I don't understand a particular step in many proofs showing the linearity of the Lebesgue integral of simple functions. Consider the canonical decomposition of a simple function $\phi = \sum_{j=1}^{N}c_jX_{E_j}$ where the $E_j$ are measurable, disjoint subsets of $\mathbb{R^d}$ with finite measure. Then, consider the canonical decomposition of a simple function $\psi = \sum_{i=1}^M r_{i}X_{G_i}$ where the $G_i$ are measurable, disjoint subsets of $\mathbb{R^d}$ with finite measure. For all $a,b \in \mathbb{R}$, it must be shown that $$\int (a\phi + b\psi) = a\int \phi + b\int \psi$$
What I have so far is that
$a\phi + b\psi = \sum_{j=1}^{N}ac_jX_{E_j} + \sum_{i=1}^M br_{i}X_{G_i}$
However, many proofs then make the "clear", "trivial", or "obvious" claim that
$$\sum_{j=1}^{N}ac_jX_{E_j} + \sum_{i=1}^M br_{j}X_{G_i} = \sum_{j=1}^N \sum_{i=1}^M (ac_j + br_{i})X_{G_i \cap E_j}$$
However, I'm having trouble seeing this. The only way I could see justifying such a step is if $$X_{G_i} = \sum_{j=1}^N X_{G_i \cap E_j} \space \text{and} \space X_{E_j} = \sum_{i=1}^M X_{E_j \cap G_i}$$
but I'm not seeing why either must necessarily be true.
For instance if $G_i = \{[1,1,\ldots,1]\}$ and the $E_j$ were non-empty sets not containing $[1,1,\ldots,1]$, then $E_j \cap G_i$ is empty for every $j$. Thus the equality on the left above would give $X_{G_i}(x) = 0 \space \forall x \in \mathbb{R^d}$, but clearly $X_{G_i} ([1,1,\ldots,1]) = 1$. So, what am I missing here? How is the above highlighted equality justified?
This is a bit of set manipulations: Let $X$ be a set. Suppose $\{G_i\}^n_{i=1}$ and $\{E_i\}^m_{i=1}$ are collections of $\textit{disjoint}$ sets, each of whose union is $X$. Then
$\tag 1 G_i=G_i\cap \bigcup^m_{j=1}E_j=\bigcup^m_{j=1}G_i\cap E_j\ \text{so}$
$\tag 2 \sum^n_{i=1}r_i\chi_{G_i}=\sum^m_{j=1}\sum^n_{i=1}r_i\chi_{G_i\cap E_j}.\ \text{Similarly,}$
$\tag 3 E_j=E_j\cap \bigcup^n_{i=1}G_i=\bigcup^n_{i=1}E_j\cap G_i\ \text{so}$
$\tag 4 \sum^m_{j=1}c_j\chi_{E_j}=\sum^n_{i=1}\sum^m_{j=1}c_j\chi_{E_j\cap G_i}=\sum^m_{j=1}\sum^n_{i=1}c_j\chi_{G_i\cap E_j}.$
To finish, add $(2)$ and $(4)$ and simplify.