Normal sample mean and residuals are independent

The mean map is continuous, hence Borel by [F2] and a statistic. Use $n$-dimensional Lebesgue measure $dx$ as base measure; it is sigma-finite by [F3]. By [F1] and the translation case of [F10], each coordinate has density $\varphi (x_i-\mu )$. Independence [F4] gives the product joint law. For $n=1$ this is already the joint density. For $n\ge 2$, repeated use of [F13] identifies ${\lambda }_n$ with the product of the one-dimensional Lebesgue measures. The measure with density $q(x)={\prod }_i\varphi (x_i-\mu )$ relative to ${\lambda }_n$ exists by [F11]. Repeated Tonelli [F4] and the rectangle formula [F12] show that its values on Borel rectangles are the products of the marginal probabilities; [F14] says these rectangles generate the Borel sigma-algebra of ${\mathbb{R}}^n$, so repeated product-measure uniqueness [F12] identifies it with the joint law. Thus $p_{\mu }(x)=(2\pi {)}^{-n\mathrm{/}2}\exp (-\frac{1}{2}{\sum }_i{x}_i^2+\mu {\sum }_i{x}_i-\frac{n}{2}{\mu }^2)$. Setting $T=\bar{X}$, $\eta =n\mu$, and $h(x)=(2\pi {)}^{-n\mathrm{/}2}{e}^{-{\sum }_i{x}_i^2\mathrm{/}2}$ writes it as $h(x)e^{\eta T(x)-A(\eta )}$ with $A(\eta )={\eta }^2\mathrm{/}(2n)$. Completing the square and using Tonelli gives $\int h{e}^{\eta T}dx=e^{{\eta }^2\mathrm{/}(2n)}$. The natural parameter set is all $\mathbb{R}$, hence open. For $b\mathrm{\ne }0$, the level set $b\bar{x}=c$ is a proper affine hyperplane, null by [F3], so $T$ has no affine constraint; it is full rank by [F5]. Thus [F6] and [F7] make $\bar{X}$ sufficient and complete.

The residual map $x\mapsto x-\bar{x}\mathbf{1}$ is continuous, hence a statistic by [F2]. Write $X_i=\mu +Z_i$ with iid $Z_i\sim N(0,1)$ by [F1]. Then $R_i=Z_i-\bar{Z}$, so the law of $R$ does not depend on $\mu$; it is ancillary by [F8]. Apply [F9] to conclude $\bar{X}$ and $R$ are independent under every $P_{\mu }$.