The one-sample Hotelling F test has exact size alpha

The statistic is Borel measurable: the sample maps $\bar{X}$ and $S$ are continuous by the model definition [F7]; by [F5], the event $S\succ 0$ is the finite intersection of strict polynomial inequalities on its leading principal minors. On this open set, $S^{-1}$ has entries equal to cofactors divided by $\det S>0$, so the quadratic-form formula for $T^2$ is continuous by [F6]. Extending that function by zero on the Borel complement remains Borel. Let $D(x)$ be the indicator that the scaled statistic exceeds the stated threshold and define a kernel on actions $\{0,1\}$ by $K(x,A)=1$ if $D(x)\in A$ and $0$ otherwise. For each $x$ this is a probability measure, and $x\mapsto K(x,A)$ is measurable because $D$ is measurable. Thus [F3] makes this a randomized procedure and [F2] its test function.

Since $p\ge 1$, ${\mathbb{R}}^p\setminus \{{\mu }_0\}$ is nonempty; hence the stated null and alternative parameter sets are nonempty and disjoint. Under AC [A1], [F4] gives a unique $q=F_{p,n-p}^{-1}(1-\alpha )>0$ satisfying $P(F_{p,n-p}\le q)=1-\alpha$. For each null covariance, [F1] gives rejection probability $P(F_{p,n-p}>q)=\alpha$, with no boundary atom because its CDF is continuous.

Taking the supremum of the same value $\alpha$ over the nonempty null set leaves it equal to $\alpha$. By [F2] this supremum is the test's size, proving exact size. The argument includes every positive-definite null covariance and uses the exact finite-sample F law; no asymptotic calibration is used.