Fisher sharp-null randomization p-values are valid

The schedule with $Y_i(z)=0$ for every unit and assignment belongs to the Fisher sharp null, so its null class is nonempty as required by [F4]. Fix any sharp-null schedule $\theta$ and choose its common response vector $y$ by [F2]. For every realized assignment $Z$, the observed vector $Y_{\theta }(Z)$ equals $y$, and every assignment score $T(z,y)$ is therefore imputable.

For each product-measurable $E$, its section $E_x:=\{u:(x,u)\in E\}$ is Borel, since the sets with Borel sections form a sigma-algebra containing all measurable rectangles. Thus the displayed finite sum defines a measure: for disjoint $E_n$, each section $({\bigcup }_n{E}_n{)}_{O_{\theta }(z)}$ is the disjoint union of the sections $(E_n{)}_{O_{\theta }(z)}$, and countable additivity of $\lambda$ passes through the finite weighted sum. Its total mass is ${\sum }_{z}p(z)\lambda ([0,1])=1$. Moreover, for $A\subseteq \mathcal{X}$ and Borel $B\subseteq [0,1]$, ${\tilde{P}}_{\theta }(A\times B)={\sum }_{z:O_{\theta }(z)\in A}p(z)\lambda (B)=P_{\theta }(A)\lambda (B)\mathrm{.}$ Hence the data coordinate has law $P_{\theta }$, $U$ is uniform and independent of the data, and therefore $U$ is independent of $Z$.

The formula $p_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(z,y)$ is a finite sum of selected nonnegative assignment weights, so it lies in $[0,1]$. It is a statistic because the data space has its full power-set sigma-algebra, so every preimage is measurable by [F5]. On the extension, each score-comparison condition defines a measurable subset of the data space, while $U$ is measurable; hence the finite formula for $p^{+}$ is measurable. Its value lies between zero and the total assignment mass one.

On this fixed schedule set $T_y(z)=T(z,y)$. Substituting $Y_{\theta }(Z)=y$ from step 1.1 into the two displayed formulas identifies $p_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(Z,Y_{\theta }(Z))$ and $p^{+}(Z,Y_{\theta }(Z),U)$ with the two finite weighted rank transforms of [F3] on $(\mathcal{Z},p)$. By steps 1.2 and 2.1, $U$ is independent of $Z$ and uniform, so the tie-randomized transform also satisfies [F3]. Hence $P_{\theta }(p_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(Z,Y_{\theta }(Z))\le u)\le u$ and ${\tilde{P}}_{\theta }(p^{+}(Z,Y_{\theta }(Z),U)\le u)=u$ for every $u\in [0,1]$.

Since the bound holds for every schedule in the sharp null, [F4] makes $p_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}$ a valid p-value for that null. The randomized value is uniform under each extended law ${\tilde{P}}_{\theta }$, and therefore is also superuniform there; its threshold rule has null rejection probability at most its threshold. No step uses an error distribution. By [F2] and [F6], the argument covers a weak average-effect schedule only when it also satisfies the sharp null.