---

Source. Hough, Section 3, printed pp. 367–372 of the published paper. These are definitions and elementary identifications used by the result pages, not an additional theorem asserted by the source.

Let $\mathcal M$ be a finite set of distinct integers greater than $M>1$, with one residue $a_m\pmod m$ for each $m\in\mathcal M$. Put $Q=\operatorname{lcm}(\mathcal M)$, with $Q=1$ for the empty set, and $v_p=v_p(Q)$. Take $1=P_{-1}<P_0<P_1<\cdots\to\infty$, with $P_0\ge2$, and define

$$Q_{-1}=1,\qquad Q_i=\prod_{p\le P_i}p^{v_p},\qquad \mathcal M_i=\{m\in\mathcal M:m\mid Q_i\},\qquad R_i=\bigcap_{m\in\mathcal M_i}(a_m\bmod m)^c.$$

Sets are regarded as periodic subsets of the integers or as their images in the indicated finite quotient. In particular $R_{-1}=\mathbb Z$. For some finite $i$, $Q_i=Q$ and $R_i$ is the final uncovered set. A nonempty subset of $\mathbb Z/Q\mathbb Z$ lifts to a periodic set of positive density, namely its cardinality divided by $Q$. No assertion about the limit of infinitely many positive densities is needed.

At step $i+1$, put

$$\mathcal N_{i+1}=\{n>1:n\mid Q_{i+1},\ p\mid n\Rightarrow P_i<p\le P_{i+1}\}.$$

Every $m\in\mathcal M_{i+1}\setminus\mathcal M_i$ factors uniquely as $m=m_0n$, where $m_0\mid Q_i$, $n\in\mathcal N_{i+1}$, and $\gcd(m_0,n)=\gcd(Q_i,n)=1$. For $r\in R_i\bmod Q_i$ define

$$A_{n,r}=(r\bmod Q_i)\cap \bigcup_{\substack{m_0\mid Q_i\\m_0n\in\mathcal M}}(a_{m_0n}\bmod m_0n), \qquad a_n(r)=|A_{n,r}\bmod nQ_i|.$$

The Chinese remainder theorem shows that a term in this union is empty unless $r\equiv a_{m_0n}\pmod{m_0}$, and otherwise is exactly one class modulo $nQ_i$. The surviving part of the fiber is

$$R_{i+1}\cap(r\bmod Q_i) =(r\bmod Q_i)\cap\bigcap_{n\in\mathcal N_{i+1}}A_{n,r}^{c}.$$

For $\lambda\ge0$, the fiber $r$ is $\lambda$-good when, for every prime $p\in(P_i,P_{i+1}]$,

$$\sum_{\substack{n\in\mathcal N_{i+1}\\p\mid n}} \frac{a_n(r)e^{\lambda\omega(n)}}n\le1-e^{-\lambda}. \tag{5}$$

Here $\omega(n)$ counts distinct prime divisors. It is $\lambda$-well distributed when its surviving part is nonempty and for every $n\in\mathcal N_{i+1}$ and every residue $b\pmod n$,

$$\frac{|R_{i+1}\cap(r\bmod Q_i)\cap(b\bmod n)\bmod Q_{i+1}|} {|R_{i+1}\cap(r\bmod Q_i)\bmod Q_{i+1}|} \le\frac{e^{\lambda\omega(n)}}n. \tag{4}$$

The same inequality for $n=1$ is the identity $1\le1$.

Let $R_{-1}^*=\mathbb Z$, and at stage $i$ select good fibers $R_i^*\subseteq S_i:=R_{i-1}^*\cap R_i\subseteq\mathbb Z/Q_i\mathbb Z$. Let $\mu_i$ be a nonnegative finite measure supported on $S_i$, with $T_i=\mu_i(S_i)>0$. For an integer $k\ge1$, let $\ell_k(m)$ be the number of ordered $k$-tuples of positive integers with least common multiple $m$, and put

$$\beta_k(i)^k= \sum_{m\mid Q_i}\ell_k(m)\max_{b\bmod m} \frac{\mu_i(S_i\cap(b\bmod m))}{T_i}.$$

The normalization makes these statistics invariant under multiplication of $\mu_i$ by a positive constant. The $m=1$ term is $1$, so every $\beta_k(i)\ge1$. Measures need not remain probability measures. The initial uniform measure and its bound are proved in the initial-stage argument; the next measure is defined in Lemma 2.

Conventions. General moduli require the full powers $p^{v_p(Q)}$. Section 2's products of primes describe only its square-free overview. Empty products are $1$ and empty sums are $0$. A stage with no new prime divisor of $Q$ has $\mathcal N_{i+1}=\varnothing$, no new exclusion, and every surviving fiber is good. This also covers $\mathcal M=\varnothing$.

Bears on. Problem 2.