Consider a real-valued random variables $X$ defined on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with cumulative distribution function $F(t):=\mathbb{P}(X\leq t)$.
I want to show that $\forall \epsilon>0$ there exists a partition $-\infty=t_0<t_1<...<t_k=\infty$ with $k<\infty$ such that $\lim_{t\rightarrow t_j^{-}} F(t)-F(t_{j-1})<\epsilon$ for $j=1,...,k$
[this is the first step in the proof of Glivenko-Cantelli Theorem in van der Vaart "Asymptotics Statistics"]
My attempt with questions:
(1) Fix $\bar{\epsilon}>0$
(2) Consider the very last continuous steps of $F$ approaching $0$ and $1$. Suppose they correspond to $(a,-\infty)$ and $[b,\infty)$ with $a<b$. By the intermediate value theorem, for there exists $t_1 \in (-\infty,a)$ such that $F(t_1)<\bar{\epsilon}$. By the intermediate value theorem, for there exists $t_{k-1} \in (b, \infty)$ such that $F(t_1)>1-\bar{\epsilon}$.
(3) Consider the interval $[b,t_{k-1}]$. It is possible to cut this interval in a finite number of points $b<t_p<t_{p+1}<...<t_{k-1}$ such that $F$ increases by less then $\bar{\epsilon}$ at each point. Which result I'm using here? How do I know that the number of points is finite?
(4) Repeat (3) for the interval $[t_1,a]$.
(5) Repeat (3) for any step of $F$.
Is this construction correct?
Since $F$ is a bounded, non decreasing function, and $F(t) \in [0,1]$ for all $t$. You need to choose $t_1,...,t_{k-1}$.
Pick $n$ such that ${1 \over n} < {1 \over 2}\epsilon$. Partition the range $[0,1]$ as $(0,{1 \over n},{2 \over n},..., 1-{1 \over n}, 1)$.
Let $J_i = [{i-1 \over n}, {i \over n})$, for $i=1,...,n-1$, $J_n = [1-{1 \over n},1]$ and $H_i = F^{-1}(J_i)$. Note that the $H_i$ form a partition of $\mathbb{R}$. If $H_i$ is non empty, let $t_i = \sup H_i$.
Note that $t_i \notin H_i$, since if $t_i \in H_i$, then $F(t_i) \in J_i = [{i-1 \over n}, {i \over n})$ and so $F(t_i ) < {i \over n}$. Since $F$ is right continuous, there is some $t' > t_i$ such that $F(t_i ) \le F(t') < {i \over n}$, which is a contradiction.
Note that $H_1, H_k$ are non empty since $\lim_{t \to -\infty}F(t) = 0$ and $\lim_{t \to \infty}F(t) = 1$.
Let $i_j$, $j=1,...,m$ be the indices for which $H_{i_j}$ is non empty, note that $i_1 = 1,i_m =k $. Let $i_0 = 0$. Then I claim that $t_{i_j}$, $j=0,...,m$ form the desired partition.
Note that $F(t_{i_1}) < {1 \over n}$, hence $\lim_{t \uparrow t_{i_1}} (F(t)-F(t_{i_0})) = 0$.
Now choose $j\ge 2$. If the interval $H_{i_j -1}$ is non empty, then we have $t_{i_j} - t_{i_{j-1}} \le {2 \over n} < \epsilon$. If not, then the intervals $H_{i_{j -1}+1},...,H_{i_{j }-1} $ are empty. Since $t_{i_{j-1}} \notin H_{i_{j-1}}$, we must have $t_{i_{j-1}} \in H_{i_{j }}$ in which case we have $F(t_{i_j}) - F(t_{i_{j-1}}) \le {1 \over n} < \epsilon$.