[proofplan] The theorem is a direct consequence of the assumed weight-level dimension-one Brun fundamental lemma. We first name the sifted indicator as an actual map on $\mathcal A$, so that the pointwise majorization and minorization statements can be summed without ambiguity. The upper and lower estimates then follow by finite summation over $\mathcal A$, substitution of the divisor-counting identity $|\mathcal A_d|=Xg(d)+r_d$, and the level-$D$ bound defining $R(D)$. The universality and limiting behavior of $F$ and $f$ are exactly the final hypotheses included in the assumed fundamental lemma. [/proofplan]

[step:Define the sifted indicator before applying the Brun weights] Fix [real numbers](/page/Real%20Numbers) $z>1$ and $D\geq 1$, and put \begin{align*} s:=\frac{\log D}{\log z}. \end{align*} Define the sifted indicator map $I_z: \mathcal A \to \{0,1\}$ by the following rule: $I_z(a)=1$ exactly when $\gcd(a,P_z)=1$, and $I_z(a)=0$ otherwise. Since $\mathcal A$ is finite, summing this indicator over $\mathcal A$ gives \begin{align*} S_{\mathcal A,\mathcal P}(z)=\sum_{a\in\mathcal A}I_z(a). \end{align*} For every divisor $d$ of $P_z$, the integer $d$ is squarefree and belongs to $\mathcal D_\infty$, so the identity $|\mathcal A_d|=Xg(d)+r_d$ is available for every divisor that can occur in the weight sums. [/step]

[step:Sum the upper majorant and isolate the remainder]Assume that $s$ lies in a compact subinterval of $(1,\infty)$. By the upper clause of the assumed weight-level dimension-one Brun fundamental lemma, there are real weights $\lambda_d^+$ supported on squarefree divisors $d\mid P_z$ with $d\leq D$, satisfying $|\lambda_d^+|\leq 1$, such that for every $a\in\mathcal A$, \begin{align*} I_z(a)\leq \sum_{d\mid \gcd(a,P_z)}\lambda_d^+. \end{align*} Summing this pointwise inequality over the finite set $\mathcal A$ and interchanging the finite sums gives \begin{align*} S_{\mathcal A,\mathcal P}(z)\leq \sum_{d\mid P_z}\lambda_d^+|\mathcal A_d|. \end{align*} Substituting $|\mathcal A_d|=Xg(d)+r_d$ for each $d\mid P_z$ gives \begin{align*} S_{\mathcal A,\mathcal P}(z)\leq X\sum_{d\mid P_z}\lambda_d^+g(d)+\sum_{d\mid P_z}\lambda_d^+r_d. \end{align*} Because $\lambda_d^+=0$ unless $d\leq D$ and $|\lambda_d^+|\leq 1$, the triangle inequality gives \begin{align*} \left|\sum_{d\mid P_z}\lambda_d^+r_d\right|\leq \sum_{d\in\mathcal D(D)}|r_d|=R(D). \end{align*} Therefore \begin{align*} S_{\mathcal A,\mathcal P}(z)\leq X\sum_{d\mid P_z}\lambda_d^+g(d)+R(D). \end{align*}[/step]

[guided]The purpose of the upper Brun weights is to replace the exact condition $\gcd(a,P_z)=1$ by a divisor sum that can be evaluated through the sieve axioms. The exact condition is already encoded by the map $I_z: \mathcal A\to\{0,1\}$, where $I_z(a)=1$ precisely when $a$ survives the sieving by primes in $\mathcal P$ below $z$. Since $s=\frac{\log D}{\log z}$ is assumed to lie in a compact subinterval of $(1,\infty)$, the upper part of the weight-level dimension-one Brun fundamental lemma applies. Its hypotheses are exactly the data in the theorem statement: the divisors are squarefree and supported on $\mathcal P$, the density map $g: \mathcal D_\infty\to\mathbb R$ is multiplicative, and the level is $D$. The conclusion gives weights $\lambda_d^+\in\mathbb R$, supported on divisors $d\mid P_z$ with $d\leq D$, with $|\lambda_d^+|\leq1$, and with the pointwise majorization \begin{align*} I_z(a)\leq \sum_{d\mid \gcd(a,P_z)}\lambda_d^+ \end{align*} for every $a\in\mathcal A$. We now sum over $a\in\mathcal A$. This causes no convergence issue because $\mathcal A$ is finite and $P_z$ has only finitely many divisors. The divisor $d$ contributes to the inner sum exactly for those $a\in\mathcal A$ such that $d\mid a$, and those elements form $\mathcal A_d$. Hence \begin{align*} S_{\mathcal A,\mathcal P}(z)=\sum_{a\in\mathcal A}I_z(a)\leq \sum_{a\in\mathcal A}\sum_{d\mid\gcd(a,P_z)}\lambda_d^+=\sum_{d\mid P_z}\lambda_d^+|\mathcal A_d|. \end{align*} For each divisor $d\mid P_z$, the divisor is squarefree and supported on $\mathcal P$, so the sieve axiom applies to the same $d$ and gives $|\mathcal A_d|=Xg(d)+r_d$. Substitution yields \begin{align*} S_{\mathcal A,\mathcal P}(z)\leq X\sum_{d\mid P_z}\lambda_d^+g(d)+\sum_{d\mid P_z}\lambda_d^+r_d. \end{align*} The last sum is the only part depending on the remainders. Since the weights vanish for $d>D$, every nonzero term has $d\in\mathcal D(D)$, and since $|\lambda_d^+|\leq1$, the triangle inequality gives \begin{align*} \left|\sum_{d\mid P_z}\lambda_d^+r_d\right|\leq \sum_{d\in\mathcal D(D)}|r_d|=R(D). \end{align*} Thus the upper majorization has reduced the sifted count to the weighted density sum, with error at most $R(D)$: \begin{align*} S_{\mathcal A,\mathcal P}(z)\leq X\sum_{d\mid P_z}\lambda_d^+g(d)+R(D). \end{align*}[/guided]

[step:Evaluate the upper weighted density sum] The upper main-term clause of the assumed weight-level dimension-one Brun fundamental lemma applies to the same weights $\lambda_d^+$. It gives \begin{align*} \sum_{d\mid P_z}\lambda_d^+g(d)=V_{\mathcal P,g}(z)\{F(s)+\varepsilon_+(D,z;s)\}. \end{align*} Substitution into the upper inequality from the preceding step gives \begin{align*} S_{\mathcal A,\mathcal P}(z)\leq X V_{\mathcal P,g}(z)\{F(s)+\varepsilon_+(D,z;s)\}+R(D). \end{align*} The asserted upper uniformity is inherited from the hypothesis that $\varepsilon_+(D,z;s)\to0$ uniformly whenever $s$ remains in a fixed compact subinterval of $(1,\infty)$. [/step]

[step:Sum the lower minorant and isolate the remainder]Assume now that $s$ lies in a compact subinterval of $(2,\infty)$. By the lower clause of the assumed weight-level dimension-one Brun fundamental lemma, there are real weights $\lambda_d^-$ supported on squarefree divisors $d\mid P_z$ with $d\leq D$, satisfying $|\lambda_d^-|\leq 1$, such that for every $a\in\mathcal A$, \begin{align*} I_z(a)\geq \sum_{d\mid \gcd(a,P_z)}\lambda_d^-. \end{align*} Summing over $\mathcal A$, interchanging finite sums, and substituting $|\mathcal A_d|=Xg(d)+r_d$ gives \begin{align*} S_{\mathcal A,\mathcal P}(z)\geq X\sum_{d\mid P_z}\lambda_d^-g(d)+\sum_{d\mid P_z}\lambda_d^-r_d. \end{align*} The support condition $d\leq D$ and the bound $|\lambda_d^-|\leq1$ imply \begin{align*} \sum_{d\mid P_z}\lambda_d^-r_d\geq -\sum_{d\in\mathcal D(D)}|r_d|=-R(D). \end{align*} Therefore \begin{align*} S_{\mathcal A,\mathcal P}(z)\geq X\sum_{d\mid P_z}\lambda_d^-g(d)-R(D). \end{align*}[/step]

[guided]The lower estimate is structurally the same finite-summation argument as the upper estimate, but the fundamental lemma supplies lower weights only in the stronger range $s\in(2,\infty)$. This is why the theorem states the lower bound on compact subintervals of $(2,\infty)$ rather than compact subintervals of $(1,\infty)$. In this range, the lower part of the assumed weight-level dimension-one Brun fundamental lemma applies to the same sieve data. It gives weights $\lambda_d^-\in\mathbb R$, supported on squarefree divisors $d\mid P_z$ with $d\leq D$, with $|\lambda_d^-|\leq1$, and with the pointwise minorization \begin{align*} I_z(a)\geq \sum_{d\mid\gcd(a,P_z)}\lambda_d^- \end{align*} for every $a\in\mathcal A$. Summing this inequality over the finite set $\mathcal A$ and reversing the order of the finite sums gives \begin{align*} S_{\mathcal A,\mathcal P}(z)=\sum_{a\in\mathcal A}I_z(a)\geq \sum_{a\in\mathcal A}\sum_{d\mid\gcd(a,P_z)}\lambda_d^- = \sum_{d\mid P_z}\lambda_d^-|\mathcal A_d|. \end{align*} Every divisor $d\mid P_z$ lies in $\mathcal D_\infty$, so the divisor-counting identity gives $|\mathcal A_d|=Xg(d)+r_d$. Hence \begin{align*} S_{\mathcal A,\mathcal P}(z)\geq X\sum_{d\mid P_z}\lambda_d^-g(d)+\sum_{d\mid P_z}\lambda_d^-r_d. \end{align*} For a lower bound, the remainder sum must be bounded from below. Since the weights vanish outside $\mathcal D(D)$ and satisfy $|\lambda_d^-|\leq1$, each summand satisfies $\lambda_d^-r_d\geq-|r_d|$. Summing over the possible nonzero terms gives \begin{align*} \sum_{d\mid P_z}\lambda_d^-r_d\geq -\sum_{d\in\mathcal D(D)}|r_d|=-R(D). \end{align*} Thus the lower combinatorial inequality is \begin{align*} S_{\mathcal A,\mathcal P}(z)\geq X\sum_{d\mid P_z}\lambda_d^-g(d)-R(D). \end{align*}[/guided]

[step:Evaluate the lower weighted density sum] The lower main-term clause of the assumed weight-level dimension-one Brun fundamental lemma applies to the lower weights $\lambda_d^-$. It gives \begin{align*} \sum_{d\mid P_z}\lambda_d^-g(d)=V_{\mathcal P,g}(z)\{f(s)+\varepsilon_-(D,z;s)\}. \end{align*} Substituting this identity into the lower inequality from the preceding step yields \begin{align*} S_{\mathcal A,\mathcal P}(z)\geq X V_{\mathcal P,g}(z)\{f(s)+\varepsilon_-(D,z;s)\}-R(D). \end{align*} The asserted lower uniformity is inherited from the hypothesis that $\varepsilon_-(D,z;s)\to0$ uniformly whenever $s$ remains in a fixed compact subinterval of $(2,\infty)$. [/step]

[step:Record the universal functions and their limiting behavior] The functions $F:(1,\infty)\to\mathbb R$ and $f:(2,\infty)\to\mathbb R$ used in the preceding estimates are the universal functions supplied by the assumed dimension-one Brun fundamental lemma. By hypothesis, their construction depends only on the dimension-one Brun combinatorics and not on the particular finite set $\mathcal A$, the parameter $X$, or the remainders $r_d$. The same assumed fundamental lemma includes the limiting relations \begin{align*} \lim_{s\to\infty}F(s)=1 \end{align*} and \begin{align*} \lim_{s\to\infty}f(s)=1. \end{align*} Together with the upper estimate on compact subintervals of $(1,\infty)$ and the lower estimate on compact subintervals of $(2,\infty)$, this proves exactly the stated theorem. [/step]