this loss; sharper bounds (without the $\log$) require deeper combinatorial techniques (e.g. Wolff's hairbrush, Davies' original two-direction trick), and for the purposes of Davies' theorem the logarithmic bound suffices. [/proof] Combining the lower bound from the lemma with the upper bound from Step 1: \begin{align*} \frac{c_4}{|\log \delta|} \le \mathcal{L}^2 \!\left( \bigcup_{i=1}^N T_{e_i}^\delta \right) \le \mathcal{L}^2(E_{2\delta}) \le C_3 \delta^{2 - \beta}. \end{align*} [/step]

[step:Derive a contradiction by sending $\delta \downarrow 0$] From Step 3 we have, for every $\delta \in (0, 1/2)$, \begin{align*} \frac{c_4}{|\log \delta|} \le C_3 \delta^{2 - \beta}. \end{align*} Equivalently, \begin{align*} \delta^{\beta - 2} \le \frac{C_3}{c_4} |\log \delta|. \end{align*} Since $\beta < 2$, we have $\beta - 2 < 0$, so $\delta^{\beta - 2} \to \infty$ as $\delta \downarrow 0$. The right-hand side, $|\log \delta|$, also tends to infinity, but only logarithmically. Comparing growth rates: for $\delta \in (0, 1/2)$, \begin{align*} \frac{\delta^{\beta - 2}}{|\log \delta|} = \frac{1}{\delta^{2-\beta} |\log \delta|} \xrightarrow[\delta \downarrow 0]{} \infty \end{align*} because $\delta^{2 - \beta} \to 0$ polynomially and $|\log \delta| \to \infty$ only logarithmically, and any polynomial decay dominates any logarithmic growth. In particular, the ratio cannot be bounded above by the constant $C_3 / c_4$, contradicting the inequality. This contradiction rules out the hypothesis $\dim_{\mathcal{H}} E < 2$. Therefore $\dim_{\mathcal{H}} E \ge 2$. Combined with the elementary upper bound $\dim_{\mathcal{H}} E \le \dim_{\mathcal{H}} \mathbb{R}^2 = 2$, we conclude \begin{align*} \dim_{\mathcal{H}} E = 2, \end{align*} which is the claim of Davies' Theorem. [/step]