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Source. Hough, Proposition 1, printed pp. 369–371 of the published paper. Use the notation and definitions of goodness and well distribution in the sieve setup.

Statement. For every $i\ge0$ and every $\lambda\ge0$, every $\lambda$-good fiber is $\lambda$-well distributed.

Complete proof. Fix a good fiber $r$ and give $F_r=(r\bmod Q_i)\subseteq\mathbb Z/Q_{i+1}\mathbb Z$ its uniform probability measure $\mathbb P_r$. Translation by $-r$ and division by $Q_i$ identifies it with $\mathbb Z/(Q_{i+1}/Q_i)\mathbb Z$. The Chinese remainder theorem decomposes this quotient into independent prime-power coordinates. Each event $A_{n,r}$ depends only on coordinates at primes dividing $n$, and

$$\mathbb P_r(A_{n,r})=\frac{a_n(r)}n.$$

Thus $A_{n,r}$ is independent of the sigma-algebra of all events with moduli coprime to $n$, not merely pairwise independent of those events. In the dependency graph join different $n,n'$ exactly when $\gcd(n,n')>1$.

If $\mathcal N_{i+1}=\varnothing$, the fiber survives in full. If $\lambda=0$, goodness forces every $a_n(r)=0$: each $n>1$ has a prime divisor and occurs with a positive coefficient in that prime's sum. Again the full fiber survives. In either case uniform CRT counting gives (4), including $\lambda=0$, directly. Hence assume $\lambda>0$.

Set $x_n=e^{\lambda\omega(n)}a_n(r)/n$. For every prime in the band, $\sum_{p\mid n}x_n\le1-e^{-\lambda}$, so $0\le x_n\le1-e^{-\lambda}<1$. Concavity of $\log(1-x)$ puts its graph above the chord joining $x=0$ to $x=1-e^{-\lambda}$; therefore

$$1-x_n\ge\exp\left(-\frac{\lambda x_n}{1-e^{-\lambda}}\right).$$

For any $n\in\mathcal N_{i+1}$, repeated factors between $0$ and $1$ can only decrease a product. Hence

$$\prod_{\substack{n'\in\mathcal N_{i+1}\\(n,n')>1}}(1-x_{n'}) \ge\prod_{p\mid n}\prod_{\substack{n'\in\mathcal N_{i+1}\\p\mid n'}}(1-x_{n'}) \ge e^{-\lambda\omega(n)}. \tag{*}$$

The last inequality follows by applying the exponential bound and each prime's goodness condition. Dropping the possible self-factor from the first product only increases it. Multiplication by $x_n$ now gives the criterion of the relative local lemma, so the event $B=\bigcap_{n'}A_{n',r}^c$ has positive probability.

Fix $b\pmod n$, and let $U=\{n'\in\mathcal N_{i+1}:(n,n')=1\}$. The event $b\pmod n$ has probability $1/n$ and is independent of the sigma-algebra generated by $(A_{n',r})_{n'\in U}$. Consequently

$$\mathbb P_r(B\cap(b\bmod n)) \le\frac1n\mathbb P_r\left(\bigcap_{n'\in U}A_{n',r}^c\right).$$

The relative lemma, including its proved empty-$U$ case, bounds $\mathbb P_r(B)$ below by the last avoidance probability multiplied by $\prod_{(n,n')>1}(1-x_{n'})$. Divide the two inequalities and use (*) to obtain $\mathbb P_r(b\bmod n\mid B)\le e^{\lambda\omega(n)}/n$. This is precisely (4), and $B$ is the required surviving fiber.

Source precision. The separate $\lambda=0$ argument avoids the printed proof's division by $1-e^{-\lambda}$ at its stated endpoint. The empty collection and empty relative subcollection are also included.

Bears on. Problem 2.