Example 19.6 in van der Vaart: show that the class of indicator functions is P-Glivenko-Cantelli and P-Donsker I have some doubts related to example 19.6 in van der Vaart "Asymptotic Statistics" which applies Theorem 19.4 (Glivenko-Cantelli) and Theorem 19.5 (Donsker) to the distribution function.
Definitions: Consider a random variable $X:\Omega \rightarrow \mathcal{X}$ defined on the probability space $(\Omega, \mathcal{A}, \mathbb{P})$ with probability distribution $P$. All functions mentioned from now on will be random functions from $\mathcal{X}$ to $\mathbb{R}$. Consider two functions $l,u$. A bracket $[l,u]$ is the set of all functions $f$ with $l\leq f\leq u$. A $\epsilon$-bracket in $L_r(P)$ is a bracket with $\int_{\mathcal{X}}(u-x)^rdP<\epsilon^r$ with $0<\epsilon<\infty$ and $r>0$.  The $L_r(P)$ norm of a function $f$ is $(\int_{\mathcal{X}}|f|^rdP)^{\frac{1}{r}}$. Let $\mathcal{F}$ be a class of functions $f$. The bracketing number $N_{[\text{ }\text{ }]}(\epsilon, \mathcal{F}, L_r(P))$ is the minimum number of $\epsilon$-brackets in $L_r(P)$ to cover $\mathcal{F}$. The entropy is $\log(N_{[\text{ }\text{ }]}(\epsilon, \mathcal{F}, L_r(P)))$. The  bracketing integral $J_{[\text{ }\text{ }]}(1, \mathcal{F}, L_2(P)):=\int_{0}^1 \sqrt{log(N_{[\text{ }\text{ }]}(\epsilon, \mathcal{F}, L_2(P)))d\epsilon}$.
Setting:
(1) $\{X_i\}_{i=1}^N$ i.i.d. defined on the probability space $(\Omega, \mathcal{A}, \mathbb{P})$; $X_i:\Omega \rightarrow \mathbb{R}$; $P$ is the probability distribution function of $X_i$; $F$ is the cdf of $X_i$
(2) Take the the class of functions $\mathcal{F}:=\{f_t \text{ s.t. } f_t=1(X_i\leq t) \text{ and } t \in \mathbb{R}\}$, where $1(\cdot)$ is the indicator function taking value $1$ if the condition inside is satisfied and $0$ otherwise.
I want to show that 
(1)  $N_{[ \text{ }]}(\epsilon, \mathcal{F}, L_1(P))<\infty$ $\forall \epsilon>0$ 
(2) $J_{[ \text{ }]} (1, \mathcal{F}, L_2(P))<\infty$  $\leftrightarrow$ $\log N_{[ \text{ }]}(\epsilon, \mathcal{F}, L_1(P))=O(\frac{1}{\epsilon^2})$
Here a summary of the proof in the book with my questions:
(a) I believe that (2) implies (1). Hence the proof is focused on showing (2)
(b) Consider a finite partition of the extended real line in $-\infty=t_0<t_1<...<t_k=\infty$ such that $\lim_{t\rightarrow t_j^{-}} F(t)-F(t_{j-1})<\epsilon$ for $j=1,...,k$ [it can be shown that it exists]. Consider the brackets $[1(X_i \leq t_{j-1}), 1(X_i \leq t_j)]$ $j=1,...,k$
(c) These brackets have $L_1(F)$-size equal to $\epsilon$, i.e. $\int_{\mathbb{R}}1(x\leq  t_{j})- 1(x \leq t_{j-1})dF=F(t_j)-F(t_{j-1})<\epsilon$. Where does this come from? The condition $\lim_{t\rightarrow t_j^{-}} F(t)-F(t_{j-1})<\epsilon$ does not imply $F(t_j)-F(t_{j-1})<\epsilon$ unless $t_{j}$ is a continuity point of $F$
(d) Choose $k<\frac{2}{\epsilon}$. How do I know that this is possible?
(e) For any function $g(X_i)$ such that $0\leq g\leq 1$ (and hence in one of the brackets) $\int_{\mathbb{R}}g^2 dF\leq \int_{\mathbb{R}}gdF$
(f) Hence, the  $L_2(F)$-size of the brackets is bounded by $\sqrt{\epsilon}$. Why?
(g) Hence, $N_{[ \text{ }]}(\sqrt{\epsilon}, \mathcal{F}, L_2(P))\leq \frac{2}{\epsilon}$. Why? 
(h) Hence, $N_{[ \text{ }]}(\epsilon, \mathcal{F}, L_2(P))\in O(\frac{1}{\epsilon^2})$. Just by taking the square of (f)
(i) Hence $\log (N_{[ \text{ }]}(\epsilon, \mathcal{F}, L_2(P)))$ is of order smaller than the order of $\log(\frac{1}{\epsilon})$. Why not $O(\log(\frac{1}{\epsilon^2}))$ [just by taking the log of the square of (g)]? 
(l) This implies $\log (N_{[ \text{ }]}(\epsilon, \mathcal{F}, L_2(P)))\in  O(\frac{1}{\epsilon^2})$.  Why? Is that because $O(\log(\frac{1}{\epsilon^2})) \in O(\frac{1}{\epsilon^2})$?
This proof is obscure to me in several parts and any hint on the topic would be really appreciated. 
 A: For (c), since [$1(X_i\leq t_{j-1}$), $1(X_i\leq t_j)$] is an $\epsilon$-bracket in the $L_1(F)$ space, this means that
$$
\epsilon\geq\|1(X_i\leq t_{j})-1(X_i\leq t_j)\|_{L_1(F)}\geq\int1(x\leq t_{j})-1(x\leq t_j)dF(x)=F(t_j)-F(t_{j-1}).
$$
For (d), just a rough reasoning: Note that $\sup_{f\in\mathcal{F}}|f|=1$, you can think of brackets as partitioning the interval [0,1] into disjoint pieces and each piece have length $\epsilon$. Then the number of pieces should be ${1/\epsilon}$.
For (f), let $f\in L_2(F)$ with $0\leq f\leq 1$, then clearly $f^2\leq f$ so that
$$
\|f\|_{L_2(F)}^2=\int f^2dF\leq\int fdF=\|f\|_{L_1(F)}.
$$
Therefore, if $\|f\|_{L_1(F)}\leq\epsilon$, then $\|f\|_{L_2(F)}\leq\sqrt{\epsilon}$, so the $L_2(F)$-size of the brackets is bounded by $\sqrt{\epsilon}$ where $\epsilon$ is the size of the $L_1(F)$ brackets.
For (g), based on (f), $N_{[]}(\sqrt{\epsilon},\mathcal{F},L_2(F))\leq N_{[]}(\epsilon, \mathcal{F}, L_1(F))\leq\frac{2}{\epsilon}$. This is because an $L_1(F)$ $\epsilon$-size bracket is an $L_2(F)$ $\sqrt{\epsilon}$-size bracket and the number of  $L_1(F)$ $\epsilon$-size brackets needed to cover $\mathcal{F}$ is $N_{[]}(\epsilon,\mathcal{F},L_1(F))$. On the other hand, the definition of bracketing number implies that $N_{[]}(\sqrt{\epsilon},\mathcal{F},L_2(F))$ is the smallest number of $L_2(F)$ $\sqrt{\epsilon}$-size brackets needed to cover $\mathcal{F}$. Therefore, we can get
$$
N_{[]}(\sqrt{\epsilon},\mathcal{F},L_2(F))\leq N_{[]}(\epsilon, \mathcal{F}, L_1(F))
$$
For (i), $O(\log\frac{1}{\epsilon^2})=O(2\log\frac{1}{\epsilon})=O(\log\frac{1}{\epsilon})$. Of course, you can say the entropy number is of order $O(\log\frac{1}{\epsilon^2})$
For (j), I think your idea is reasonable.
Hope this helps.
