The split-sample univariate trimmed mean is total and Borel measurable

Since $0<\epsilon \le 1\mathrm{/}4$, both $\epsilon$ and $1-\epsilon$ lie in $(0,1]$. Thus $1\le r(\epsilon ),r(1-\epsilon )\le N$, and monotonicity of the ceiling gives $r(\epsilon )\le r(1-\epsilon )$. The corresponding sorted values therefore satisfy $\alpha \le \beta$, so every clipped value in [F1] is a finite real and the output is defined for every input, including ties.

For a scalar list $y_1,\dots ,y_N$, its $k$th order statistic equals ${\min }_{J\subseteq \{1,\dots ,N\},\mathrm{\mid }J\mathrm{\mid }=k}{\max }_{j\in J}{y}_j$: every $k$-subset has a maximum at least the $k$th smallest value, while the indices of the first $k$ sorted values attain it. This is a finite minimum of finite maxima of coordinate functions, hence is continuous in the list. The two cutoff maps in [F1] are therefore continuous; ${\varphi }_{\alpha ,\beta }(x)$ is continuous in $(\alpha ,\beta ,x)$ because it is a composition of minimum and maximum, and the final finite average is continuous. Hence the estimator is a continuous map on ${\mathbb{R}}^{2N}$.

For any Borel target set, its preimage under a continuous map is Borel: preimages of open sets are open, and the sets whose preimages are Borel form a sigma-algebra containing all open sets. By [F2] this is exactly Borel measurability, so the total estimator from step 1.1 is Borel.