Uniform convergence of the empirical right-censoring risk set

Put $V_i=-U_i$, let $F(x)=P(V_i\le x)$, and let $F_n(x)=n^{-1}{\sum }_{i=1}^n\mathbf{1}\{V_i\le x\}$. Then ${\bar{Y}}_n(u)=F_n(-u)$ and $s_0(u)=F(-u)$. Both CDFs are right-continuous, so their supremum error over $\mathbb{R}$ is the supremum over rational $x$; over $[-t,0]$ it is the supremum over $(\mathbb{Q}\cap [-t,0))\cup \{0\}$ (with $\{0\}$ alone when $t=0$). Thus both suprema are measurable. Fix $0<\epsilon <1$. Choose the finitely many distinct quantiles $x_k=\inf \{x:F(x)\ge k\epsilon \}$ for $1\le k<1\mathrm{/}\epsilon$, together with $x_0=-\mathrm{\infty }$ and $x_{m+1}=+\mathrm{\infty }$, where $m$ is the number of distinct finite cutpoints. Between consecutive distinct quantiles $x_j<x_{j+1}$, the quantile construction gives $F(x_{j+1}-)-F(x_j)\le \epsilon$; any jump at a cutpoint is isolated as its singleton point. At each finite cutpoint, both $F_n(x_j)$ and $F_n(x_j-)$ converge in probability to their expectations by Chebyshev's inequality applied to the IID indicators $\mathbf{1}\{V_i\le x_j\}$ and $\mathbf{1}\{V_i<x_j\}$. For $x\in (x_j,x_{j+1})$, monotonicity gives $F_n(x_j)\le F_n(x)\le F_n(x_{j+1}-)$ and $F(x_j)\le F(x)\le F(x_{j+1}-)$; hence $\mathrm{\mid }{F}_n(x)-F(x)\mathrm{\mid }$ is at most the largest one-sided endpoint error plus $\epsilon$. At singleton cutpoints, direct endpoint convergence applies. Taking the maximum over the finite partition, then $n\to \mathrm{\infty }$ and $\epsilon \downarrow 0$, proves the first convergence.

Let $Z_n={\sup }_{0\le u\le t}\mathrm{\mid }{\bar{Y}}_n(u)-s_0(u)\mathrm{\mid }$. Since $Z_n\le 1$, for every $\eta >0$, $E[Z_n^2]\le {\eta }^2+P(Z_n>\eta )$. Step 1.1 makes the second term tend to zero; then letting $\eta \downarrow 0$ proves the expected-square convergence.