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Source: arXiv v2, p. 5, Lemma 3.2 and its proof.

Statement

For $x\in\mathbb Z/Q\mathbb Z$ and $1\le j\le J$,

$$\alpha_j(x)\le \sum_{r=1}^{\nu_j}\ \sum_{g\mid Q_{j-1}}\quad \sum_{\substack{1\le i\le n\\d_i=gp_j^r}} \frac{\mathbf1_{x\subseteq a_i+g\mathbb Z}}{p_j^r}. \tag{1}$$

Here $x=c+Q\mathbb Z\subseteq a_i+g\mathbb Z$ means $c\equiv a_i\pmod g$.

Full proof

Write $x=c+Q\mathbb Z$. Since $|F_{j-1}(x)|=Q/Q_{j-1}$, the union bound gives

$$\alpha_j(x)\le\frac{Q_{j-1}}Q \sum_{\substack{1\le i\le n\\P^+(d_i)=p_j}} \sum_{\substack{a\pmod Q\\a\equiv c\pmod {Q_{j-1}}\\ a\equiv a_i\pmod {d_i}}}1. \tag{2}$$

For every index in (2), factor uniquely

$$d_i=gp_j^r, \qquad g\mid Q_{j-1}, \qquad 1\le r\le\nu_j.$$

The two congruences in the inner sum are compatible exactly when

$$c\equiv a_i\pmod{\gcd(Q_{j-1},d_i)} =a_i\pmod g. \tag{3}$$

Condition (3) is precisely the indicator condition in (1). When it holds, the combined congruence has modulus

$$[Q_{j-1},d_i]=Q_{j-1}p_j^r,$$

so it has $Q/(Q_{j-1}p_j^r)$ solutions modulo $Q$. Multiplying this count by the prefactor $Q_{j-1}/Q$ in (2) leaves $p_j^{-r}$. Summing over $r,g,i$ proves (1).

In the displayed regrouping on source p. 5, the prefactor is printed as $Q_{j-1}/Q_j$ rather than the unchanged $Q_{j-1}/Q$, and the following compatibility sentence says modulo $d_i$ rather than modulo $g$. The count in the next sentence and the lemma's stated bound require exactly the corrected relations (2)--(3). These are compilation corrections, not an author-issued erratum.