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Source. Hough, Lemma 5, printed p. 375 of the published paper. Use the notation in the sieve setup.

Statement. Let $k\ge1$ be an integer, $B>0$, and $w_n\ge0$ for $n\in\mathcal N_{i+1}$. Then

$$\frac1{T_i}\mu_i\left(\left\{r\in S_i: \sum_nw_na_n(r)>B\right\}\right) \le\frac{\beta_k(i)^k}{B^k}\left(\sum_nw_n\right)^k.$$

Complete proof. If $W=\sum_nw_n=0$, the weighted sum vanishes identically and both sides are zero. Otherwise put $u_n=w_n/W$. The convexity of $x\mapsto x^k$ on $[0,\infty)$ gives

$$\left(\sum_nu_na_n(r)\right)^k\le\sum_nu_na_n(r)^k.$$

Average over the probability measure $\mu_i/T_i$ and apply Lemma 4 to every $n$. The result is $\mathbb E(\sum_nu_na_n)^k\le\beta_k(i)^k$. For a nonnegative random variable $X$, the pointwise inequality $1_{\{X>B\}}\le X^k/B^k$ implies $\mathbb P(X>B)\le\mathbb E(X^k)/B^k$ by summation. Apply this with $X=W\sum_nu_na_n$ to obtain the claim.

Scope. The source excludes the all-zero weight family. Its trivial extension above includes primes not dividing $Q$, whose weights in the next application are all zero.

Bears on. Problem 2.