diagonalization procedure for stochastic processes This is a question about a proof I saw in a script about stochastic processes. First I state a theorem which is needed in the proof. After that there are two questions, which are highlighted. Between the questions I provide my thoughts / work so far.

$\mathbf{Theorem}$ For any sequence $(X_n)$ of positive r.v. there exists a sequence $(\tilde{X}_n)$ with $\tilde{X}_n\in\operatorname{conv}(X_k;k\ge n)$ (the convex hull) such that $\tilde{X}_n\to X$ $P$-a.s. The r.v. variable $X$ can take values in $[0,\infty].$

This theorem is also known als Komlos theorem. The convex hull is the set $\operatorname{conv}(X_k;{k\ge n})=\{\sum_{k=n}^\infty \lambda_k X_k| \sum_{k=n}^\infty \lambda_k=1,\forall k\ge n:\lambda_k\ge 0,\lambda_k\not=0 \mbox{ for finitely many } \}$ and does not depend on $\omega$, i.e. it is deterministic.
Suppose I have a sequence of r.v. $g_n$ all positive which converge in probability to $g$. Furthermore, I know that for every $n$ there is a stochastic process $Z^n$, such that $g_n\le Z^n_T$. The stochastic processes $Z^n$ are positive, RCLL and indexed by $t\in[0,T]$.

$\mathbf{Question}$ They claim, using a diagonal argument one can find convex combinations $(\tilde{g}_n)$ and $(\tilde{Z}^n_q)$ for all rationals $q\in[0,T]$ converging $P$-a.s. to $g$ and $Z^\infty_q$ respectively. 

Is it meant, that we use the same sequence $\Lambda\subset\mathbb{N}$ for $(\tilde{g}_n)$ and $(\tilde{Z}^n)$? Clearly I can find such a sequence for $(g_n)$ using the theorem above. For $(Z^n)$ I would take a numeration of $\mathbb{Q}$, i.e. $q_1,q_2,\dots$ Then looking at $(Z^n_{q_1})$, we can find a $(\tilde{Z}^n_{q_1})$, where $n\in\Lambda_1$, such that $\tilde{Z}^n_{q_1}\to Z^\infty_{q_1}$. For $q_2$ I take a subsequence $\Lambda_2\subset\Lambda_1$ such that $\tilde{Z}^n_{q_2}\to Z^\infty_{q_2}$ and so on. Therefore a general $\tilde{Z}^n_{q_j} \in \operatorname{conv}(Z^k_{q_j};k\ge n)$, for $n\in \Lambda_j$, right? 
If it is meant that we use the same sequence $\Lambda$ for both $(\tilde{g}_n)$ and $(\tilde{Z}^n)$, then I would just start with $(g_n)$ and the go to $(Z^n_{q_1})$ etc. 

$\mathbf{Question}$ They state, we have $g_n\le Z^n_T$ and we use same convex combinations for $(\tilde{g}_n)$ and all $\tilde{Z}^n_q$, so $g\le Z^\infty_T$.

Why we are using the same convex combinations? If so, then then we can conclude:
$$\tilde{g}_n=\sum_{k=n}^\infty\lambda_kg_k\le\sum_{k=n}^\infty\lambda_kZ_T^k=\tilde{Z}^n_T$$
hence $g\le Z^\infty_T$. But what guarantees that we can use the same convex combinations, i.e. the same $\lambda_k$'s? I'm very thankful for your help!
 A: It looks like the argument is a slight twist on the standard diagonalization argument, because the text is talking about convex combinations and not subsequences. 
Your approach is pretty much spot on except that you take a subsequence $\Lambda_2\subset\Lambda_1$ instead of a sequence of convex combinations, which leads to the wrong conclusions. There's a couple of extra arguments you need to add for the convex combinations.
So you have to be a little bit careful constructing the sequence, once you do so it's obvious that you can use the "same convex combinations" 
of $g_n$ and $Z^n_t$.
I'll start the argument from the beginning but try to follow your notation.
First as $g_n$ converges in probability we may choose a subsequence $g_{n_k}$ that converges almost surely with $g_{n_k} \leq Z^{n_k}_t$. 
Therefore we may assume without loss of generality that $g_n$ converges almost surely.
Now consider a stochastic process $\hat Z_t\in \operatorname{conv}(Z_t^k:k\geq n)$ indexed by $t\in[0,T]$ with 
$$\hat Z_t = \sum_{k=n}^\infty \hat\lambda_k Z^k_t$$ 
Then by definition for every $s\in[0,T]$ we have $\hat Z_s = \sum_{k=n}^\infty \hat\lambda_k Z^k_s$. 
In particular for fixed $s\in[0,t]$ and every random variable $X\in\operatorname{conv}(Z_s^k:k\geq n)$ we may choose a stochastic process 
$\hat Z_t\in\operatorname{conv}(Z_t^k:k\geq n)$ with $\hat Z_s = X$ almost surely.
Now, 
order the rationals $\{q_i:i\in\mathbb N\}$ , 
I want to choose a sequence of stochastic processes $\tilde Z_t^n \in \operatorname{conv}(Z_t^k:k\geq n)$ and a stochastic process $Z^\infty_t$ such that for every $i \in\mathbb N$ we have    $\tilde Z_{q_i}^n \to Z_{q_i}^\infty$ almost surely.
First by Komlos theorem we may choose a sequence of random variables $X^1_n\in\operatorname{conv}(Z_{q_i}^k:k\geq n)$ and a random variable $Z^\infty_{q_1}$ such that $X^1_n\to Z^\infty_{q_1}$ almost surely.
Hence we may choose a sequence of stochastic processes $\tilde Z_t^{n,1} \in \operatorname{conv}(Z_t^k:k\geq n)$ such that $\tilde Z_{q_1}^{n,1} = X^1_n\to Z^\infty_{q_1}$ almost surely.
Now let us suppose inductively that I have constructed a sequence $\tilde Z_t^{n,i} \in \operatorname{conv}(Z_t^k:k\geq n)$ such that $\tilde Z_{q_j}^{n,i}\to Z^\infty_{q_j}$ as $n\to\infty$ almost surely for every $j=i$.
As $\tilde Z_{q_{i+1}}^{n,i}$ is a sequence of random variables 
 I may choose 
$X^{i+1}_n \in\operatorname{conv}(\tilde Z_{q_i+1}^{k,i}:k\geq n)$ such that
$X^{i+1}_n\to Z^\infty_{q_{i+1}}$  almost surely as $n\to\infty$ for some random variable $Z^\infty_{q_{i+1}}$ and a sequence of stochastic processes
$$\tilde Z_{t}^{n,i+1} \in \operatorname{conv}(\tilde Z_t^{k,i}:k\geq n)$$
Such that $\tilde Z_{q_{i+1}}^{n,i+1} = X^{i+1}_n\to Z^\infty_{q_{i+1}}$ as $n\to\infty$ almost surely. 
Now $\tilde Z_{t}^{n,i+1}$ is a convex combination of a finite number of elements of $\operatorname{conv}(Z_t^k:k\geq n)$. Hence it is itself a convex combination of a finite number of elements of $\{Z_t^k:k\geq n\}$. 
Therefore $\tilde Z_{t}^{n,i+1}\in \operatorname{conv}(Z_t^k:k\geq n)$ and it remains to show that $\tilde Z_{q_j}^{n,i+1}$ converges almost surely for $j\leq i+1$.
For every $\varepsilon>0$ $\mathbb P$ almost every $\omega$ and each $j\leq i$ I may choose $N(\omega)$ large enough that $\left| Z^\infty_{q_j}(\omega) -\tilde Z_{q_{j}}^{N+k,i}(\omega)\right| < \varepsilon$
hence $\left|Z_{q_j} - Z^\infty_{q_j}\right|$ for every $Z_t\in  \operatorname{conv}(\tilde Z_t^{k,i}:k\geq N)$.
In particular $\tilde Z_{q_j}^{n,i+1}\to Z^\infty$ for every $j\leq i+1$.
Now we may consider the diagonal sequence 
 $$\tilde Z_t^{n,n} = \sum_{k=n}^{\infty}  \lambda_{n,k} Z_t^n\in\operatorname{conv}(Z_t^k:k\geq n).$$
And set
$$\tilde g_n = \sum_{k=n}^{\infty}  \lambda_{n,k} g_{k}\in\operatorname{conv}(g_k:k\geq n).$$
So $\tilde g_n \leq \tilde Z_t^{n,n} $ and $\tilde g_n$ converges almost surely to $g$ so we have $g\leq Z^\infty_{t}$ for every rational $t$.
