Is the following proof valid?
preliminaries:
Define a step function as any function that can be written as:
$x = \sum_i^m \lambda_i f_i$
where $f_k = 1$ on $[a_k, b_k)$ and $0$ elsewhere (the “characteristic function” of that interval)
the proof
Let $f$ be a step function. If it has only one representation, say $\sum_i^n \lambda_i f_i$, then clearly the integral can take only one value:
$\int f = \sum_i^n \lambda_i (b_i - a_i)$
If there is not a unique representation, then any two representations $\sum_i^l a_i f_i$ and $\sum_i^m b_i f_i$ can be written as the basic representation, $\sum_i^n c_i f_i$.
So we have
$\sum_i^l a_i f_i = \sum_i^n c_i f_i = \sum_i^m b_i f_i$
And
$\int f = \int \sum_i^l a_i f_i = \int \sum_i^n c_i f_i = \int \sum_i^m b_i f_i$
The point of this exercise is to explain why defining the integral of a step function $f:=\sum_{i=1}^l a_i \chi_{A_i}$ with respect to measure $\mu$ as $$\int f\,d\mu:=\sum_{i=1}^l a_i \mu(A_i)\tag1$$ doesn't lead to inconsistency. If $f$ can be written also as $\sum_{j=1}^m b_j\chi_{B_j}$, where the $B_j$'s might differ from the $A_j$'s and the $b_j$'s might differ from the $a_i$'s, and maybe $m\ne l$, the definition would state that the integral of $f$ is now $\sum_{j=1}^m b_j\mu(B_j)$. How do we know that this second number is the same as (1)? The concept of integral is not well-defined if formula (1) yields different numbers for two functions that are equal (pointwise).
Your proof needs improvement because it has to consider the possibility that the second representation might not involve the $f_i$'s that appear in the first representation. As for the idea of a 'basic' representation, you need to explain why we can always find such a thing. And it remains to prove why the definition of the integral yields the same number for this basic representation as it does for the other two representations. It is not enough to say "these functions are equal, therefore their integrals are equal": this statement is what you are being asked to prove.