[proofplan] The proof first uses the Fourier integral formula for the representation count to identify $R_{s,k}(N;P)$ with the full unit-interval integral of the Weyl-sum integrand. Since the major arcs and minor arcs form a disjoint measurable partition of $[0,1)$, finite additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) splits this full integral into the two displayed pieces. For the asymptotic statement, we write the major arc integral as the proposed main term plus an error and the minor arc integral as another error. Both errors are $o(M(N))$, so division by the eventually nonzero function $M(N)$ gives the desired ratio limit. [/proofplan]

[step:Split the Fourier integral over the major and minor arcs]Define the integrand \begin{align*} F_{N,P}:[0,1)&\to\mathbb C \end{align*} by \begin{align*} F_{N,P}(\alpha):=S_k(\alpha;P)^s e(-N\alpha). \end{align*} Since $S_k(\alpha;P)$ is a finite sum of continuous functions of $\alpha$, the function $F_{N,P}$ is continuous on $[0,1)$ and therefore Lebesgue measurable and bounded. By [citetheorem:9067], applied with the same integers $k,s,P,N$ and the same Weyl sum $S_k(\alpha;P)$, we have \begin{align*} R_{s,k}(N;P)=\int_0^1 F_{N,P}(\alpha)\,d\mathcal L^1(\alpha). \end{align*} The hypotheses give a disjoint measurable decomposition $[0,1)=\mathfrak M\sqcup\mathfrak m$. Since $F_{N,P}$ is bounded and measurable on a set of finite [Lebesgue measure](/page/Lebesgue%20Measure), it is Lebesgue integrable over $[0,1)$. Finite additivity of the Lebesgue integral over disjoint measurable sets gives \begin{align*} \int_0^1 F_{N,P}(\alpha)\,d\mathcal L^1(\alpha)=\int_{\mathfrak M}F_{N,P}(\alpha)\,d\mathcal L^1(\alpha)+\int_{\mathfrak m}F_{N,P}(\alpha)\,d\mathcal L^1(\alpha). \end{align*} Substituting the definition of $F_{N,P}$ gives \begin{align*} R_{s,k}(N;P)=\int_{\mathfrak M} S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)+\int_{\mathfrak m} S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*}[/step]

[guided]We want to separate the full circle method integral into the part supported on the major arcs and the part supported on the minor arcs. The object being integrated is the function \begin{align*} F_{N,P}:[0,1)&\to\mathbb C \end{align*} defined by \begin{align*} F_{N,P}(\alpha):=S_k(\alpha;P)^s e(-N\alpha). \end{align*} This is a legitimate Lebesgue integrand: the Weyl sum $S_k(\alpha;P)$ is a finite sum of continuous functions $\alpha\mapsto e(\alpha x^k)$, and products and powers of continuous complex-valued functions are continuous. Hence $F_{N,P}$ is continuous, measurable, and bounded on $[0,1)$. The full integral formula is supplied by [citetheorem:9067]. Its hypotheses match the present notation: $k,s,P\in\mathbb N$, $k\ge 2$, $N\in\mathbb Z$ in that theorem and here $N\in\mathbb N\subset\mathbb Z$, and the Weyl sum is exactly \begin{align*} S_k(\alpha;P)=\sum_{x=1}^{P} e(\alpha x^k). \end{align*} Therefore \begin{align*} R_{s,k}(N;P)=\int_0^1 F_{N,P}(\alpha)\,d\mathcal L^1(\alpha). \end{align*} Now we use the defining property of the major and minor arcs in this theorem: $\mathfrak M$ and $\mathfrak m$ are Lebesgue measurable, disjoint, and their union is $[0,1)$. Since the integrand is integrable over $[0,1)$, finite additivity of the Lebesgue integral over a disjoint measurable union gives \begin{align*} \int_0^1 F_{N,P}(\alpha)\,d\mathcal L^1(\alpha)=\int_{\mathfrak M}F_{N,P}(\alpha)\,d\mathcal L^1(\alpha)+\int_{\mathfrak m}F_{N,P}(\alpha)\,d\mathcal L^1(\alpha). \end{align*} Replacing $F_{N,P}$ by its definition yields \begin{align*} R_{s,k}(N;P)=\int_{\mathfrak M} S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)+\int_{\mathfrak m} S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} This is the exact major-minor arc decomposition: no estimate has entered yet, only the Fourier representation and the measurable partition of the unit interval.[/guided]

[step:Combine the major arc and minor arc errors] For $N\in\mathcal N$, define \begin{align*} I_{\mathfrak M}(N):=\int_{\mathfrak M_N} S_k(\alpha;P(N))^s e(-N\alpha)\,d\mathcal L^1(\alpha) \end{align*} and \begin{align*} I_{\mathfrak m}(N):=\int_{\mathfrak m_N} S_k(\alpha;P(N))^s e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} By the first part of the theorem, applied with $P=P(N)$ and with the partition $[0,1)=\mathfrak M_N\sqcup\mathfrak m_N$, we have \begin{align*} R_{s,k}(N;P(N))=I_{\mathfrak M}(N)+I_{\mathfrak m}(N). \end{align*} The two asymptotic hypotheses state that there are functions $\varepsilon_{\mathfrak M}:\mathcal N\to\mathbb C$ and $\varepsilon_{\mathfrak m}:\mathcal N\to\mathbb C$ such that \begin{align*} I_{\mathfrak M}(N)=M(N)+\varepsilon_{\mathfrak M}(N), \end{align*} \begin{align*} I_{\mathfrak m}(N)=\varepsilon_{\mathfrak m}(N), \end{align*} and \begin{align*} \frac{\varepsilon_{\mathfrak M}(N)}{M(N)}\to 0, \end{align*} \begin{align*} \frac{\varepsilon_{\mathfrak m}(N)}{M(N)}\to 0 \end{align*} as $N\to\infty$ through $\mathcal N$. Since $M(N)$ is eventually nonzero, the quotient by $M(N)$ is defined for all sufficiently large $N\in\mathcal N$. Combining the identities gives \begin{align*} R_{s,k}(N;P(N))=M(N)+\varepsilon_{\mathfrak M}(N)+\varepsilon_{\mathfrak m}(N). \end{align*} Dividing by $M(N)$, we obtain \begin{align*} \frac{R_{s,k}(N;P(N))}{M(N)}=1+\frac{\varepsilon_{\mathfrak M}(N)}{M(N)}+\frac{\varepsilon_{\mathfrak m}(N)}{M(N)}. \end{align*} The two error ratios tend to $0$, so the right-hand side tends to $1$. Hence \begin{align*} R_{s,k}(N;P(N))\sim M(N), \end{align*} which is the asserted asymptotic formula. [/step]