Integral of simple functions in standard and non-standard representation Some definitions
Let $(X,\mathbb X,\mu)$ be a measure space. 
A real-valued function is simple if it has only a finite number of values. A simple $\mathbb X$-measurable function $\varphi$ can be represented in the form
  $$\varphi=\sum_{j=1}^n a_n\chi_{E_j}$$
where $a_j\in\mathbb R$ and $\chi_{E_j}$ is the characteristic function of a set $E_j\in\mathbb X$. If we add the restriction that the $a_j$ be distinct and the $E_j$ form a partition of X, then the representation is unique and is called the standard representation of $\varphi$.
If $\varphi$ is a simple function in $M^+(X,\mathbb X)$ with the standard representation above, we define the integral of $\varphi$ with respect to $\mu$ to be the extended real number
  $$\int\varphi\,d\mu=\sum_{j=1}^n a_j\mu(E_j)$$
My question
If the simple function $\varphi\in M^+(X,\mathbb X)$ has the (not necessarily standard) representation
  $$\varphi=\sum_{k=1}^m b_k\chi_{F_k}$$
where $b_k\in\mathbb R$ and $F_k\in\mathbb X$, it can be shown that
  $$\int\varphi\,d\mu = \sum_{k=1}^m b_k\,\mu(F_k).$$
My problem is that I cannot find a clean yet rigorous step-by-step proof of that result.
My idea is to rewrite the function $\varphi$ as $\sum_{k=1}^n a_k\chi_{\phi^{-1}(a_k)}$ where $a_k$ is the sum of some $b_k$ terms and $\phi^{-1}(a_k)$ is the union of intersections of some $F_k$ terms. After some manipulations I note that I can put back together all the "pieces" and find $\sum_{k=1}^m b_k\,\mu(F_k)$. Unfortunately, that very last passage is left to the reader. Is there a way to make all the process explicit and yet clean and easy to follow?
 A: Suppose $m$ is finite. Using the linearity of the integral, we have
$$
  \int \left(\sum_{k=1}^m b_k1_{F_k}(x)\right)\mu(\mathrm dx) = \sum_{k=1}^m
b_k\int1_{F_k}(x)\mu(\mathrm dx) = \sum_{k=1}^mb_k\mu(F_k)
$$
regardless of the shape of the collection $F_k$. 
If you don't want to use linearity, note that given some finite collection $\{F_k\}_{k=1}^m$ of measurable sets there is a unique coarsest partition $\mathscr G = \{G_i\}_{i=1}^n$ such that $F_k$ are unions of some elements in $\mathscr G$. Let $g:\{1,\dots,n\}\to2^{\{1,\dots,m\}}$ be the index function uniquely defined by
$$
  k\in g(i)\quad\Leftrightarrow\quad G_i\subset F_k
$$
for any $k\in\{1,\dots,m\}$ and any $i\in \{1,\dots,n\}$. Furthermore, note that for any $k\in \{1,\dots,m\}$ the inverse of $g$ satisfies
$$
  \{i:k\in g(i)\} = \{i:G_i\subset F_k\}
$$
is partition of $F_k$ in $\mathscr G$. In particular, $\mu(F_k) = \sum_{i:k\in g(i)}\mu(G_i).$ Then we have:
$$
  \varphi(x) = \sum_{k=1}^m b_k1_{F_k}(x) = \sum_{i=1}^n\left(\sum_{k\in g(i)}b_k\right)1_{G_i}(x)
$$
where the first equality is the definition of $\varphi$ and the latter function is standard simple one. Thus
$$
  \int\varphi\;\mathrm d\mu = \sum_{i=1}^n\left(\sum_{k\in g(i)}b_k\right)\mu(G_i) = \sum_{k=1}^n b_k\left(\sum_{i:k\in g(i)}\mu(G_i)\right) = \sum_{k=1}^n b_k\mu(F_k)
$$
where we passed from the summation over $G_i$ to the summation over $b_k$.
A: A somewhat more constructive proof.
$$
f = \sum_{k = 1}^{m} b_{k}\chi_{B_{k}} \implies \int f d\mu = \sum_{k = 1}^{m} b_{k}\mu({B_{k}})
$$
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} $. Seja $ 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:
$$
B_{k} = \bigsqcup_{\gamma \in I(k \in \gamma)}C_{\gamma} \implies \mu(B_{k}) = \sum_{\gamma \in I(k \in \gamma)} \mu(C_{\gamma})
$$
$$
 \sum_{i = 1}^{m}b_{i}m(B_{i}) = \sum_{i = 1}^{m}\sum_{\gamma \in I(i \in \gamma)} b_{i} \mu(C_{\gamma}) = \sum\{b_{i}\mu(C_{\gamma}):i \in \{1,...,m\} \land \gamma \in I \land i \in \gamma \} = 
$$
$$
 = \sum_{\gamma \in I} (\sum_{i \in \gamma} b_{i}) C_{\gamma} = \sum_{\bar{\alpha} \in I/\sim}\sum_{\gamma \in \bar{\alpha}}\sum_{i \in \gamma} b_{i}C_{\gamma} = \int f d_{\mu \quad \square}
$$
