Fact 2.1 (Artin–Schreier, see [Reference Jacobson7, p. 674])

Let C be an algebrically closed field and F be a proper subfield of C such that $C/F$ is finite. Then F is real closed and $C = F \left ( \sqrt {- 1} \right )$ .

Fact 2.2 (“Lying-over,” see [Reference Jacobson7, p. 411])

Let E be a commutative ring and R a subring such that E is integral over R (for every element $\alpha $ of E, there is a monic polynomial $f(x) \in R[x]$ such that $f(\alpha )=0$ ). Then for each prime ideal p of R, there is a prime ideal P of E such that $p = P \cap R$ . In particular, if E is a field, so is R.

Proof. (i) follows from Fact 2.1. For (ii), consider the normal closure K of $\mathbb {Q}_p/F$ ; K is finite over $\mathbb {Q}_p$ and hence any automorphism of K is continuous. Hence any automorphism of K restricts to the identity on $\mathbb {Q}_p$ , which implies that $\mathbb {Q}_p=F=K$ .

Proof. (i) By our assumption, $\mathbb {R}$ is finite over A and hence by Fact 2.2, A is a field since $\mathbb {R}$ is a field. Then Lemma 2.3(i) implies $A =\mathbb {R}$ .

(ii) By Fact 2.2 and our assumption, A is a subfield of $\mathbb {Q}_p$ . If $A\neq \mathbb {Q}_p$ , then $[\mathbb {Q}_p:A]\neq 1$ is finite. Then Lemma 2.3(ii) implies $A =\mathbb {Q}_p$ .

Fact 2.5 (See [Reference Jech8, Theorem 11.18])

Every analytic subset of $\mathbb {R}$ is measurable.

Proof. Assume otherwise, then F is a maximal subfield of K avoiding x. This implies that $K/F$ is algebraic. Indeed, if x is not algebraic over F, then $x \notin F(x^2)$ . Now if $K/F$ is not algebraic, then K contains a copy of $F(X)$ for X transcendental over F, which does not contain x.

So each $y\in K\setminus F$ satisfies a polynomial over F. Let $D_n$ denote the set of elements in K whose minimal polynomial over F is of degree at most n, this is an analytic set and $\bigcup _{n\in \mathbb {N}}D_n=K$ . In particular, $D_n$ is measurable by Fact 2.5. It is clear that $D_n$ is closed under multiplication by $-1$ . Note further that the difference of 2 elements of degree at most n over F has degree at most $n^2$ . Moreover, there must be n such that $\mu (D_n)>0$ where $\mu $ denote the Haar measure on K. So $D_{n^2}=K$ by the Steinhaus theorem for Haar measure on locally compact groups (see [Reference Stromberg19]). This means that any element in $K/F$ is of degree at most n for some $n\in \mathbb {N}$ . By the primitive element theorem, we have that $[K:F]$ is at most n. Then by Lemma 2.3, $K=F$ , a contradiction.

Proof. (i) By the definition of a subgroup.

(ii) Suppose that $p = \frac {m}{n}$ for some integers m and $n \neq 0$ . Then for any $c \in \mathbb {R}$ , there are $a, b \in A$ such that $a + \frac {m b}{n} = c$ . So $n c = n a + m b \in A$ . Thus $\mathbb {R}= \frac {1}{n} A$ . Since $\mathrm {Char}(\mathbb {R})=0$ , we have $\mathbb {R}=n\mathbb {R}=A$ .

(iii) Suppose that $p_i = \frac {m_i}{n_i}$ for some integers $m_i$ and $n_i \neq 0$ . Let n be the least common multiple of those $n_i$ . Then $\mathbb {R}=A + \sum _{i = 0}^k p_i A \subseteq \frac {1}{n}A$ , so $A =\mathbb {R}$ .

(iv) Note that $\mathbb {Q}\otimes _{\mathbb {Z}}A \subseteq \bigcup _{n \ge 1}\frac {1}{n}A$ . But it is clear that $\mathrm {dim_H} (A) = \mathrm {dim_H} (\frac {1}{n} A)$ for any $n \ge 1$ since $f:\mathbb {R} \to \mathbb {R},x\mapsto nx$ is bi-Lipschitz. So

$$\begin{align*}\mathrm{dim_H} (A) \leqslant \mathrm{dim_H} \left ( \mathbb{Q}\otimes_{\mathbb{Z}}A \right) \leqslant \mathrm{dim_H} \left ( \bigcup_{n \ge 1} \frac{1}{n} A \right) = \sup_i \mathrm{dim_H}(\frac{1}{n} A) = \mathrm{dim_H} (A).\end{align*}$$

(v) Note that

$$ \begin{align*}A+mxA=m(\frac{A}{m}+x A)\supseteq m(A+xA)=m \mathbb{R}.\end{align*} $$

Since $\mathrm {Char}(\mathbb {R})=0$ , we have $A+mxA= \mathbb {R}$ .

(vi) Note that

$$ \begin{align*}A+(x+\frac{n}{m})A=A+(m x+n) \frac{A}{m} \supseteq A+(mx+n)A = A+mxA.\end{align*} $$

By (v), we have $A+(x+\frac {n}{m})A= \mathbb {R}$ .

Fact 3.3 (Marstrand [Reference Marstrand13])

Let $F \subseteq \mathbb {R}^2$ be an analytic set with $\mathrm {dim_H}(F)>1$ . Then for almost all $\theta \in (-\frac {\pi }{2},\frac {\pi }{2})$ , $\mathrm {proj_\theta }(F)$ (the projection of F on the line $l_\theta :y=\mathrm {tan}\theta \cdot x$ ) has positive length.

Proof. By the product formula of Hausdorff dimension (see [Reference Falconer4, Chapter 7]),

$$ \begin{align*}\mathrm{dim_H}(A \times A) \ge\mathrm{dim_H}(A)+\mathrm{dim_H}(A)>1.\end{align*} $$

Hence for almost all $\theta \in (-\frac {\pi }{2},\frac {\pi }{2})$ , $\mathrm {proj_\theta }(A\times A)$ has positive length. Let $f(t)=\arctan t$ for all $t\in \mathbb {R}$ . Define $\Theta =\{\theta \in (-\frac {\pi }{2},\frac {\pi }{2}):\mathrm {proj_\theta }(A\times A) \text { has positive length}\}$ . Since $\Theta $ is co-null (and hence measurable), $f^{-1}(\Theta )$ is measurable. We claim that $f^{-1}(\Theta )$ is co-null. Suppose not, then $\mathbb {R} \backslash f^{-1}(\Theta )$ has positive measure. Then there is $N \in \mathbb {R}$ such that $\mathbb {R} \backslash f^{-1}(\Theta ) \cap [-N,N]$ has positive measure. But ${\inf _{t\in [-N,N]}f'(t) >0}$ , then $f(f^{-1}(\Theta ) \cap [-N,N])$ has positive measure, a contradiction.

Suppose $(a,b)$ , $(c,d) \in A \times A$ , $a \neq c$ and $b \neq d$ . Then the line passing through these two points is

$$ \begin{align*}l_{(a,b),(c,d)}:\frac{y-d}{x-c}=\frac{y-b}{x-a}.\end{align*} $$

The slope of $l_{(a,b),(c,d)}$ is $\frac {b-d}{a-c}$ . And if $t\neq 0$ and $a+tb=c+td$ , then $t=-\frac {a-c}{b-d}$ . So for $(a,b),(c,d),$ and t as above, $a+tb=c+td$ if and only if $l_{(a,b),(c,d)}$ is perpendicular to $l_{\arctan t}: y= t\cdot x$ . In particular, $l_{\arctan t}$ and $A+tA$ are homeomorphic via:

$$ \begin{align*}(a,b) \mapsto a+bt.\end{align*} $$

And

$$ \begin{align*}|a+bt|=(1+t^2)|a|=\sqrt{1+t^2}\sqrt{a^2+b^2}=\sqrt{1+t^2}|\overrightarrow{0,(a,b)}|.\end{align*} $$

Hence the measure of $A+t A$ is equal to the measure of $\mathrm {proj_{\arctan t}}(A \times A)$ times $\sqrt {1+t^2}$ provided $t \neq 0$ . By Proposition 3.1(i), $A+tA$ is a group. Then by Corollary 1.2, $A+tA= \mathbb {R}$ if $\mathrm {proj}_{\arctan t}(A\times A)$ has positive length. Since $f^{-1}(\Theta )$ is co-null, $\mathrm {proj}_{\arctan t}(A\times A)$ has positive length for almost all t.

Proof of Theorem 1.4

As a convention, we consider a real a and its binary expansion $m + 0. a_1 a_2 \dots a_n \dots $ with infinitely many 0s to be equivalent, where $m \in \mathbb {Z}$ is the integer part of a and the numbers $a_i \in \{0, 1\}$ express its binary decimal part.

We first prove a weaker result: there is an $F_\sigma $ subgroup A of $(\mathbb {R},+)$ and $x \in \mathbb {R}$ such that $A + x A =\mathbb {R}$ and A is null.

Define $Q_n =\{m \in \mathbb {N}: 3^{n - 1} \leqslant m < 3^n \} $ for $n \ge 1$ . Then each $Q_n$ contains $2 \cdot 3^{n - 1}$ numbers. Divide $\{1, 2,. . .\}$ into two parts $P_1, P_2$ such that for all $n \ge 1$ , the first $3^{n - 1}$ numbers of $Q_n$ are in $P_1$ , and the other numbers of $Q_n$ are in $P_2$ . Formally,

$$\begin{align*}P_1 = \{ k : \exists n \exists i \in [0,3^n) (k = 3^n + i) \}, \end{align*}$$

$$\begin{align*}P_2 = \{ k : \exists n \exists i \in [3^n, 2 \cdot 3^n) (k = 3^n + i) \}. \end{align*}$$

Define

$$\begin{align*}A_0 = \{ m + 0. a_1 a_2 \cdot \cdot \cdot a_n \dots : m \in \mathbb{Z} \wedge \forall n \in P_2 (a_n = 0) \}. \end{align*}$$

$A_0$ consists of reals with segments of 0s of length $3^k (k \in \mathbb {N})$ , and ahead of each segment of 0s there is an arbitrary segment of the same length. Define for each n,

$$\begin{align*}A_n = \overbrace{A_0 \pm A_0 \pm \cdots \pm A_0}^n = \{ \alpha_1 \pm \alpha_2 \pm \cdots \pm \alpha_n : \alpha_i \in A_0, 1 \leqslant i \leqslant n \}. \end{align*}$$

Fix $n \ge 1$ and $\varepsilon>0$ . Choose $k \ge 1$ such that

(i) $2^{- \frac {1}{2} \cdot 3^k + 1} < \varepsilon $ ;

(ii) for all $a = m + 0. a_1 a_2 \dots \in A_n$ ,

$$ \begin{align*}\forall j \in \left[ 2 \cdot 3^{k - 1}, \frac{5}{2} \cdot 3^{k - 1} \right](a_j=0) \lor \forall j \in \left[ 2 \cdot 3^{k - 1}, \frac{5}{2} \cdot 3^{k - 1} \right](a_j=1).\end{align*} $$

Then the measure of $A_n$ is no more than $2 \cdot 2^{- \frac {1}{2} \cdot 3^k} < \varepsilon $ . Since $\varepsilon $ is arbitrary, $A_n$ is null. Let A be the subgroup of $(\mathbb {R},+)$ generated from $A_0$ . Then $A = \bigcup _{n \ge 1} A_n$ is null. Since $A_0$ is compact, A is $F_{\sigma }$ . Define $x = 0. x_1 x_2 \dots $ such that $x_n = 1$ if and only if $\exists k(n = 3^k)$ . Given any real $y = m + 0. y_1 y_2 \dots $ , we shall define a real $b = 0. b_1 b_2 \dots \in A_0$ such that $b \cdot x$ and y are equal on the nth binary place for each $n \in P_2$ . Then we can pick $c \in A_0$ such that $c + b x = y$ , so $A + x A =\mathbb {R}$ .

Intuitively, in the calculation $b \cdot x$ , $x_{3^k} = 1$ causes the binary places of b to shift to the right $3^k$ places, i.e.,

$$\begin{align*}0. \overbrace{0 \dots 0}^{3^k - 1} 1 \times 0. b_1 b_2 \dots = 0. \overbrace{0 \dots 0}^{3^k} b_1 b_2 \dots. \end{align*}$$

Denote $b x = 0. c_1 c_2 \dots .$ By recursion on n, one can choose $\{b_m : m \in Q_n \cap P_1 \}$ such that $c_m = y_m$ for all $m \in Q_n \cap P_2$ as follows: suppose $\{b_m : m \in Q_k \cap P_1, k < n\}$ have been chosen so that $c_m = y_m$ for all $k < n$ and $m \in Q_k \cap P_2$ . Whatever $\{b_m : m \in Q_k \cap P_1, k \ge n + 1\}$ are chosen, we have

$$\begin{align*}0. \overbrace{0 \dots 0}^{3^n - 1} b_{3^n} b_{3^n + 1} b_{3^n + 2} \dots \times x < 0. \overbrace{0 \dots 0}^{3^n-1} 1. \end{align*}$$

Hence $\{c_m : m \in Q_n \cap P_2 \}$ is determined only by $\{b_m : m \in Q_k \cap P_1, k \leqslant n\}$ . Now it is clear that we can choose $\{b_m : m \in Q_n \cap P_1 \}$ to ensure $c_m = y_m$ for all $m \in Q_n \cap P_2$ . This ends the proof of the weaker result.

Achieving $\mathrm {dim_H} (A) = \frac {1}{2}$ requires a bit more effort. Let $\{p_n\}_{n \ge 1}$ be an increasing sequence of even numbers such that $\sum _{k<n}p_k=o(p_n)$ and $3^n=o(p_n)$ . Define

$$ \begin{align*}Q_n=\left\{m \in \mathbb{N}:1+ \sum_{k<n}p_k \le m <1+ \sum_{k \le n}p_k\right\}.\end{align*} $$

Divide $\{1, 2,. . .\}$ into two parts $P_1, P_2$ such that for all $n \ge 1$ , the first half of the numbers of $Q_n$ are in $P_1$ , and the other numbers of $Q_n$ are in $P_2$ . Define

$$\begin{align*}A_0 = \{ m + 0. a_1 a_2 \cdot \cdot \cdot a_n \dots : m \in \mathbb{Z} \wedge \forall n \in P_2 (a_n = 0) \}. \end{align*}$$

Define $A_n = \overbrace {A_0 \pm A_0 \pm \cdots \pm A_0}^n$ and $A = \bigcup _{n \ge 1} A_n$ . As before, A is an $F_\sigma $ subgroup of $(\mathbb {R},+)$ and there is x such that $A+x A=\mathbb {R}$ . It remains to show that $\mathrm {dim_H}(A_n)=\frac {1}{2}$ for all n. Fix n. For all $k>n$ and $a=m+0.a_1a_2 \dots \in A_n$ ,

$$ \begin{align*}\forall j \in\! \left(\! \sum_{j<k}p_j+\frac{p_k}{2}, \sum_{j \le k}p_j-n \!\right]\!(a_j=0) \lor \forall j \in \!\left(\! \sum_{j<k}p_j+\frac{p_k}{2}, \sum_{j \le k}p_j-n \!\right]\!(a_j=1).\end{align*} $$

Let $\delta =2^{-\sum _{j \le k}p_j}$ . If $s>\frac {1}{2}$ , then

$$ \begin{align*}\mathcal{H}^s_\delta(A_n) \le 2^{1+\sum_{j<k}p_j+\frac{p_k}{2}}2^{n+3}(2^{-\sum_{j \le k}p_j})^s \le 2^{n+4+\sum_{j<k}p_j-(s-\frac{1}{2})p_k}.\end{align*} $$

Since $\sum _{j<k}p_j=o(p_k)$ , $\mathcal {H}^s(A_n)=0$ . So $\mathrm {dim_H}(A_n) \le \frac {1}{2}$ .

Proof. We only include the proof for the existence of such A, the calculation of the Hausdorff dimension of A is left to the reader as it follows from the same argument as in $\mathbb {R}$ .

Note that every element $a \in \mathbb Z_p$ has a unique expansion

$$\begin{align*}a=\sum_{n=0}^\infty a_n p^n, \qquad a_n\in\{0,\dots,p-1\}. \end{align*}$$

Define

$$\begin{align*}Q_n=\{m\in\mathbb N:p^{n-1}\le m<p^n\}. \end{align*}$$

Put

$$\begin{align*}q_n=p^{n-1},\qquad N_n=|Q_n|=(p-1)p^{n-1}, \end{align*}$$

and let

$$\begin{align*}L_n=\left\lfloor \frac{N_n}{2} \right\rfloor, \qquad s_n=N_n-L_n=\left\lceil \frac{N_n}{2}\right\rceil. \end{align*}$$

Thus the last part of $Q_n$ has length $L_n$ , while the first part has length $s_n$ , and $s_n+L_n=N_n$ . Partition the coordinates by putting the first $s_n$ indices of $Q_n$ into $P_1$ and the remaining $L_n$ indices into $P_2$ . Formally,

$$\begin{align*}P_1\cap Q_n=\{q_n,q_n+1,\dots,q_n+s_n-1\}, \qquad P_2=\mathbb{N}\setminus P_1. \end{align*}$$

Define

$$\begin{align*}A_0= \left\{ \sum_{n=0}^\infty a_n p^n: \forall n\in P_2\ (a_n=0) \right\}. \end{align*}$$

Thus elements of $A_0$ have arbitrary digits on $P_1$ and vanish on $P_2$ . For each $m\ge 1$ , define

$$\begin{align*}A_m= \underbrace{A_0\pm A_0\pm\cdots\pm A_0}_{m}. \end{align*}$$

Let

$$\begin{align*}A=\bigcup_{m\ge1}A_m. \end{align*}$$

Since $A_0$ is compact, every $A_m$ is compact, hence A is an $F_\sigma $ subgroup of $(\mathbb Z_p,+)$ .

We now make explicit the carry estimate used in the measure calculation. Fix m. In any signed sum of at most m elements of $A_0$ , the carry sequence is bounded in absolute value by a constant $B_m$ depending only on m and p. Indeed, at each digit the non-carry part is between $-m(p-1)$ and $m(p-1)$ , so the usual base-p carrying recursion keeps all carries in a fixed finite interval. Now suppose that the summand digits are all zero on a consecutive interval $J\subseteq P_2$ . After at most

$$\begin{align*}d_m=\left\lceil \log_p(B_m+1)\right\rceil+1 \end{align*}$$

positions, the carry has become either $0$ or $-1$ . Consequently all later output digits in J are, respectively, all $0$ or all $p-1$ .

Apply this to the interval $P_2\cap Q_k$ , whose length is $L_k$ . For all sufficiently large k, every element of $A_m$ has, on a terminal subinterval $I_k\subseteq P_2\cap Q_k$ of length at least $L_k-d_m$ , either only digits $0$ or only digits $p-1$ . Therefore $A_m$ can be covered by at most

$$\begin{align*}2p^{p^k-(L_k-d_m)} \end{align*}$$

balls of radius $p^{-p^k}$ . Since such a ball has Haar measure $p^{-p^k}$ , we get

$$\begin{align*}\mu(A_m)\le 2p^{d_m}p^{-L_k}\to0. \end{align*}$$

Hence every $A_m$ is Haar null, and therefore A is Haar null.

Define

$$\begin{align*}x=\sum_{k=1}^\infty p^{s_k}. \end{align*}$$

Notice that the shift used in each k-th block is $s_k$ , the length of the first part of $Q_k$ ; this is exactly what sends the first $L_k$ free coordinates of $P_1\cap Q_k$ onto $P_2\cap Q_k$ .

Let

$$\begin{align*}y=\sum_{n=0}^\infty y_n p^n\in\mathbb Z_p. \end{align*}$$

We recursively construct

$$\begin{align*}b=\sum b_n p^n\in A_0 \end{align*}$$

such that $bx$ agrees with y on all coordinates in $P_2$ . Suppose that the coordinates of b in $P_1\cap Q_i$ have already been chosen for all $i<k$ so that $bx$ agrees with y on $P_2\cap Q_i$ . Later choices have indices at least $q_k$ and all shifts in x are non-negative, so they cannot change any digit below $q_k$ ; hence the previously treated blocks remain fixed. It remains to choose the free coordinates in $P_1\cap Q_k$ . The term $p^{s_k}$ in x gives the direct shift

$$\begin{align*}p^{s_k}\sum_{i=0}^{L_k-1} b_{q_k+i}p^{q_k+i} = \sum_{i=0}^{L_k-1} b_{q_k+i}p^{q_k+s_k+i}, \end{align*}$$

whose exponents are precisely the coordinates in $P_2\cap Q_k$ . Since the digits of $bx$ in $P_2\cap Q_k$ depend only on the choice of $\{b_i\}_{i \le q_k+L_k-1}$ , we may choose

$$\begin{align*}\{b_n:n\in P_1\cap Q_k\} \end{align*}$$

so that $(bx)_n=y_n$ for all $n\in P_2\cap Q_k$ . Proceeding block by block gives $b\in A_0$ with $bx$ agreeing with y on $P_2$ .

Define $c=y-bx$ . Since $bx$ already agrees with y on $P_2$ , the element c has vanishing digits on $P_2$ , hence $c\in A_0$ . Therefore $y=c+bx\in A+xA$ . Since $y\in \mathbb Z_p$ was arbitrary, $A+xA=\mathbb Z_p$ .

Fact 3.7 (Le Gac [Reference Le Gac5])

If G and H are analytic subgroups of $(\mathbb {R},+)$ such that $G+H=\mathbb {R}$ and $G \cap H =\{0\}$ , then either $G=\mathbb {R}$ or $G=\{0\}$ .

Fact 3.9 (Jankov–von Neumann, see [Reference Kechris9, Theorem 18.1])

Let X, Y be standard Borel spaces and $P \subseteq X \times Y$ be analytic. Then there is a $\Sigma $ -measurable function $f:X \to Y$ such that $(x,f(x)) \in P$ for all $x \in proj_X(P)$ (the projection from P to X), where $\Sigma $ is the $\sigma $ -algebra generated by the analytic sets.

Proof of Theorem 3.8

For (i), suppose that A is an analytic subgroup of $(\mathbb {R},+)$ and $x \in X_A$ . Then the graph of the function $\varphi :A\times A \to \mathbb {R}$ so that $(a,b)\mapsto a+xb$ is an analytic subset of $\mathbb {R}^3$ . By Fact 3.9, there is a Lebesgue measurable function $f:\mathbb {R} \to A\times A$ such that for any $z \in \mathbb {R}$ , $\varphi (f(z))=z$ . By the Lusin theorem, there is a compact set $P\subseteq \mathbb {R}$ with positive measure and a continuous function $\psi : P\to A\times A$ so that for any $z\in P$ , $\varphi (\psi (z))=z$ . Let

$$ \begin{align*}C_0=\{a \in \mathbb{R}:\exists z \in P \exists b \in \mathbb{R}(\psi(z)=(a,b))\},\end{align*} $$

and

$$ \begin{align*}C_1=\{b \in \mathbb{R}:\exists z \in P \exists a \in \mathbb{R}(\psi(z)=(a,b))\}.\end{align*} $$

Both $C_0$ and $C_1$ are compact and so is $C_0\cup C_1$ . Let B be the group generated by $C_0\cup C_1 $ . Then B is $F_{\sigma }$ and $B \subseteq A$ . Moreover, the image $\varphi (B \times B)\supseteq P$ has positive measure. Since $\varphi (B \times B)$ is a subgroup of $\mathbb {R}$ , by Corollary 1.2, $\varphi (B \times B)=\mathbb {R}$ . Hence $x \in X_B$ .

For (ii), suppose that A is an analytic subgroup of $(\mathbb {R},+)$ and $X_A$ is co-null. By the regularity of Lebesgue measure, there is a co-null $F_{\sigma }$ subset $F \subseteq X_A$ . It suffices to show that there is an $F_\sigma $ subgroup $B \subseteq A$ such that for almost all $x \in F$ , we have $x\in X_B$ .

The graph of the function

$$\begin{align*}\varphi : A \times A \times F \rightarrow \mathbb{R} \times F, (a, b, x) \mapsto (a + x b, x) \end{align*}$$

is an analytic subset of $\mathbb {R}^5$ . By Fact 3.9, there is a Lebesgue measurable function

$$ \begin{align*}f : \mathbb{R} \times F \rightarrow A \times A \times F\end{align*} $$

such that for any $z \in \mathbb {R} \times F$ , $\varphi (f (z)) = z$ . By the Lusin theorem, for all n, there is a compact subset $K_n$ of $\mathbb {R} \times F$ such that $\psi _n := f \upharpoonright K_n$ is continuous and

$$ \begin{align*}\mu ((\mathbb{R} \times F) \cap B_n (0) \backslash K_n) < \frac{1}{n},\end{align*} $$

where $\mu $ is the Lebesgue measure on $\mathbb {R}^2$ and $B_n (0) \subseteq \mathbb {R}^2$ is the closed ball with radius n and center $0$ . Let

$$ \begin{align*}C_0=\{a \in \mathbb{R}:\exists n\exists z \in K_n \exists b,x \in \mathbb{R}(\psi_n(z)=(a,b,x))\},\end{align*} $$

and

$$ \begin{align*}C_1=\{b \in \mathbb{R}:\exists n\exists z \in K_n \exists a,x \in \mathbb{R}(\psi(z)=(a,b,x))\}.\end{align*} $$

Both $C_0$ and $C_1$ are $F_\sigma $ and so is $C_0\cup C_1$ . Let B be the additive group generated by $C_0\cup C_1 $ . Then B is an $F_{\sigma }$ set and $B \subseteq A$ . Moreover, the projection of the image $\varphi (B \times B \times \{x\})$ to the first coordinate has positive measure for almost all $x \in F$ by the Fubini theorem since $(\mathbb {R}\times F) \backslash \cup _n K_n$ is null. So $X_B$ is conull.

Fact 4.2 (Hölzl et al. [Reference Hölzl, Merkle, Miller, Stephan and Liang6])

There is a constant c such that for every generic real x and $i \in \{0, 1\}$ , there are infinitely many n such that

$$\begin{align*}\forall m \in \left[ n, 2^{2^{2^n}} \right] (K (G \upharpoonright m) \leqslant K (m) + c \wedge G (m) = i). \end{align*}$$

Fact 4.3 (The point-to-set principle, Lutz and Lutz [Reference Lutz and Lutz10])

For every set $E \subseteq \mathbb {R}$ ,

$$\begin{align*}\mathrm{dim_H} (E) = \min_{A \subseteq \omega} \sup_{x \in E} \liminf_{n \rightarrow \infty} \frac{K^A (x \upharpoonright n)}{n}. \end{align*}$$

Proof. Note that $K (n) \leqslant \log n + 2 \log \log n + O (1)$ for every $n \in \omega $ (see [Reference Downey and Hirschfeldt2]). Then by Fact 4.2, we have $ \liminf _{n \rightarrow \infty } \frac {K (x \upharpoonright n)}{n} = 0 $ for every generic real x. Hence by Fact 4.3, $\mathrm {dim_H} (G) = 0$ .

Proof. We proof that $g+b$ is $a \oplus b \oplus x$ -generic first. Fix a $\Sigma _1^0 (a \oplus b \oplus x)$ dense set S and $\{S_s\}_{s \in \omega }$ an $a \oplus b \oplus x$ -recursive enumeration of S ( $S_n \subseteq S_{n+1}$ for each n and there is exactly one element in $S_{n+1} \backslash S_{n}$ ). We inductively define $S {}^{\prime }_s$ at stage $s+1$ as follows:

Suppose $\sigma \in S_{s+1} \backslash S_{s}$ . Since there are infinitely many 0s in the sequence b, there is $\tau $ such that $\forall y \succ \tau (y + b \succ \sigma )$ . Choose $\tau $ of the least length and enumerate it into $S{}'$ .

It is clear that $S{}'$ is $\Sigma _1^0 (a \oplus b \oplus x)$ . Fix $\gamma \in 2^{<\omega }$ . Choose $\gamma _1$ such that $\{y+b:y \succ \gamma \} \supseteq I_{\gamma _1}$ . Since S is dense, there is $\rho \in S$ such that $\rho \succ \gamma _1$ . By the definition of $S'$ , there is $\tau \in S'$ such that $\forall y \succ \tau (y+b \succ \rho )$ . Then $\tau \succeq \gamma $ . Hence $S'$ is dense. Since g is $a \oplus b \oplus x$ -generic, there is $\tau \prec g$ such that $\tau \in S{}'$ . By the definition of $S{}'$ , there is $\sigma \in S$ such that $\sigma \prec g+b$ . So $g+b$ is $a \oplus b \oplus x$ -generic.

Similarly one can prove that $a \cdot g$ , $g^{-1}$ and $g^{2}$ are $a \oplus b \oplus x$ -generic, which implies $g^{n}$ and $a \cdot g + b$ also are.

Proof of Theorem 1.5

Fix $\{ (x_{\alpha }, y_{\alpha })\}_{\alpha < \aleph _1}$ an enumeration of $ (\mathrm {dom}(F)\backslash F^{-1}(\mathbb {Q})) \times \mathrm {dom}(F)$ (F is defined at the beginning of this section). We construct a sequence of pairs of reals $\{ (g_{\alpha }, h_{\alpha })\}_{\alpha < \aleph _1}$ by induction on $\alpha < \aleph _1$ .

Stage $\boldsymbol {\alpha }$ . Define

$$\begin{align*}G_{\alpha}=\{ x_{\beta} \oplus y_{\beta} \oplus g_{\beta} \oplus h_{\beta} \oplus x_{\alpha} \oplus y_{\alpha} : \beta < \alpha \}. \end{align*}$$

Let

$$\begin{align*}I_{\alpha}=\{x:(\exists n)(\exists a_0,\ldots,a_n \in G_{\alpha})[x \leq_T a_0 \oplus a_1 \oplus\cdots\oplus a_n]\},\end{align*}$$

the Turing ideal generated by $G_{\alpha }$ . Since $I_{\alpha }$ is countable, there is a real $g_{\alpha }$ that is z-generic for all $z \in I_{\alpha }$ . Let $h_{\alpha } = \frac {y_{\alpha } - g_{\alpha }}{x_{\alpha }}$ .

Let $A_0$ be the group generated by $\{g_{\alpha } : \alpha < \aleph _1 \} \cup \{h_{\alpha } : \alpha < \aleph _1 \}$ . We claim that $A_0$ contains only generic reals. For any $g \in A_0$ , there are finite sequences $\{g_{\alpha _i}, h_{\alpha _i}, s_i, t_i \}_{0 \leqslant i \leqslant n}$ , where $s_i$ , $t_i \in \mathbb {Z}$ , and ordinals $\alpha _0 < \alpha _1 < \cdots < \alpha _n$ such that $s_n^2+t_n^2\neq 0$ and

$$\begin{align*}g = \sum_{i = 0}^n (s_i g_{\alpha_i} + t_i h_{\alpha_i}). \end{align*}$$

Define $c = (\oplus _{0 \leqslant i \leqslant n - 1} (g_{\alpha _i} \oplus h_{\alpha _i})) \oplus x_{\alpha _n} \oplus y_{\alpha _n}$ . By the construction, $g_{\alpha _n}$ is c-generic. Also

$$\begin{align*}s_n g_{\alpha_n} + t_n h_{\alpha_n} = t_n \frac{y_{\alpha_n}}{x_{\alpha_n}} + \left ( s_n - \frac{t_n}{x_{\alpha_n}} \right) g_{\alpha_n}, \end{align*}$$

so

$$\begin{align*}g = \left ( s_n - \frac{t_n}{x_{\alpha_n}} \right) g_{\alpha_n} + \left ( t_n \frac{y_{\alpha_n}}{x_{\alpha_n}} + \sum_{i = 0}^{n - 1} (s_i g_{\alpha_i} + t_i h_{\alpha_i}) \right). \end{align*}$$

Since $x_{\alpha _n}$ is irrational, we have that $s_n - \frac {t_n}{x_{\alpha _n}} \neq 0$ . By Lemma 4.5, g is c-generic and hence generic. Thus $\mathrm {dim_H} (A_0) = 0$ by Lemma 4.4. Let A be the subgroup of $\mathbb {R}$ generated by $F(A_0)$ . Then $\mathrm {dim_H} (A) = 0$ . For any pair $ (x, y) \in [0,1] \times [0,1]$ with $x \not \in \mathbb {Q}$ , there are members $g, h \in A$ so that $h = \frac {y - g}{x}$ and so $g + x \cdot h = y$ . So $\mathbb {R}\backslash \mathbb {Q} \subseteq X_A$ . By Proposition 3.1(ii), $\mathbb {Q} \cap X_A= \emptyset $ .

Question 2. (i) Can $\mathrm {CH}$ in Theorem 1.5 be removed?

(ii) Can A in Theorem 1.5 be Borel?

Fact 4.7. Let $\{I_i\}_{i\leq n}$ be a finite set of open intervals and J an open interval, then

$$ \begin{align*}\mu((\bigcup_{i\leq n}I_i)+J)\leq n\mu(J)+\sum_{k\leq n}\mu(I_i).\end{align*} $$

Proof of Proposition 4.6

Suppose $A = \bigcup _{n \ge 1} A_n$ is an $F_{\sigma }$ null subset of $\mathbb {R}$ with each $A_n$ compact. We claim that for each $m, n \ge 1$ , the set

$$ \begin{align*}D_{m, n} = \left\{ x \in \mathbb{R}: \mu (A_n + x A_n) < \frac{1}{m} \right\}\end{align*} $$

contains a dense open subset of $\mathbb {R}$ .

Using this claim, the set $\{x : \mu (A+xA)=0\}$ is comeager by the Baire category theorem. It reclaims to prove the claim.

Given $m, n \ge 1$ , we shall show that $D_{m, n}$ contains a dense open set in $(0, 1)$ . It is not hard to generalize the proof to show that $D_{m, n}$ contains a dense open set in $\mathbb {R}$ .

Let $\sigma $ be a binary string and $q \in \mathbb {N}$ . Fix a real $x = 0. \sigma 0^q x^{\ast } \in (0, 1)$ , where $x^{\ast } \in 2^{\omega }$ is the tail of the binary expansion of x. Define $y = 0. x^{\ast } \in (0, 1)$ . Then

$$\begin{align*}A_n + x A_n \subseteq A_n + \sum_{0 \leqslant i \leqslant | \sigma | - 1} \sigma (i) 2^{- i - 1} A_n + 2^{- | \sigma | - q} y A_n. \end{align*}$$

The set $A_{n,\sigma }:=A_n+\sum _{0 \leqslant i \leqslant | \sigma | - 1} \sigma (i) 2^{- i - 1} A_n$ is compact and

$$ \begin{align*}A_{n,\sigma}\subseteq A+\sum_{0 \leqslant i \leqslant | \sigma | - 1} \sigma(i) 2^{- i - 1} A\end{align*} $$

is null by Proposition 3.1(i) and Corollary 1.2. So there is a finite open interval cover $\{ I_j \}_{1 \leqslant j \leqslant k}$ of $A_{n,\sigma }$ such that $\sum _{1 \leqslant j \leqslant k} | I_j | < \frac {1}{2 m}$ . Suppose that $A_n \subseteq [- N, N]$ and so $yA_n \subseteq [- N, N]$ . Then if $2^{- q+1} Nk < \frac {1}{2m}$ , by Fact 4.7, we have

$$ \begin{align*}\mu (A_n + x A_n) \le\mu ( A_{n,\sigma} + 2^{- | \sigma | - q} y A_n ) \le \sum_{0\leq j\leq k}| I_j|+k 2^{- q-|\sigma|} 2N < \frac{1}{2m}+\frac{1}{2m}=\frac{1}{m}.\end{align*} $$

Since $\sigma $ is arbitrary, $D_{m,n}$ contains a dense open set in $(0,1)$ .

Fact 4.8 (Exchange of algebraic closure)

For a set $A \subseteq \mathbb {R}$ and reals a and b, if $b \in \mathrm {acl}_{\mathbb {R}}(A \cup \{a\})$ and $b \notin \mathrm {acl}_{\mathbb {R}}(A)$ , then $a \in \mathrm {acl}_{\mathbb {R}}(A \cup \{b\})$ .

Proof. Fix $\{ y_{\alpha } \}_{\alpha < \aleph _1}$ an enumeration of $\mathbb {R}$ . Fix x a transcendental number. Define $A = \bigcup _{\alpha < \aleph _1} A_{\alpha }$ by induction on stages $\alpha < \aleph _1$ :

Stage 0. Define $A_0=\mathbb {Q}$ . $x \notin \mathrm {acl}_{\mathbb {R}}(A_0)$ since it is transcendental.

Stage $\boldsymbol {\alpha>0}$ . Recall that G is the set of generic reals (see Lemma 4.4). Suppose by induction that we have a countable field

$$ \begin{align*}B_{\alpha} = \bigcup_{\beta < \alpha} A_{\beta} \subseteq G \cup \mathbb{Q}\end{align*} $$

such that $x \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha )$ . Choose the least $\gamma < \aleph _1$ such that $y_{\gamma } \notin \mathrm {acl}_{\mathbb {R}}(B_{\alpha })$ and $x \notin \mathrm {acl}_{\mathbb {R}}(B_{\alpha } \cup \{y_{\gamma }\})$ (if such a $\gamma $ does not exist, end the whole construction). And if such a $\gamma $ exists, we say that $\gamma $ acts at stage $\alpha $ . Let $I_\alpha $ be the Turing ideal generated by $B_\alpha \cup \{y_\gamma ,x\}$ . Choose a real $g_\alpha $ such that

(i) $g_\alpha $ is y-generic for all $y \in I_\alpha $ ;

(ii) $g_\alpha \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{y_\gamma ,x\} )$ .

Let $A_\alpha $ be the subfield of $\mathbb {R}$ generated by $B_{\alpha } \cup \{ g_\alpha , \frac {g_\alpha -x}{y_\gamma }\}$ .

We first show that $x \notin \mathrm {acl}_{\mathbb {R}}(A_\alpha )$ if there is an ordinal $\gamma $ which acts at stage $\alpha $ . By induction, we assume that $x \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha )$ . Let $h_\alpha =\frac {g_\alpha -x}{y_\gamma }$ . Then $x\in \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{g _\alpha ,h_\alpha \})$ but $x \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{g _\alpha \})$ ( $x \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha )$ by the induction hypothesis, and $g_\alpha \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{x\} )$ by its choice, then by Fact 4.8, $x \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{g _\alpha \})$ ). So ${h_\alpha \in \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{g _\alpha ,x \})}$ by Fact 4.8. Then $y_\gamma \in \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{g _\alpha ,x \})$ . And by Fact 4.8 again, since $y_\gamma \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha )$ , we have $y_\gamma \notin \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{x\})$ . So by Fact 4.8 again, $g_\alpha \in \mathrm {acl}_{\mathbb {R}}(B_\alpha \cup \{y_\gamma ,x\})$ , a contradiction.

By Lemma 4.5, $A_{\alpha } \subseteq G \cup \mathbb {Q}$ for all $\alpha $ . So the induction hypothesis holds at any stage before the construction ends.

Now we define $A = \bigcup _{\alpha < \aleph _1} A_{\alpha }$ and verify that A satisfies the conditions in the statement. We may assume the construction does not end at any stage $\alpha <\aleph _1$ (otherwise A is countable and it is clear that the conditions hold). Since ${A = \bigcup _{\alpha < \aleph _1} A_{\alpha } \subseteq G \cup \mathbb {Q}}$ , $\mathrm {dim_H} (A)=0$ . If $y\notin \mathrm {acl}_{\mathbb {R}}(A)$ , choose $\gamma < \aleph _1$ such that $y=y_\gamma $ . Then $x \in \mathrm {acl}_{\mathbb {R}}(A\cup \{y\})$ . Otherwise, $\gamma $ acts at some stage $\alpha < \aleph _1$ . So $g_\alpha $ , $\frac {g_\alpha -x}{y}$ and y are in $\mathrm {acl}_{\mathbb {R}}(A\cup \{y\})$ . Then x is also in $\mathrm {acl}_{\mathbb {R}}(A\cup \{y\})$ , a contradiction.