A simple function can be defined as $s=\sum_i^na_iA_i$, $x\in X$, where $n$ is a positive integer, $a_1,a_2,...,a_n$ are extended real numbers and $A_i$ $\subseteq X$ for every $i$.
We have the following proposition:
Let it be $s_1$ and $s_2$ two simple measurable functions, and $(X,\mathcal{S})$ a measurable space. Then the maximum function $(s_1\vee s_2)=max\{s_1(x),s_2(x)\},x\in X$ should also be a measurable function.
There's a demonstration of this that I found here on the minute 44 of the video.
He says that
$$max\{s_1(x),s_2(x)\}=\sum_i\sum_jmax\{a_i,b_j\}\chi_{\left(A_i\bigcap B_j\right)}$$
But the maximum sum value is different of the sum of the maximum values. So is this method correct?
Another doubt that I have is on minute 47 when he tries to demonstrate that $\left |s \right|$ is measurable if $s$ is a measurable function.
He follows a similiar method as on the previous demonstration when he says that:
$$\left |s\right|=\sum_i^n \left|a_i\right| \chi_{A_i}$$
The thing is that the absolute value of a sum is different of a sum of absolute values. So is $|s|$ a different operation that is only relative to the elements of the sum?
It's important to assume that (WLOG) $A_i$'s are mutually exclusive and also that $B_j$'s are mutually exclusive.
$\bigcup\limits_{i=1}^{M} A_{i} = X$, $\bigcap\limits_{i=1}^{M} A_{i} = \phi$
$\bigcup\limits_{i=1}^{N} B_{i} = X$, $\bigcap\limits_{i=1}^{N} B_{i} = \phi$
Now, let $x \in X$ which means that $x$ lies only in one of $A_i$'s and one of $B_j$'s. Say, $x \in A_i \cap B_j$
$max\{s_1(x),s_2(x)\} = max\{a_i,b_j\} $
Hence, $$max\{s_1(x),s_2(x)\}=\sum_i\sum_jmax\{a_i,b_j\}\chi_{\left(A_i\bigcap B_j\right)}$$
Along the similar lines, its easy to prove the other identity. Hope it helps!