Approximation of stochastic processes in Protter I'm reading Stochast integration and stochastic differential equation by Protter. In particular I have a question about Theorem 10 in chapter 2.4. Here Protter defines a simple predictable processes $H$ if it has a representation
$$H_t=H_0\mathbf1_{\{0\}}+\sum_{i=1}^nH_i\mathbf1_{(T_i,T_{i+1}]}(t)$$
where $0=T_1\le \dots\le T_n<\infty$ are stopping times and $H_i\in\mathcal{F}_{T_i}$ with $|H_i|<\infty$ a.s. The collection of all such processes is denoted with S. In Theorem 10 he wants to prove that S is dense in the space of all adapted caglad processes, denoted with $\mathbb{L}$, w.r.t. ucp convergence.
The goal is therefore to approximate $Y\in\mathbb{L}$ with a sequence in S. First one can reduce the claim to $Y$ which are additionally bounded. He then looks at the right continuous "version" of $Y$, i.e.
$$Z:=\lim_{u\downarrow t}Y_u$$
Clearly this is an adapted and cadlag process. Now the key point of the proof is to define the following sequence of stopping times for $\epsilon>0$:
$$T^\epsilon_0:=0$$
$$T^\epsilon_{n+1}:=\inf\{t:t>T^\epsilon_n \text{ and }|Z_t-Z_{T^\epsilon_n}|>\epsilon\}$$
Moreover $$Z^\epsilon:=\sum_n Z_{T^\epsilon_n}\mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})}$$
Protter says $Z^\epsilon$ converges uniformly to $Z$ as $\epsilon\to 0$. In what sense does he mean uniformly and how is it established?
Assuming this he defines the left continuous process $$ U^\epsilon=Y_0\mathbf1_{\{0\}}\sum_n Z_{T^\epsilon_n}\mathbf1_{(T^\epsilon_n,T^\epsilon_{n+1}]}$$
How does the proceeding implies $U^\epsilon\to Y_0\mathbf1_{\{0\}}+Z_{-}=Y$ in the ucp sense?
 A: First question:

Protter says $Z^\epsilon$ converges uniformly to $Z$ as $\epsilon\to 0$. In what sense does he mean uniformly and how is it established?

$$Z^\epsilon=\sum_n Z_{T^\epsilon_n}\mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})}
\\
|Z^\epsilon - Z|
=\sum_n |Z_{T^\epsilon_n}-Z|
\mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})}$$
now using the definition of $T^\epsilon_{n+1}$, 
you get 
$$
|Z^\epsilon-Z|
\mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})} \le
\epsilon\mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})}
$$
and then, the convergence is uniform in the strong sense:
$$
\sup_{t>0} |Z_t^\epsilon -Z_t| \le \epsilon
$$
Second question:
$Z,Y$ only differ on their (random) discontinuities. 
Let us fix $T$, and work on $[0,T]$. Because of the uniform convergence of
$Z^\epsilon$ to $Z$, and because every discontinuity of $Y$, big enough to 
kill the inequality
$$
\sup_{0\le t\le T} |Z_\epsilon-Z| \le \epsilon
$$
is one of the $T^\epsilon_n$ on which the value has been changed from
$Z^\epsilon$ to $U^\epsilon$,
then the convergence $U^\epsilon\to Y$
is uniform on everysubset $[0,T]$.
Actually it seems to me that the convergence is stronger than the convergence stated, so there is maybe something I am missing.
