On proving that the integral of a simple function does not depend on the chosen form of the simple function. EDIT: to attract an answerer I simplify the question by asking for a proof of the "easily seen fact" stated in the image below

.
I leave the original question for the sake of completeness:
As a definition of simple function I am using: a simple function is a finite linear combination of indicator functions of measurable sets.
I am working with the integral of a simple function from a measurable set $A$ to $[0, +\infty[$ defined as follows: Let $g$ be a simple function with $Img(g) = \{ \lambda_1, \dots, \lambda_m \}$ the integral is then
$$\int_A f(x) dx = \sum_{k =1}^m \lambda_m u(\{x \in A | f(x) = \lambda_k  \})$$
where $u$ is the Lebesgue measure.
I would like to prove that given a measurable set $A \subset R^n$ and a simple function $f:A \rightarrow [0, \infty[$ If there exist $\lambda_1, \lambda_2 \dots \lambda_p \in \ [0, \infty[, \quad A_1, \dots A_p$ measurable subsets of $A$ s.t.
$$f = \sum_{k = 1}^p \lambda_k \chi_{A_k}$$ then
$$\int_A f(x) dx = \sum_{k = 1}^p \lambda_k u({A_k})$$
I tried to approach proving this by induction but I am having troubles rigorously proving the obvious, even in the base case when I have two generic simple functions
$$ V \chi_{A_k} = \lambda \chi_{B_k}  $$
I am having troubles being rigorous in showing that their integrals are equal. Could someone show me in detail how it's done?
 A: I came here for some critique for my proof and found this question so...
The general idea is defining the integral as the sum on the canonical representation and, given another representation, finding it's can. rep. and applying the formula, and then we must get to the same number in the end, somehow.
Definition: $ (X,\Sigma,\mu) $ a measurable space and $ f:X \rightarrow \mathbb{K} \in \{\mathbb{R} ,\mathbb{C} \}$, say $ f $ is simple iff $ f(X) $ is finite and $ \forall a \in f(X) \left(f^{-1}(\{a\}) \in \Sigma \right) $.
Definition: The canonical representation of a simple $ f $ is 
$$ 
f = \sum_{a \in f(X)}a \chi_{f^{-1}(\{a\})} 
$$ Every $ f $ have a can. rep. and it is unique, assuming non-existence we lose simplicity. Assuming non-uniqueness we lose $ f $ being a function.
Definition: Let $ f $ be simple, the integral of $ f $ relative to $ \mu $, given $ f = \sum a_{k}\chi_{A_{k}} $ can. rep., is 
$$
 \int f d\mu \doteq \sum a_{k}\mu(A_{k}) 
$$
Proposition: The same formula may be used for non-canonical representations. Hence, $ \int f d\mu $ is well defined since all of the calculations give the same number of the formula aplied to the can. rep.
$$
f = \sum_{k = 1}^{m} b_{k}\chi_{B_{k}} \implies \int f d\mu = \sum_{k = 1}^{m} b_{k}\mu({B_{k}})
$$
Let $ \mathcal{I} = \{1,...,m\} $, $ I \doteq \mathcal{P}(\mathcal{I})\backslash\{\emptyset\} $.
Let $ \alpha \in I $, define:
$$
C_{\alpha} = 
\left(
\bigcap_{i \in \alpha} B_{i}
\right)
\cap 
\left( 
\bigcap_{j \in \mathcal{I}\backslash \alpha} X\backslash B_{j} 
\right) 
$$
in fact $ \alpha \neq \beta \implies C_{\alpha} \cap C_{\beta} = \emptyset $.
Let $ \sim $ equivalence relation over $ I $ defined as $ \alpha \sim \beta \iff \sum_{k \in \alpha} b_{k} = \sum_{k \in \beta}b_{k} $. Let $ a_{\bar{\alpha}} \doteq \sum_{k \in \alpha}b_{k} $ well defined in $ I/\sim $, let $ A_{\bar{\alpha}} \doteq \sqcup_{\gamma \in \bar{\alpha}}C_{\gamma} $. 
Statement: the can. rep. of $ f $ is:
$$
f = \sum_{\bar{\alpha} \in I/\sim}a_{\bar{\alpha}}\chi_{A_{\bar{\alpha}}}
$$
hence:
$$
\int f d\mu = \sum_{\bar{\alpha} \in I/\sim}a_{\bar{\alpha}}\mu(A_{\bar{\alpha}})
$$
but $ a_{\bar{\alpha}}\mu(A_{\bar{\alpha}}) = a_{\bar{\alpha}}\sum_{\gamma \in \bar{\alpha}} \mu(C_{\gamma})  = \sum_{\gamma \in \bar{\alpha}}a_{\bar{\alpha}}\mu(C_{\gamma}) = \sum_{\gamma \in \bar{\alpha}}a_{\bar{\gamma}}\mu(C_{\gamma}) = \sum_{\gamma \in \bar{\alpha}}(\sum_{i \in \gamma}b_{i})\mu(C_\gamma)$, thus:
$$
\int f d\mu = \sum_{\bar{\alpha} \in I/\sim}\sum_{\gamma \in \bar{\alpha}}\sum_{i \in \gamma}b_{i}\mu(C_\gamma)
$$
on the other hand:
$$
B_{k} = \bigsqcup_{\gamma \in I(k \in \gamma)}C_{\gamma} \implies \mu(B_{k}) = \sum_{\gamma \in I(k \in \gamma)} \mu(C_{\gamma})
$$
$$
 \sum_{i = 1}^{m}b_{i}m(B_{i}) = \sum_{i = 1}^{m}\sum_{\gamma \in I(i \in \gamma)} b_{i} \mu(C_{\gamma}) = \sum\{b_{i}\mu(C_{\gamma}):i \in \{1,...,m\} \land \gamma \in I \land i \in \gamma \} = 
$$
$$
 = \sum_{\gamma \in I} (\sum_{i \in \gamma} b_{i}) C_{\gamma} = \sum_{\bar{\alpha} \in I/\sim}\sum_{\gamma \in \bar{\alpha}}\sum_{i \in \gamma} b_{i}C_{\gamma} = \int f d_{\mu \quad \square}
$$
