The sum of two elementary processes is an elementary process We work with respect to a filtered probability space $(\Omega,\mathcal F,\{\mathcal{F_t\}_{t\ge 0},P})$
An elementary process is of the form
\begin{equation} \label{eq:1}
\xi(t) = Z_01_{\{t=0\}}+\sum_{k=1}^n Z_k1_{\{s_k<t\le t_k\}}
\end{equation}
for ${n\ge 0}$, times $0 \le s_1 < t_1 \le s_2 < t_2 \le \dots \le s_n < t_n \,$, ${\mathcal{F}_0}$-measurable random variable ${Z_0}$ and ${\mathcal{F}_{s_k}}$-measurable random variables ${Z_k}$. 
If we take 
$$\xi_1(t) = Z_0 1_{\{t=0\}} +\sum\limits_{k=1}^{n_1} Z_k 1_{\{s_k < t \le t_k\}}$$ and $$\xi_2(t) = Y_01_{\{t=0\}}+\sum_{k=1}^{n_2} Y_k1_{\{\tilde{s}_k<t\le \tilde{t}_k\}}$$
two elementary processes, how to show that $\xi_1(t) + \xi_2(t)$ is also elementary ? It is really a question of notation, I don't find a good rigorous way of writing it.

A first observation is that the part with the indicator of $\left\{0\right\}$ is not a problem as we choose $W_0:=Y_0+Z_0$. Therefore we will assume that $Y_0=Z_0=0$.
Define the intervals $I_{k,1}:=\left(s_k,t_k\right]$, $I_{k,2}:=\left(\bar{s_k},\bar{t_k}\right]$. Let $J_{k_1,k_2}:=I_{k_1,1}\cap I_{k_2,2}$, where $1\leqslant k_1\leqslant n_1,1\leqslant k_2\leqslant n_2$. Then $\left(J_{k_1,k_2}\right)_{\substack{1\leqslant k_1\leqslant n_1,\\1\leqslant k_2\leqslant n_2}}$ is a collection of disjoint interval (some can be empty). Therefore, 
\begin{equation}
\xi_1(t)+\xi_2(t)=\sum_{k_1=1}^{n_1}\sum_{k_2=1}^{n_2}W_{k_1,k_2}\mathbf 1_{J_{k_1,k_2}}\left(t\right)
\end{equation}
where $W_{k_1,k_2}=Z_{k_1}+Y_{k_2}$ if $J_{k_1,k_2}$ is non-empty and $W_{k_1,k_2}=0$ otherwise. 
Since $Z_{k_1}$ is $\mathcal F_{s_{k_1}}$-measurable and $J_{k_1,k_2}$ has the form $\left(a_{k_1,k_2},b_{k_1,k_2}\right]$ with $a_{k_1,k_2}=\max\left\{ s_{k_1},\bar{s_{k_2}}\right\}$, $Z_{k_1}$ is also $\mathcal F_{a_{k_1,k_2}}$-measurable. A similar argument shows that $Y_{k_2}$ is $\mathcal F_{a_{k_1,k_2}}$-measurable hence so is $W_{k_1,k_2}$.
