[proofplan] We first convert the representation count into an integral of the $s$th power of the Weyl sum by exponential orthogonality on the unit interval. We then split this integral over the measurable major arcs $\mathfrak M_N$ and minor arcs $\mathfrak m_N$. The major-arc contribution is exactly the assumed main term, while the minor-arc contribution is bounded in absolute value by the assumed minor-arc $L^s$ estimate. Finally, the identity $P=N^{1/k}$ converts $P^{s-k}$ into $N^{s/k-1}$. [/proofplan] [step:Express the representation count as a Weyl-sum integral] Fix $N\in\mathcal N$ and put $P=N^{1/k}$. Let $M:=\lfloor P\rfloor\in\mathbb N$. Since $x\le P$ for an integer $x\in\mathbb N$ is equivalent to $1\le x\le M$, the definition of $S_k(\alpha;P)$ is the same as the Weyl sum with cutoff $M$. By the representation integral formula for $k$th-power sums, applied with cutoff $M$ and target integer $N$, or equivalently by [citetheorem:9067] after replacing its integer parameter $P$ by $M$, we have \begin{align*} R_{s,k}(N;P)=\int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} The hypotheses of that formula are satisfied because $k,s,M,N\in\mathbb N$, and the set counted by the formula is precisely the set in the definition of $R_{s,k}(N;P)$. [guided] The only small point here is that the cutoff $P=N^{1/k}$ need not be an integer. To remove that ambiguity, define \begin{align*} M:=\lfloor P\rfloor. \end{align*} For an integer $x\in\mathbb N$, the inequality $x\le P$ is equivalent to $1\le x\le M$. Hence the Weyl sum appearing in the theorem is exactly \begin{align*} S_k(\alpha;P)=\sum_{x=1}^{M}e(\alpha x^k). \end{align*} We now apply the representation integral formula for $k$th-power sums, namely [citetheorem:9067], with integer cutoff $M$ and target integer $N$. Its hypotheses require $k,s,M\in\mathbb N$ and $N\in\mathbb Z$. Here $k,s\in\mathbb N$ by assumption, $M\in\mathbb N$ because $P=N^{1/k}\ge 1$, and $N\in\mathbb N\subset\mathbb Z$. Therefore the formula gives \begin{align*} \#\left\{(x_1,\dots,x_s)\in\{1,\dots,M\}^s:x_1^k+\cdots+x_s^k=N\right\} = \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} The set on the left is exactly the set defining $R_{s,k}(N;P)$, because $\{1,\dots,M\}$ is the set of positive integers at most $P$. Thus \begin{align*} R_{s,k}(N;P)=\int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} [/guided] [/step] [step:Decompose the representation integral into major and minor arcs] Since $\mathfrak M_N\subset[0,1]$ is Lebesgue measurable and $\mathfrak m_N=[0,1]\setminus\mathfrak M_N$, the sets $\mathfrak M_N$ and $\mathfrak m_N$ are disjoint measurable subsets whose union is $[0,1]$. Therefore finite additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} R_{s,k}(N;P) = \int_{\mathfrak M_N}S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha) + \int_{\mathfrak m_N}S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} [/step] [step:Insert the assumed major-arc asymptotic] By the major-arc hypothesis, as $N\to\infty$ through $\mathcal N$ with $P=N^{1/k}$, \begin{align*} \int_{\mathfrak M_N}S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha) = \mathfrak S_{s,k}(N)J_{s,k}(1)P^{s-k}+o(P^{s-k}). \end{align*} [/step] [step:Bound the minor-arc contribution by the minor-arc hypothesis] For every $\alpha\in\mathbb R$, $|e(-N\alpha)|=1$. Hence the triangle inequality for the Lebesgue integral gives \begin{align*} \left|\int_{\mathfrak m_N}S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)\right| \le \int_{\mathfrak m_N}|S_k(\alpha;P)|^s\,d\mathcal L^1(\alpha). \end{align*} By the minor-arc hypothesis, the right-hand side is $o(P^{s-k})$. Therefore \begin{align*} \int_{\mathfrak m_N}S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=o(P^{s-k}). \end{align*} [/step] [step:Convert the cutoff scale into the stated power of $N$] Combining the decomposition with the major-arc and minor-arc estimates yields \begin{align*} R_{s,k}(N;P) = \mathfrak S_{s,k}(N)J_{s,k}(1)P^{s-k}+o(P^{s-k}). \end{align*} Since $P=N^{1/k}$, we have \begin{align*} P^{s-k}=(N^{1/k})^{s-k}=N^{s/k-1}. \end{align*} Substituting this identity into the preceding asymptotic gives \begin{align*} R_{s,k}(N;P) = \mathfrak S_{s,k}(N)J_{s,k}(1)N^{s/k-1}+o(N^{s/k-1}). \end{align*} This is the desired asymptotic along the unbounded set $\mathcal N$. [/step]