A nontrivial affine copy of a set $A \subseteq \mathbb{R}$ is a set

$$
x + sA = \{x + sa : a \in A\}
$$

with $x \in \mathbb{R}$ and $s \in \mathbb{R} \setminus \{0\}$. The set $A$ is measure universal when every Lebesgue-measurable $E \subseteq \mathbb{R}$ of positive measure contains such a copy. The Erdos similarity conjecture, recorded as Problem 4.33.7* in a list assembled at the Fifth Balkan Mathematical Congress, asserts that no infinite $A$ is measure universal. Erdos states the problem as follows: for every infinite set $E$ of real numbers, there is a set $S$ of positive measure containing no image of $E$ similar to it in the sense of elementary geometry, with $E$ assumed denumerable and its only limit point $0$ outside $E$.

refs/erdos-1974-problems.pdf

Every nonempty finite pattern $F$ is measure universal. If a measurable $E_0$ has finite positive measure, translation continuity in $L^1$ gives $m(\bigcap_{a \in F} (E_0 - sa)) \to m(E_0) > 0$ as $s \to 0$, so a small nonzero shift carries a copy of $F$ inside $E_0$. For an infinite pattern, infinitely many conditions on a single point must hold at once, and the obstruction is to have a positive-measure set satisfying every condition simultaneously.

For positive sequences $a_n \downarrow 0$ with $a_{n+1}/a_n \to 1$, Falconer and, independently, Eigen proved that $\{a_n\}$ is not measure universal. Humke and Laczkovich recast the problem in terms of finite hitting sets and gave relative-gap criteria that reproduce these examples.

refs/falconer-1984-erdos-sequence-measurable.pdf
refs/humke-laczkovich-1998-visit-erdos-problem.pdf

Kolountzakis added a finite-gap criterion: a set of positive measure avoiding every affine copy of a positive sequence $a_n$ exists provided the minimum gaps of finite subsets, divided by the largest elements of these subsets, have negative logarithms that grow sublinearly in the cardinality. Chlebik gives a translation-invariant form of the criterion, normalised by the diameter of the finite subsets. Dyadic subsets $a_1 > \cdots > a_n$ taken from $D$ violate both criteria: the minimum gap is at most $a_{n-1} \le 2^{-(n-2)} a_1$ while the diameter is at least $a_1/2$, so the normalised minimum gap is at most $2^{-(n-3)}$.

refs/kolountzakis-1997-infinite-patterns.pdf
refs/arxiv-1512.05607.pdf

Additive structure gives a second family. Bourgain proved that a sum of three infinite sets is never measure universal. Kolountzakis handled sums of two copies of certain sequences, including $D + D$. Mora Cuellar, Iosevich, Kulkarni, Rojas Aravena and Yavicoli settled the two-fold sumset case with a geometric summand: for any infinite $A$ and any $0 < |r| < 1$, neither $\{a r^n : n \ge 1\} + A$ nor $\{a r^n : n \ge 1\} - A$ is measure universal. A single geometric summand is not covered, since avoiding a larger pattern does not imply avoiding a subset of it.

refs/arxiv-2607.03584.pdf

A third family exploits smoothness of measures supported on $A$. Iosevich, Kulkarni, Mora Cuellar, Rojas Aravena and Yavicoli prove nonuniversality for any $A$ that supports a probability measure whose Fourier transform tends to zero at infinity. A countable set cannot support such a Rajchman measure, so this does not cover $D$. The uniform-density criterion in the same work also fails for $D \cup \{0\}$, whose image modulo one is invariant under every dyadic dilation and stays a distance $1/4$ from $3/4$.

refs/arxiv-2609.04456.pdf

For certain sequences decreasing to zero, Cruz, Lai and Pramanik construct sets of Hausdorff dimension one avoiding affine copies of the sequence together with its limit point; the avoiding sets in that construction have Lebesgue measure zero. In the other direction, the bi-Lipschitz embedding theorem of Feng, Lai and Xiong applies to $D$, which is compatible with affine nonuniversality because bi-Lipschitz maps need not be affine. An earlier preprint of Cruz, Lai and Pramanik claiming the full similarity conjecture was withdrawn after a gap in its Proposition 3.3. Only the dyadic sequence is settled; for most fast-decaying sequences, including $2^{-n^2}$, the question remains open.

refs/arxiv-2204.12720.pdf
refs/arxiv-2312.01319.pdf
refs/arxiv-2001.02395.pdf

Random-cell discretisation of the dilation parameter at a fixed center, integration over centers, and periodic blocking sets with summable measure budgets appear in Kolountzakis, Chlebik, Kolountzakis-Papageorgiou, and Iosevich et al. Exceptional centers are repaired by an open cover in Kolountzakis and, in a related form, in Chlebik. Solymosi's summer report records an earlier attempt to extend the random-cell method to $D$, without a completed construction.

refs/arxiv-2208.02637.pdf
refs/solymosi-2011-patterns-sparse.pdf

For every $\eta \in (0,1)$ there exists a compact set $E \subseteq [0,1]$ with $m(E) > 1 - \eta$ such that no nontrivial affine copy of $D = \{2^{-n} : n \ge 1\}$ is contained in $E$. Equivalently, for every $x \in \mathbb{R}$ and every $s \in \mathbb{R} \setminus \{0\}$, some $n \ge 1$ satisfies $x + s 2^{-n} \notin E$.

The argument has two layers. A periodic version, valid for normalised scales $t \in [1,2]$ at every real center, reduces the problem to a finite tree of independent coin flips. A dyadic rescaling then converts the periodic statement into one valid for every nonzero real scale, at a measure cost that can be made arbitrarily small.

For each $p \in (0,1)$ there exists an open, 1-periodic set $H \subseteq \mathbb{R}$ with $\rho(H) = m(H \cap [0,1]) \le 6p$ such that for every $x \in \mathbb{R}$ and every $t \in [1,2]$ there is an integer $n \ge 1$ with $x + t 2^{-n} \in H$.

To prove the periodic statement, fix $p \in (0,1)$ and choose integers $M \ge 2$ and $d \ge 1$ satisfying

$$
\frac{p(M-1)}{2} \ge 10, \qquad (1 - 2^{1-M})^d < p.
$$

The first inequality is a lower bound on M. Once M is fixed, $1 - 2^{1-M}$ lies strictly between 0 and 1, and the second inequality becomes a lower bound on d.

Let T be the complete ordered tree of height d in which every internal node has M children labeled $P_1, \ldots, P_M$ from left to right; leaves have height 0, and an edge $e = (P,i)$ connects P to its i-th child. The total number of edges is

$$
K = M + M^2 + \cdots + M^d.
$$

List edges in the child-respecting preorder: at a node P, list first $(P,1)$ and every edge below $P_1$, then $(P,2)$ and every edge below $P_2$, and so on through $(P,M)$. Each edge, together with all edges below its child, forms one consecutive block of this list. Pick an integer $g \ge 1$ with

$$
K 2^{2-g} < p.
$$

The integer g will be the number of unused indices left between consecutive windows.

Give every edge of height h the same window length $r_h$, computed bottom-up. First choose $r_* \ge 1$ so that, for every integer $r \ge r_*$,

$$
20 M r (1 + 2^{2r+2}) \exp\left( - \frac{p(M-1)}{2} r \right) < p.
$$

Such $r_*$ exists: the left side is at most $100 M e^{-7 r}$ for $r \ge 1$, because $p(M-1)/2 \ge 10$, $1 + 2^{2r+2} \le 5 e^{2r}$ and $r \le e^r$, and $100 M e^{-7 r}$ decreases to zero. Set $\sigma_0 = 0$ and, for $h = 1, \ldots, d$,

$$
r_h = \max(r_*, g + \sigma_{h-1}),
$$

$$
\sigma_h = M r_1 + (M-1) g \quad \text{if } h = 1, \qquad \sigma_h = M (r_h + g + \sigma_{h-1}) + (M-1) g \quad \text{if } h \ge 2.
$$

The recursion is well defined because the right side mentions only smaller heights. Read $\sigma_h$ as the total span, in indices including the gaps, of a subtree of height h.

Place integer windows in the preorder, starting at 3. Let $f$ be the edge preceding $e$ in the preorder, and, when $e$ has height h, set

$$
W_e = \{a_e, \ldots, b_e\}, \qquad b_e = a_e + r_h - 1, \qquad a_e = b_f + g + 1,
$$

with $a_e = 3$ when $e$ is initial. Let $\mathcal{N}$ be the union of all $W_e$. Every element of $\mathcal{N}$ is an integer at least 3.

For an edge $e = (P,i)$, let $b_e^*$ be the largest endpoint among the window of $e$ and every window in the subtree below $P_i$. The whole child block fits in twice the incoming window:

$$
b_e^* - a_e + 1 \le r_h + g + \sigma_{h-1} \le 2 r_h.
$$

The right inequality is $r_h \ge g + \sigma_{h-1}$. At $h = 1$ the child $P_i$ is a leaf, so $b_e^* = b_e$ and the bound is immediate.

The dyadic grid at resolution b has boundaries on the lattice $\Gamma_b = 2^{-(b+2)} \mathbb{Z}$, and the half-open cell of a real number z is indexed by

$$
J_b(z) = \left\lfloor 2^{b+2} \{z\} \right\rfloor, \qquad \{z\} = z - \lfloor z \rfloor.
$$

Grids nest: for $B \ge b$,

$$
J_b(z) = \left\lfloor \frac{J_B(z)}{2^{B-b}} \right\rfloor,
$$

so a fine key determines every coarser key.

Call $x \in \mathbb{R}$ stable if, for every noninitial window $W_e$ with preceding window ending at $b'$,

$$
(x, x + \delta_e] \cap \Gamma_{b'} = \varnothing, \qquad \delta_e = 2^{1 - a_e}.
$$

The value $\delta_e$ is the largest displacement, measured from x, produced by an index inside $W_e$ and a scale $t \in [1,2]$.

Lemma. The set G of stable centers is measurable and 1-periodic, its complement has density $\rho(G^c) < p$, and for every $x \in G$, every window $W_e$, every $n \in W_e$, every $t \in [1,2]$ and every window $W_f$ earlier than $W_e$,

$$
J_{b_f}(x + t 2^{-n}) = J_{b_f}(x).
$$

Proof. For a fixed noninitial window $W_e$, the centers that violate stability are precisely the set $E_e = \bigcup_{\beta \in \Gamma_{b'}} [\beta - \delta_e, \beta)$, since $\beta \in (x, x + \delta_e]$ if and only if $x \in [\beta - \delta_e, \beta)$. Because $a_e = b' + g + 1$,

$$
2^{b'+2} \delta_e = 2^{b'+3 - a_e} = 2^{2 - g} < 1.
$$

The intervals $[\beta - \delta_e, \beta)$ therefore have disjoint interiors modulo one, and $E_e$ has density $\rho(E_e) = 2^{2-g}$. The complement of G is the union of $E_e$ over the $K - 1$ noninitial windows, so $\rho(G^c) \le (K - 1) 2^{2-g} < K 2^{2-g} < p$. Measurability and periodicity are immediate.

For the key equality, take $x \in G$ and a noninitial $W_e$ (there is no earlier window when $W_e$ is initial). Then $0 < t 2^{-n} \le \delta_e$, so $(x, x + t 2^{-n}]$ contains no point of $\Gamma_{b'}$, and both endpoints share the same $J_{b'}$ key. The same key holds at every coarser resolution $b_f \le b'$ by nesting. QED

The tree, all windows and G are now fixed. All randomness introduced from this point is independent of this deterministic data.

For each nondefault edge $e = (P,i)$ with $i < M$, draw a table

$$
S_e: \{0, 1, \ldots, 2^{b_e + 2} - 1\} \to \{0,1\}
$$

whose entries are independent fair bits. There is no selector table for default edges $(P,M)$. For a leaf $L$ with incoming window ending at $b(L)$, draw an independent terminal table

$$
T_L: \{0, 1, \ldots, 2^{b(L) + 2} - 1\} \to \{0,1\}
$$

whose entries are independent coins of bias p. All entries over all tables are mutually independent, and the set of tables is finite.

Define a periodic random set $B$. Start from the root. At an internal node P, choose the first child $P_i$ with $i < M$ for which $S_{(P,i)}(J_{b_{(P,i)}}(z)) = 1$; if all $M-1$ selector values vanish at z, take the default child $P_M$. Continue until reaching a leaf, denoted $L(z)$. Set

$$
B = \{z \in \mathbb{R} : T_{L(z)}(J_{b(L(z))}(z)) = 1\}.
$$

Because both routing and membership are read from periodic keys on nested grids, membership in B is constant on each cell of the finest grid used, so B is a union of finitely many half-open dyadic cells per unit period. In particular B is 1-periodic and measurable.

Average density. Let $\mathcal{S}$ be the sigma-field generated by all selector tables. Condition on $\mathcal{S}$ and fix z. The leaf L(z) and the key of its terminal entry are functions of $\mathcal{S}$; the terminal entry at that address is independent of $\mathcal{S}$ with $\mathbb{P}(z \in B \mid \mathcal{S}) = p$. Integrating over one period and using the finite weighted sum over outcomes,

$$
\mathbb{E}[\rho(B)] = \int_0^1 \mathbb{P}(z \in B)\, dz = p.
$$

For a fixed center x, expose one selector entry per nondefault edge, taken at the address $J_{b_e}(x)$:

$$
\mathcal{F}_x = \sigma\left( S_e(J_{b_e}(x)) : e = (P,i),\, i < M \right) \subseteq \mathcal{S}.
$$

An atom $\xi$ of $\mathcal{F}_x$ fixes a full assignment of exposed selector values; every atom has positive probability. On each atom the remaining selector entries and all terminal entries are still independent with the original laws, because the exposed addresses are pairwise distinct coordinates of the finite table family and each atom conditions only on those coordinates.

The exposed selector values determine the route of x. If that route hits a default child somewhere, let U be the first internal node at which this happens. The identity and height of U are functions of $\mathcal{F}_x$ alone.

Lemma. For every fixed real x,

$$
\mathbb{P}\left(\text{path of x never chooses the default child}\right) = (1 - 2^{1-M})^d < p.
$$

Proof. The default child at an internal node is chosen exactly when all $M-1$ exposed selector values at x vanish for that node, which has probability $2^{-(M-1)} = 2^{1-M}$. Along the path, reveal the entries of a node only when the path first reaches it. Tables at strict ancestors are disjoint from tables at the current node, and were already revealed, so the current node's own entries at x remain independent fair bits given the preceding route. At each of the d internal-node decisions the conditional probability of a nondefault choice is therefore $1 - 2^{1-M}$. Iterating over d decisions gives $(1 - 2^{1-M})^d$. The inequality is the choice of d. QED

Fix a stable center x in G and an atom xi of F_x on which the first default node U exists. Let h be the height of U and write r = r_h. For each i < M put

$$
e_i = (U,i), \qquad b_i^* = b_{e_i}^*, \qquad y_n(t) = x + t 2^{-n}.
$$

The exposed selector values at U are all zero on the nondefault children. For a point z, let L_i(z) be the leaf reached by routing from the child U_i by the same rule (with L_i(z) = U_i when U_i is a leaf). Define the local predicate

$$
Q_i(z) = S_{e_i}(J_{b_{e_i}}(z)) \cdot T_{L_i(z)}(J_{b(L_i(z))}(z)).
$$

The predicate is one exactly when the selector on e_i accepts and the terminal entry at its leaf accepts.

Lemma (routing preservation). On xi, for every i < M, every n in W_{e_i} and every t in [1,2],

$$
Q_i(y_n(t)) = 1 \implies y_n(t) \in B.
$$

Proof. Consider a strict ancestor P of U and let $P_a$ be the child of P lying on the route of x to U. Since U is the first default node, this choice has a < M. The routing decision at P depends on the vanishing of selectors at edges $(P,1), \ldots, (P,a-1)$ and on the value 1 at $(P,a)$. Every edge $(P,j)$ with $j \le a$ precedes every edge in the subtree below $P_a$, hence precedes $e_i$ in the preorder. Applying the stability lemma with $x \in G$, the window $W_{e_i}$ and the exponent n shows that at each of these ancestor edges, the selector evaluates the same at x and at $y_n(t)$. Inducting down the ancestor chain gives the same choices, so $y_n(t)$ reaches U.

At U itself, each edge $(U,j)$ with j < i precedes $e_i$, so its selector evaluates as at x, namely zero, and $y_n(t)$ rejects children 1 through i-1. If $Q_i(y_n(t)) = 1$, the selector at $e_i$ evaluates to one at $y_n(t)$, so the route chooses child i; the continuation from that child uses only tables in the subtree below U_i, and by construction matches the definition of $L_i(y_n(t))$. The terminal entry at that leaf is one, so $y_n(t) \in B$. QED

Independence at a fixed scale. Fix t in [1,2] and let

$$
I = \{(i,n) : 1 \le i < M,\, n \in W_{e_i}\},
$$

so $|I| = (M-1) r$. For each (i,n) in $I$, define the selector bit

$$
A_{i,n} = S_{e_i}(J_{b_{e_i}}(y_n(t))).
$$

For fixed i, write $a = a_{e_i}$ and $b = b_{e_i}$. For any two distinct exponents n < n' in the same window,

$$
y_n(t) - y_{n'}(t) = t 2^{-n'}(2^{n'-n} - 1) \ge 2^{-b},
$$

because $t \ge 1$, $n' \le b$ and $2^{n'-n} \ge 2$. Similarly $y_n(t) - x = t 2^{-n} \ge 2^{-b}$. All these points stay in $[x, x + 2^{1-a}]$, an interval of length at most $2^{1-3} = 1/4$ because a is at least 3. On such an interval the ordinary distance equals the circular distance. The circular separation between any two of the points $x, y_n(t)$ for $n \in W_{e_i}$ is therefore at least $2^{-b}$.

Two points of the same half-open b-cell have circular distance less than $2^{-(b+2)}$. Since $2^{-b} > 2^{-(b+2)}$, points of $x, y_n(t)$ land in distinct b-cells. Hence the b-keys $J_b$ of these points are pairwise distinct.

For different values of i, the tables $S_{e_i}$ are distinct. The bits $A_{i,n}$ for (i,n) in $I$ are therefore entries of the table family outside the exposed coordinates of F_x, so conditional on the atom xi these bits remain independent fair bits.

Condition next on the full selector sigma-field S. For each (i,n) in $I$, the selected leaf $L_{i,n} = L_i(y_n(t))$ and the terminal address

$$
\kappa_{i,n} = \left(L_{i,n}, J_{b(L_{i,n})}(y_n(t))\right)
$$

are then fixed. The addresses are distinct. When $L_{i,n}$ and $L_{i,n'}$ are different leaves, the corresponding terminal tables are different. If the two addresses name the same leaf L below U_i, the earlier separation shows $y_n(t)$ and $y_{n'}(t)$ have distinct b-keys at resolution $b = b_{e_i}$, and $b(L) \ge b_{e_i}$ (with equality when U_i is itself a leaf, and strict inequality otherwise since the incoming edge of L occurs later in the preorder). Nesting preserves the distinction at every finer resolution.

Terminal tables are independent of S. Conditional on S, the addresses $\kappa_{i,n}$ are fixed distinct coordinates, so the terminal bits

$$
V_{i,n} = T_{L_{i,n}}(J_{b(L_{i,n})}(y_n(t)))
$$

have product law with parameter p. Since $Q_i(y_n(t)) = A_{i,n} V_{i,n}$,

$$
\mathbb{P}\left(Q_i(y_n(t)) = 0 \text{ for all } (i,n) \in I \,\middle|\, \mathcal{S}\right) = (1-p)^{\sum_{(i,n) \in I} A_{i,n}}.
$$

Averaging over the exposed-fair bits on xi, each $A_{i,n}$ contributes a factor $\tfrac{1}{2} \cdot 1 + \tfrac{1}{2}(1-p) = 1 - p/2$. There are exactly $(M-1) r$ independent factors, giving

$$
\mathbb{P}\left(Q_i(y_n(t)) = 0 \text{ for all } (i,n) \in I \,\middle|\, \text{xi}\right) = (1 - p/2)^{(M-1) r}.
$$

No independence between scales is required; the estimate concerns a single fixed scale t.

The fixed-scale estimate controls one t at a time. The remaining estimate handles every t in [1,2] simultaneously, using that the local predicates change only when a test point crosses a grid boundary. The number of such crossings is finite.

Lemma (representative scales). Fix x in G, an atom xi of F_x on which U exists, let h be the height of U, r = r_h, and put e_i = (U,i), b_i^* = b_{e_i}^*. There is a finite set R(x,U) of scales in [1,2], depending only on x, U, and the fixed tree and windows, such that

$$
|R(x,U)| \le 20 M r (1 + 2^{2r+2}),
$$

and such that for every realization of the tables consistent with xi and every t in [1,2], some t' in R(x,U) satisfies

$$
Q_i(x + t 2^{-n}) = Q_i(x + t' 2^{-n}) \quad \text{for all } 1 \le i < M,\, n \in W_{e_i}.
$$

Proof. On a fixed realization of the tables, $Q_i(z)$ depends only on $J_{b_i^*}(z)$. The selector on $e_i$ has resolution $b_{e_i} \le b_i^*$; every selector below U_i lies on an edge whose endpoint is at most $b_i^*$; every possible terminal leaf has its incoming edge ending at most at $b_i^*$. Nesting then produces every key used by $Q_i$ from $J_{b_i^*}$.

For each (i,n) in $I$, let $D_{i,n}(x)$ be the set of scales t in [1,2] at which $x + t 2^{-n}$ meets a boundary of $\Gamma_{b_i^*}$, that is, the scales at which the continuous point crosses a grid line. It is enough to control the union of these sets together with the two endpoints 1 and 2, because between two consecutive elements of the union every b_i^*-key is constant, and therefore every $Q_i$ is constant.

For fixed (i,n), as t runs over [1,2] the point $x + t 2^{-n}$ covers a closed interval of length $2^{-n}$, and the lattice $\Gamma_{b_i^*}$ has spacing $2^{-(b_i^*+2)}$. Counting lattice points in the interval,

$$
|D_{i,n}(x)| \le 1 + 2^{-n} / 2^{-(b_i^*+2)} = 1 + 2^{b_i^* + 2 - n}.
$$

The extra one allows both endpoints to fall on the lattice. The span bound gives $b_i^* - a_{e_i} + 1 \le 2 r$, and since $n \ge a_{e_i}$,

$$
b_i^* + 2 - n \le 2 r + 1 \le 2r + 2,
$$

so $|D_{i,n}(x)| \le 1 + 2^{2r+2}$.

Set

$$
D(x,U) = \{1, 2\} \cup \bigcup_{1 \le i < M} \bigcup_{n \in W_{e_i}} D_{i,n}(x).
$$

List the distinct elements of D(x,U) as $d_1 = 1 < d_2 < \cdots < d_L = 2$, and let

$$
R(x,U) = D(x,U) \cup \left\{ \tfrac{d_j + d_{j+1}}{2} : 1 \le j < L \right\}.
$$

Each open interval $(d_j, d_{j+1})$ lies between consecutive crossings, so every relevant key is constant there and the midpoint of the interval reproduces every value on the interval. Each element of D(x,U) is included and represents itself. The same holds at the endpoints 1 and 2.

Counting sizes. There are exactly $(M-1) r$ indices in $I$, so

$$
L \le 2 + (M-1) r (1 + 2^{2r+2}), \qquad |R(x,U)| = 2L - 1 \le 2(M-1) r (1 + 2^{2r+2}) + 3 \le 20 M r (1 + 2^{2r+2}).
$$

QED

Proposition (raw failure). For every x in G,

$$
\mathbb{P}\left(\exists\, t \in [1,2] \text{ such that } x + t 2^{-n} \notin B \text{ for every } n \in \mathcal{N} \right) \le 2p.
$$

Proof. Condition on an atom xi of F_x on which U exists. On xi the node U and the finite set R(x,U) are fixed. Fix a representative t' in R(x,U). The fixed-scale estimate, applied with that t', gives

$$
\mathbb{P}\left(Q_i(x + t' 2^{-n}) = 0 \text{ for all } (i,n) \in I \,\middle|\, \text{xi}\right) = (1-p/2)^{(M-1) r} \le \exp\left( - \frac{p(M-1)}{2} r \right).
$$

Assume some t in [1,2] misses B at every index in N. The routing preservation lemma forces every local predicate $Q_i(y_n(t))$ to vanish for that t. By the representative lemma, some t' in R(x,U) carries the same values, so all predicates vanish at that representative t' as well.

A union bound over R(x,U) at the exposed atom gives

$$
\mathbb{P}\left(\exists\, t \text{ missing} \mid \text{xi}\right) \le |R(x,U)| \cdot \exp\left( - \frac{p(M-1)}{2} r \right) \le 20 M r (1 + 2^{2r+2}) \exp\left( - \frac{p(M-1)}{2} r \right) < p,
$$

using the choice of r_* and r = r_h >= r_*.

Average over atoms. The estimate holds on every atom with a first default. On atoms without a first default the path of x never takes the default child, an event of probability less than p by the previous lemma; on that event the failure indicator is bounded by 1. Combining,

$$
\mathbb{P}(\exists\, t \text{ missing}) \le 1 \cdot p + p \cdot 1 = 2p.
$$

QED

Completion of the periodic statement. Enlarge every realization $B_\omega$ of $B$ to an open periodic set $B_\omega^+ \supseteq B_\omega$. Since $B_\omega$ is a union of finitely many half-open dyadic cells of a grid with $Q$ cells per period, taking the closure of the union and adding open neighborhoods of circular radius p/(4Q) produces an open periodic set $B_\omega^+$ with

$$
\rho(B_\omega^+) \le \rho(B_\omega) + p.
$$

For the empty set use the empty enlargement. Every realization is a finite object, so no measurable-selection issue arises. Averaging,

$$
\mathbb{E}[\rho(B^+)] \le 2p.
$$

Define the exceptional-center set

$$
R_\omega = \{ x \in \mathbb{R} : \exists t \in [1,2]\, \forall n \in \mathcal{N},\, x + t 2^{-n} \notin B_\omega^+ \}.
$$

This R_\omega is 1-periodic and closed. To see closedness, work on the circle $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ and let $\pi$ be the quotient map. The map $(\pi(x), t) \mapsto \pi(x + t 2^{-n})$ is continuous for each fixed n, and only finitely many indices occur in the finite set N. The pairs $(\pi(x), t)$ in $\mathbb{T} \times [1,2]$ whose images under all these maps avoid the open set $\pi(B_\omega^+)$ form a closed subset of a compact set, hence compact. Its projection onto $\mathbb{T}$ is compact and therefore closed. Its inverse image under $\pi$ is precisely $R_\omega$. Periodicity is immediate because $\pi(x + 1) = \pi(x)$ and $B_\omega^+$ is 1-periodic.

Measure bound. For every $x \in G$ the raw failure proposition gives $\mathbb{P}(x \in R_\omega) \le 2p$, since enlarging $B_\omega$ to $B_\omega^+$ can only decrease the exceptional set. For $x \notin G$ the probability is at most 1. Exchanging the finite sum over outcomes with the integral over one period,

$$
\mathbb{E}[\rho(R_\omega)] = \int_0^1 \mathbb{P}(x \in R_\omega)\, dx \le 2p \cdot \rho(G) + 1 \cdot \rho(G^c) \le 2p + p = 3p.
$$

Adding, $\mathbb{E}[\rho(B^+) + \rho(R_\omega)] \le 5p$. Some single outcome $\omega_*$ satisfies

$$
\rho(B^+) + \rho(R_\omega) \le 5p
$$

with $B^+ = B_{\omega_*}^+$ and $R = R_{\omega_*}$.

Open neighborhood of R. Let $R_T = \pi(R)$ be the image of R in T. If $R_T$ is empty, set $V$ = empty. Otherwise, for eps > 0, the metric neighborhoods

$$
V_\varepsilon = \{ z \in T : \operatorname{dist}_T(z, R_T) < \varepsilon \}
$$

are open and decrease to $R_T$ as eps decreases to zero. Because measure on the compact circle has finite total mass, continuity from above provides some eps > 0 such that the periodic lift $V = \pi^{-1}(V_\varepsilon)$ satisfies

$$
R \subseteq V, \qquad \rho(V) \le \rho(R) + p.
$$

Set H = B^+ union V. This is open and 1-periodic, and

$$
\rho(H) \le \rho(B^+) + \rho(V) \le \rho(B^+) + \rho(R) + p \le 6p.
$$

Hits for every center and every normalised scale. Fix $x \in \mathbb{R}$ and $t \in [1,2]$. If $x \notin R$, negating the definition of R directly gives some $n \in \mathcal{N}$ with $x + t 2^{-n} \in B^+ \subseteq H$. If $x \in R \subseteq V$, choose $n_0 \ge 1$ with $2^{1 - n_0} < \varepsilon$. For $n \ge n_0$,

$$
\operatorname{dist}_T(\pi(x + t 2^{-n}), R_T) \le \operatorname{dist}_T(\pi(x + t 2^{-n}), \pi x) \le t 2^{-n} \le 2^{1 - n_0} < \varepsilon,
$$

so $x + t 2^{-n} \in V \subseteq H$. Both cases give an $n \ge 1$ with $x + t 2^{-n} \in H$, completing the proof of the periodic statement.

Fix $\eta$ in $(0,1)$ and set $p_j = (\eta/24) 2^{-j}$ for $j = 1, 2, \ldots$. Each $p_j$ belongs to (0,1). By the periodic statement there is an open 1-periodic set $H_j$ with $\rho(H_j) \le 6 p_j$ such that for every $x \in \mathbb{R}$ and every $t \in [1,2]$ some integer $n \ge 1$ satisfies $x + t 2^{-n} \in H_j$. Fix all $H_j$ before any center or scale is chosen.

Define

$$
C = \bigcup_{j \ge 1} (2^{-j} H_j \cup -2^{-j} H_j), \qquad E = [0,1] \setminus C.
$$

Every factor is open, so C is open; the definition gives $-C = C$. Hence E is closed inside [0,1], hence compact, and disjoint from C.

Measure of the removed set. Periodicity plus a change of variables gives, for each j,

$$
m(2^{-j} H_j \cap [0,1]) = 2^{-j} m(H_j \cap [0,2^j]) = \rho(H_j),
$$

with the analogous identity for the reflected copy. Countable subadditivity gives

$$
m(C \cap [0,1]) \le 2 \sum_{j \ge 1} \rho(H_j) \le 12 \sum_{j \ge 1} p_j = \frac{\eta}{2}.
$$

Therefore

$$
m(E) = 1 - m(C \cap [0,1]) \ge 1 - \frac{\eta}{2} > 1 - \eta.
$$

Hits on every signed copy. Fix $x \in \mathbb{R}$ and $s \in \mathbb{R} \setminus \{0\}$. When $s > 0$, choose an integer k with $t = 2^k s$ in $[1,2)$, and set $j = \max(1, k)$. The periodic statement applied to the center $2^j x$ and the scale t gives an integer $n \ge 1$ with

$$
2^j x + t 2^{-n} \in H_j.
$$

Multiplying by $2^{-j}$,

$$
x + s 2^{-(n + j - k)} = 2^{-j}\left(2^j x + t 2^{-n}\right) \in 2^{-j} H_j \subseteq C.
$$

The index $N = n + j - k$ is an integer at least 1, because $n \ge 1$ and $j \ge k$. When $s < 0$, apply the same construction to the center $-x$ and the positive scale $-s$, obtaining $N \ge 1$ with

$$
-(x + s 2^{-N}) = (-x) + (-s) 2^{-N} \in C.
$$

Since $-C = C$, the point $x + s 2^{-N}$ lies in $C$. In both cases the copy $x + sD$ contains a point of $C$ and hence escapes $E$. The compact set $E$ of measure greater than $1 - \eta$ therefore contains no nontrivial affine copy of $D$, settling the dyadic case of the Erdos similarity conjecture.