Prove $\int(\phi+\psi)\,d\mu = \int\phi\,d\mu + \int\psi\,d\mu$ I need help with understanding of this please
Lemma 4.4.
Let $(X, \mathbb{X}, \mu)$ be a measure space. Let $\phi, \psi$ be simple functions
in $M^{+}(X, \mathbb{X})$ and let $c \geq 0 .$ Then
$$
\begin{array}{l}{\text { (i) } \int c \phi d \mu=c \int \phi d \mu, \text { and }} \\ {\text { (ii) } \int(\phi+\psi) d \mu=\int \phi d \mu+\int \psi d \mu}\end{array}
$$
Proof

I don't understand how the lemma follows from the last sentence. Another clearer proof would help a lot.
 A: It is not entirely clear what your definition of a simple function is.
I will call $s$ simple if it has a finite range and $\mu(s^{-1}(\{0\}^c) $ is finite.
Let the range of $s$ be $r_1,..,r_n$ and $A_k = s^{-1}(\{r_k\})$, then we can write $s= \sum_k r_k 1_{A_k}$. 
Now suppose $\phi= \sum_i f_i 1_{F_i}$ and $\psi = \sum_j u_j 1_{U_j}$, where the
representations are as with $s$ above.
It is clear that $\phi+\psi$ has finite range and $\mu((\phi+\psi)^{-1}(\{0\}^c) $ is finite.
Let $I = \{ (i,j) | F_i \cap U_j \neq \emptyset \}$, note that $\{ F_i \cap U_j \}_{(i,j) \in I }$ is a partition of the domain.
Let $v_1,...,v_m$ be the range of $\phi+\psi$ and for each $k$, let
$I_k= \{ (i,j) \in I | f_i+u_j = v_k \}$, and $V_k = (\phi+\psi)^{-1}(\{v_k\})$.
Note that $V_k = \cup_{(i,j) \in I_k} (F_i \cap U_j)$ and $\mu V_k = \sum_{(i,j) \in I_k} \mu(F_i \cap U_j)$.
Then
\begin{eqnarray}
\int (\phi+\psi) d \mu &=& \sum_k v_k \mu V_k \\
&=& \sum_k v_k \sum_{(i,j) \in I_k} \mu(F_i \cap U_j) \\
&=& \sum_k  \sum_{(i,j) \in I_k} (f_i+u_j)\mu(F_i \cap U_j) \\
&=& \sum_{(i,j) \in I} (f_i+u_j)\mu(F_i \cap U_j) \\
&=& \sum_{(i,j) \in I} f_i\mu(F_i \cap U_j) + \sum_{(i,j) \in I} u_j \mu(F_i \cap U_j)\\
&=& \sum_{i,j} f_i\mu(F_i \cap U_j) + \sum_{i,j} u_j \mu(F_i \cap U_j)\\
&=& \sum_{i} f_i\mu F_i + \sum_j u_j \mu U_j\\
&=& \int \phi d \mu + \int \psi d \mu
\end{eqnarray}
