PG23 Source Clarifications

Cutoff probabilities and support arguments

Equal Cutoffs (including differential access): connected support → also zero probability of every exact score level. Connected support alone permits atoms: half point mass at zero plus half uniform mass on $[0,1]$ has $\Pr(Z=0)=1/2$ . This refutes only the inference to zero boundary mass.
Probability Formula and Theorems 1–2: add atomless noise to identify the source’s strict-tail formulas with its weak matching events: $\Pr(v+X\geq P)=\Pr(v+X>P).$ Without atomlessness, their difference is $\Pr(X=P-v)$ . The formalized Theorems 1–2 state the weak events and prove this identification.
Theorem 3: add atomless values to derive the equality of baseline and differential monoculture cutoffs from market clearing. Necessity for the source conclusion is unresolved; no counterexample is known here.
Corollary 4 and the strict Theorem 1–2 comparisons: add nondegenerate noise. The proof uses two distinct support points; necessity of the full regularity package for the source conclusions is unproved.
Unqualified appendix intervals/endpoints → nonempty open intervals meeting the support interior. A singleton has zero uniform mass, while an endpoint atom can make the lower-endpoint CDF positive. The interior argument also permits unbounded support.

Theorem 1: add absolute integrability of value and atomless values for the welfare limit and strict comparison. Necessity for the broader theorem is unproved. Theorem 2’s eventual-advantage branch also uses atomless values.
Differential-access Nash sentence → ex-ante expected payoffs over same-cardinality application sets, with ranking-monotone utility and the equal-cutoff success law. The current incentive result does not cover unrestricted application counts or different information timing.

Footnote means $1-n/(n+1)$ and $(n-1)/n$ → $\mathbb E[X^{(n)}]=n/(n+1)$ ; Chebyshev denominator $\varepsilon$ → $\varepsilon^2$ . Here $X^{(n)}$ is the maximum of $n$ iid uniform- $[0,1]$ draws. Its variance is $n/[(n+1)^2(n+2)]$ , so the corrected tail bound is $n/[\varepsilon^2(n+1)^2(n+2)]\to0$ for every $\varepsilon>0$ . The illustrative maximum-concentration conclusion follows; this is not a premise of the named results.