Proof of Theorem 1.9, Jun Shao' Mathematical Statistics I am trying to understand the proof of Theorem 1.9 of Jun Shao's Mathematical Statistics.
Overall I do not have big problems, but I am not able to understand a few minor points.
(I) (Green highlight) For part (i), Shao is trying to prove that, if $X, X_1, X_2, \cdots$ are random k-vectors, "$E[h(X_n)] \rightarrow E[h(X)]$ for every bounded continuous function h" implies "$\lim \sup_n P_{X_n}(C) \leq P_X(C)$ for any closed set $C \subset R^k$". In his proof, he let $f_C(x) = \inf\{\|x-y\|: y\in C\}$, where $C$ is a closed set. Then he defined $\varphi_j(t) = I_{(-\infty, 0]} + (1-jt)I_{(0, j^{-1}]}$ for $j = 1, 2, \cdots,$ and claim that $h_j(x) = \varphi_j(f_C(x)) \rightarrow I_C(x)$ as $j \rightarrow \infty$.
This is the part I do not understand. Why does $h_j(x) = \varphi_j(f_C(x)) \rightarrow I_C(x)$ as $j \rightarrow \infty$?
(II.1) (Blue highlight) Shao claimed that since $\phi_X$ is continuous at 0 and $\phi_X(0) = 1$, for any $\epsilon > 0$, there is a $u > 0$ such that $u^{-1} \int_{-u}^u[1-\phi_X(t)]dt < \epsilon/2$. Why is this?
(II.2) (Blue highlight) Following the preceeding question, he claimed that since $\phi_{X_n} \rightarrow \phi_X$, by the dominated convergence theorem, $\sup_n\{u^{-1} \int_{-u}^u[1-\phi_{X_n}(t)]dt\} < \epsilon$. Why is this $\epsilon$ instead of $\epsilon / 2$? And what exactly is the integrable function bounding it?
Thank you very much in advance!
Here's the full statements and the proofs of the theorem.


 A: (I) For this part, we can consider cases. In the first case, $x\in C$, so $I_C(x) = 1$, and $f_C(x) = 0$, so we need to show that $\varphi_j(f_C(x))=\varphi_j(0)\to 1$ as $j\to\infty$. By the definition of $\varphi_j$, it is clear that for every $j$, $\varphi_j(0) = 1$, so the claim holds in this case.
If $x\notin C$, then $I_C(x) = 0$, and we need to show that $\varphi_j(f_C(x))\to 0$. Since $C$ is closed and $x\notin C$, there is some $j_0\ge 1$ such that $f_C(x) > 1/j_0$. Hence for all $j\ge j_0$, $I_{(0,j^{-1}]}(f_C(x)) = 0$, and of course $I_{(-\infty,0]}(f_C(x)) = 0$. Therefore, $\varphi_j(f_C(x)) = 0$ for all $j\ge j_0$, so the claim holds in this case as well, which completes the argument for this part.
(You should try to draw the graphs of $\varphi_j(f_C(x))$ for the case when $C = [a,b]\subset\mathbb R$ to get some intuition for this construction.)
(II.1) By assumption $\phi_X$ is continuous at $0$, and $\phi_X(0) = 1$. Thus, $(1-\phi_X(t))\to 0$ as $t\to 0$. Therefore, let $\epsilon>0$ be given, and choose $\delta > 0$ so small that if $|t|<\delta$, $1-\phi_X(t) < \epsilon/4$. Then if $u\in(0,\delta)$,
$$
\frac{1}{u}\int_{-u}^u(1-\phi_X(t))\,dt < \frac{1}{u}\int_{-u}^u\epsilon/4\,dt = \epsilon/2.
$$
(Note that the $1/4$ is arbitrary, and in general a bound of the form $C\epsilon$ for any universal constant $C$ independent of $\epsilon$ would work just as well as long as the overall proof is consistent.)
(II.2) Fix $u$ so small that $\frac{1}{u}\int_{-u}^u(1-\phi_X(t))\,dt<\epsilon/2$ by (II.1). Since $\phi_{X_n}\to \phi_X$, and $|1-\phi_{X_n}|\le 2$, which is a bounded (and hence integrable) function on $[-u,u]$, by the bounded (or dominated) convergence theorem, there is $N\ge 1$ so that if $n\ge N$, we have $|\frac{1}{u}\int_{-u}^u(1-\phi_{X_n}(t))\,dt - \frac{1}{u}\int_{-u}^u(1-\phi_X(t))\,dt|<\epsilon/2$ for all $n\ge N$. By the triangle inequality, if $n\ge N$,
$$
\frac{1}{u}\int_{-u}^u(1-\phi_{X_n}(t))\,dt < \epsilon.
$$
To get $\sup_n\frac{1}{u}\int_{-u}^u(1-\phi_{X_n}(t))\,dt < \epsilon$ (i.e., to handle $n$ possibly less than $N$), we only need to choose $u$ small enough ahead of time so that
$$
\sup_{n\le N}\frac{1}{u}\int_{-u}^u(1-\phi_{X_n}(t))\,dt<\epsilon,
$$
which is possible since each of the $\phi_{X_n}$ are ch.f.'s (hence continuous at $0$), and this is a finite number of inequalities we need to satisfy with our choice of $u$.
