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Source. Hough, Lemma 4, printed pp. 374–375 of the published paper. Use the notation in the sieve setup.

Statement. For every integer $k\ge1$ and $n\in\mathcal N_{i+1}$,

$$\frac1{T_i}\sum_{r\in S_i}\mu_i(r)a_n(r)^k\le\beta_k(i)^k.$$

Complete proof. Let $D_n=\{m_0\mid Q_i:m_0n\in\mathcal M\}$. The CRT identification of the excluded events gives

$$a_n(r)\le\sum_{m_0\in D_n} 1_{\{r\equiv a_{m_0n}\pmod{m_0}\}}.$$

Raise to the integer power $k$, expand the finite sum into ordered tuples $(m_1,\ldots,m_k)\in D_n^k$, and average with $\mu_i/T_i$. Each simultaneous system $r\equiv a_{m_jn}\pmod{m_j}$ is either inconsistent or specifies exactly one class modulo $m=\operatorname{lcm}(m_1,\ldots,m_k)$: two solutions differ by a multiple of every $m_j$, hence of $m$, and any one solution generates the whole class. Its mass is therefore at most $\max_{b\bmod m}\mu_i(S_i\cap(b\bmod m))/T_i$.

For a fixed $m\mid Q_i$, there are at most $\ell_k(m)$ such tuples, because $D_n$ is a subset of the divisors of $Q_i$ and $\ell_k(m)$ counts all ordered tuples with that least common multiple. Summing this bound over $m\mid Q_i$ gives exactly $\beta_k(i)^k$.

Source precision. Restricting the tuple sum to $D_n$ avoids assigning a residue $a_{m_0n}$ to a modulus absent from $\mathcal M$, which is implicit in the printed sum over all divisors. This proof needs the moduli to be distinct; arbitrary repetitions would add multiplicities.

Bears on. Problem 2.