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Source. Hough, equation (8) and Lemma 2, printed pp. 372–373 of the published paper. Use $S_i,T_i,\mu_i$ from the sieve setup.

Statement. Let $R_i^*\subseteq S_i$ consist of good fibers and have positive $\mu_i$-mass. Put $\pi_i^{\rm good}=\mu_i(R_i^*)/T_i$. Define $\mu_{i+1}$ on $\mathbb Z/Q_{i+1}\mathbb Z$ by zero off $S_{i+1}=R_i^*\cap R_{i+1}$, and, for $s\in S_{i+1}$, by

$$\mu_{i+1}(s)= \frac{\mu_i(s\bmod Q_i)} {|R_{i+1}\cap(s\bmod Q_i)\bmod Q_{i+1}|}. \tag{8}$$

This measure is constant on each surviving fiber over $R_i^*$, and

$$T_{i+1}=\mu_{i+1}(S_{i+1}) =\mu_i(R_i^*)=\pi_i^{\rm good}T_i>0.$$

Complete proof. Proposition 1 makes every denominator in (8) a positive integer. Both numerator and denominator depend only on the underlying residue $r\pmod{Q_i}$, which proves constancy. If that fiber has $h_r$ surviving residues, their total mass is $h_r\mu_i(r)/h_r=\mu_i(r)$. Summing over the disjoint good fibers gives the displayed identity. Thus unequal surviving cardinalities do not change the relative incoming masses of the good fibers.

Bears on. Problem 2.