[proofplan] We expand the $2s$-th power of the polynomial exponential sum as $s$ ordinary copies and $s$ conjugate copies. Since all sums are finite, we may interchange the resulting finite sum over integer tuples with the integral over $[0,1)^k$. Each term is then evaluated by the elementary orthogonality identity for one-dimensional additive characters, applied independently in the $k$ coordinates. The only tuples that survive are exactly those satisfying the Vinogradov power-sum equations, so the integral equals $J_{s,k}(X)$. [/proofplan]

[step:Expand the absolute value into ordinary and conjugate exponential sums] Define the finite set $A:=\{1,\dots,N\}$. Define the function $S:[0,1)^k\to\mathbb C$ by \begin{align*} S(\theta):=\sum_{n=1}^{N}e(\theta_1n+\cdots+\theta_kn^k), \end{align*} where $\theta=(\theta_1,\dots,\theta_k)\in[0,1)^k$. Since $\theta_1n+\cdots+\theta_kn^k$ is real, the identity $\overline{e(t)}=e(-t)$ gives \begin{align*} |S(\theta)|^{2s}=S(\theta)^s\overline{S(\theta)}^s. \end{align*} Expanding the two finite products over $A^s$, we obtain \begin{align*} |S(\theta)|^{2s}=\sum_{(x_1,\dots,x_s)\in A^s}\sum_{(y_1,\dots,y_s)\in A^s}e\left(\sum_{j=1}^k\theta_j\left(\sum_{i=1}^s x_i^j-\sum_{i=1}^s y_i^j\right)\right). \end{align*} [/step]

[step:Move the finite expansion through the integral]For each tuple $(x,y)=(x_1,\dots,x_s,y_1,\dots,y_s)\in A^{2s}$ and each integer $j$ with $1\le j\le k$, define the integer \begin{align*} A_j(x,y):=\sum_{i=1}^s x_i^j-\sum_{i=1}^s y_i^j. \end{align*} By finite linearity of the [Lebesgue integral](/page/Lebesgue%20Integral), \begin{align*} \int_{[0,1)^k}|S(\theta)|^{2s}\,d\mathcal L^k(\theta)=\sum_{(x,y)\in A^{2s}}\int_{[0,1)^k}e\left(\sum_{j=1}^k\theta_jA_j(x,y)\right)\,d\mathcal L^k(\theta). \end{align*}[/step]

[guided]The point of introducing $A_j(x,y)$ is to isolate the obstruction to a tuple surviving the integral. For fixed $(x,y)\in A^{2s}$, the quantity $A_j(x,y)$ is an integer because it is a difference of sums of integer powers. The expansion from the previous step is a finite sum indexed by the finite set $A^{2s}$, so no convergence theorem is needed: ordinary finite additivity and linearity of the Lebesgue integral allow us to integrate term by term. Thus \begin{align*} \int_{[0,1)^k}|S(\theta)|^{2s}\,d\mathcal L^k(\theta)=\sum_{(x,y)\in A^{2s}}\int_{[0,1)^k}e\left(\sum_{j=1}^k\theta_jA_j(x,y)\right)\,d\mathcal L^k(\theta). \end{align*} This formula reduces the theorem to evaluating one elementary integral for each fixed tuple.[/guided]

[step:Evaluate each coordinate integral by additive character orthogonality] We use the elementary identity that, for every integer $m$, \begin{align*} \int_0^1 e(mt)\,d\mathcal L^1(t)=\begin{cases}1,&m=0, \end{align*} \begin{align*} 0,&m\ne 0.\end{cases} \end{align*} Indeed, if $m=0$, the integrand is identically $1$. If $m\ne 0$, then \begin{align*} \int_0^1 e(mt)\,d\mathcal L^1(t)=\frac{e(m)-1}{2\pi im}=0, \end{align*} because $e(m)=1$ for every integer $m$. Fix $(x,y)\in A^{2s}$. The integrand factors as \begin{align*} e\left(\sum_{j=1}^k\theta_jA_j(x,y)\right)=\prod_{j=1}^k e(\theta_jA_j(x,y)). \end{align*} Since this product is bounded and measurable on $[0,1)^k$, repeated one-dimensional integration gives \begin{align*} \int_{[0,1)^k}e\left(\sum_{j=1}^k\theta_jA_j(x,y)\right)\,d\mathcal L^k(\theta)=\prod_{j=1}^k\int_0^1 e(\theta_jA_j(x,y))\,d\mathcal L^1(\theta_j). \end{align*} By the one-dimensional identity, this product equals $1$ if $A_j(x,y)=0$ for every $1\le j\le k$, and equals $0$ otherwise. [/step]

[step:Identify the surviving tuples with the Vinogradov mean value count] Substituting the preceding evaluation into the finite sum gives \begin{align*} \int_{[0,1)^k}|S(\theta)|^{2s}\,d\mathcal L^k(\theta)=\#\{(x,y)\in A^{2s}:A_j(x,y)=0\text{ for every }1\le j\le k\}. \end{align*} By the definition of $A_j(x,y)$, the condition $A_j(x,y)=0$ is precisely \begin{align*} \sum_{i=1}^s x_i^j=\sum_{i=1}^s y_i^j. \end{align*} for every integer $j$ with $1\le j\le k$. Therefore the cardinality on the right-hand side is exactly $J_{s,k}(X)$, and hence \begin{align*} J_{s,k}(X)=\int_{[0,1)^k}\left|\sum_{n=1}^{N} e(\theta_1 n+\cdots+\theta_k n^k)\right|^{2s}\,d\mathcal L^k(\theta). \end{align*} This is the desired identity. [/step]