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Source. Hough, Proposition 3, printed pp. 373–374 of the published paper. Use the sieve setup and the positive-mass reweighting of Lemma 2.

Statement. If $R_i^*$ is $\lambda$-good and $\pi_i^{\rm good}>0$, then for every integer $k\ge1$,

$$\beta_k(i+1)^k\le \frac{\beta_k(i)^k}{\pi_i^{\rm good}} \prod_{P_i<p\le P_{i+1}} \left(1+e^\lambda\sum_{j=1}^{v_p(Q)} \frac{(j+1)^k-j^k}{p^j}\right).$$

Complete proof. Factor $m\mid Q_{i+1}$ uniquely as $m=m_0n$ with $m_0\mid Q_i$ and $n\in\{1\}\cup\mathcal N_{i+1}$. For any $b\pmod m$, fibrewise constancy of $\mu_{i+1}$ gives

$$\mu_{i+1}(S_{i+1}\cap(b\bmod m)) =\sum_{\substack{r\in R_i^*\bmod Q_i\\r\equiv b\pmod{m_0}}} \mu_i(r) \frac{|R_{i+1}\cap(r\bmod Q_i)\cap(b\bmod n)\bmod Q_{i+1}|} {|R_{i+1}\cap(r\bmod Q_i)\bmod Q_{i+1}|}.$$

By Proposition 1, the ratio is at most $e^{\lambda\omega(n)}/n$; for $n=1$ this remains true with ratio $1$. Since $R_i^*\subseteq S_i$ and $T_{i+1}=\pi_i^{\rm good}T_i$, it follows that

$$\max_{b\bmod m} \frac{\mu_{i+1}(S_{i+1}\cap(b\bmod m))}{T_{i+1}} \le\frac{e^{\lambda\omega(n)}}{n\pi_i^{\rm good}} \max_{c\bmod m_0}\frac{\mu_i(S_i\cap(c\bmod m_0))}{T_i}.$$

Multiply by $\ell_k(m)=\ell_k(m_0)\ell_k(n)$ and sum independently over $m_0$ and $n$. The first sum is $\beta_k(i)^k$ and the second is

$$\sum_{n\in\{1\}\cup\mathcal N_{i+1}} \frac{\ell_k(n)e^{\lambda\omega(n)}}n =\prod_{P_i<p\le P_{i+1}} \left(1+e^\lambda\sum_{j=1}^{v_p(Q)}\frac{\ell_k(p^j)}{p^j}\right).$$

This uses the multiplicativity established in the initial-stage proof. It proves the claim, including primes with $v_p(Q)=0$ and empty bands.

Bears on. Problem 2.