Real Analysis, Folland Proposition 2.13 Integration of Nonnegative Functions Question:

Proposition 2.13 - Let $\phi$ and $\psi$ be simple functions in $L^+$.
a.) If $c\geq 0$, $\int c\phi = c\int \phi$.
b.) $\int(\phi + \psi) = \int \phi + \int \psi$.
c.) If $\phi\leq \psi$, then $\int \phi\leq \int \psi$.
d.) The map $A\rightarrow \int_{A}d\mu$ is a measure on $M$.

Attempted proof a.) - If $c\geq 0$ then $$\int c\phi = \int c\sum_{j}a_j \chi_{E_j} = c\int \sum_{j}a_j \chi_{E_j} = c\int \phi$$
Attempted proof b.) Let $\phi = \int_{j}a_j\chi_{E_j}$ and $\psi = \sum_{k}b_k\chi_{f_k}$, then $$\int \phi + \int \psi = \int \sum_{j}a_j\chi_{E_j} + \int \sum_{k}b_k\chi_{F_k} = \sum_{j}a_j\mu(E_j) + \sum_{j}b_k\mu(F_k)$$
Before I go on with c and d I saw that after this last step we can say $$\sum_{j}a_j\mu(E_j) + \sum_{j}b_k\mu(F_k) = \sum_{j}a_k\sum_{k}\mu(E_j\cap F_k) + \sum_{k}b_k\sum_{j}\mu(E_j\cap F_k)$$
I don't understand how they are equal. So this is a more of an algebra question I guess more than a measure theory question. Any suggestions is greatly appreciated.
Background Information:
We fix a measure space $(X,M,\mu)$, and we define $$L^+ = \ \ \text{the space of all measurable functions from} \ X \ \text{to} \ [0,\infty]$$
If $\phi$ is a simple function in $L^+$ with standard representation $\phi = \sum_{1}^{n}a_j\chi_{E_j}$, we define the integral of $\phi$ with respect to $\mu$ by $$\int \phi d\mu = \sum_{1}^{n}a_j\mu(E_j)$$
 A: Let $(X,M, \mu)$ be a measure space. 
Question:

Proposition 2.13 - Let $\phi$ and $\psi$ be simple functions in $L^+$.
a.) If $c\geq 0$, $\int c\phi = c\int \phi$.
b.) $\int(\phi + \psi) = \int \phi + \int \psi$.
c.) If $\phi\leq \psi$, then $\int \phi\leq \int \psi$.
d.) The map $A\rightarrow \int_{A}d\mu$ is a measure on $M$.

Proof a.) - If $c\geq 0$ then 
\begin{align*}
\int c\phi &= \int c\sum_{j}a_j \chi_{E_j} = \int \sum_{j}ca_j \chi_{E_j} = \sum_{j}ca_j \mu(E_j)= \\ &= c\left(\sum_{j}a_j \mu(E_j) \right)= c\int \sum_{j}a_j \chi_{E_j} = c\int \phi
\end{align*}
Proof b.) Let $\phi = \sum_{j}a_j\chi_{E_j}$, $\psi = \sum_{k}b_k\chi_{F_k}$ and $\phi +\psi = \sum_{i}c_i\chi_{H_i}$ be the standard representations of $\phi$,  $\psi$ and $\phi + \psi$. 
Then $\{E_j\}_j$ is a finite family of disjoint measurable sets, such that $X= \bigcup_jE_j$. We also have that $\{F_k\}_k$ is a finite family of disjoint measurable sets, such that $X= \bigcup_kF_k$. And $\{H_i\}_i$ is a finite family of disjoint measurable sets, such that $X= \bigcup_kH_i$.
So we have that $\{E_j\cap F_k \cap H_i\}_{j,k,i}$ is a finite family of disjoint measurable sets, such that, for all $j$, $E_j =\bigcup_{k,i}(E_j\cap F_k \cap H_i)$; for all $k$, $F_k =\bigcup_{j,i}(E_j\cap F_k\cap H_i)$ and  $H_i =\bigcup_{j,k}(E_j\cap F_k \cap H_i)$. So we have 
$$ \mu(E_j) =\sum_{k,i}\mu(E_j\cap F_k \cap H_i)$$
$$ \mu(F_k) =\sum_{j,i}\mu(E_j\cap F_k\cap H_i)$$
and 
$$\mu(H_i) = \sum_{j,k}\mu(E_j\cap F_k \cap H_i)$$
Moreover, for all $j$, $k$, $i$, if $E_j\cap F_k \cap H_i \neq \emptyset$, then $a_j+b_k=c_i$.
\begin{align*}
\int(\phi + \psi) &= \sum_{i}c_i\mu(H_i)= \sum_{i,j,k}c_i\mu(E_j\cap F_k \cap H_i)= \sum_{\substack{i,j,k \\ E_j\cap F_k \cap H_i \neq \emptyset}}c_i\mu(E_j\cap F_k \cap H_i)= \\
& =  \sum_{\substack{i,j,k \\ E_j\cap F_k \cap H_i \neq \emptyset}}(a_j+b_k)\mu(E_j\cap F_k \cap H_i)=  \sum_{i,j,k }(a_j+b_k)\mu(E_j\cap F_k \cap H_i)=\\
&=  \sum_{i,j,k }a_j\mu(E_j\cap F_k \cap H_i)+   \sum_{i,j,k }b_k\mu(E_j\cap F_k \cap H_i)= \\
&=  \sum_j a_j\sum_{i,k }\mu(E_j\cap F_k \cap H_i)+   \sum_k b_k\sum_{i,j }\mu(E_j\cap F_k \cap H_i)= \\
&=  \sum_j a_j\mu(E_j)+   \sum_k b_k\mu( F_k )= \\
&=\int \phi + \int \psi
\end{align*}
Proof c.) Let $\phi = \sum_{j}a_j\chi_{E_j}$ and$\psi = \sum_{k}b_k\chi_{F_k}$ be the standard representations of $\phi$ and  $\psi$. 
Then $\{E_j\}_j$ is a finite family of disjoint measurable sets, such that $X= \bigcup_jE_j$. And $\{F_k\}_k$ is a finite family of disjoint measurable sets, such that $X= \bigcup_kF_k$. 
So we have that $\{E_j\cap F_k \}_{j,k}$ is a finite family of disjoint measurable sets, such that, for all $j$, $E_j =\bigcup_k(E_j\cap F_k)$ and for all $k$, $F_k =\bigcup_j(E_j\cap F_k)$. So we have 
$$ \mu(E_j) =\sum_{k,i}\mu(E_j\cap F_k \cap H_i)$$ 
and
$$ \mu(F_k) =\sum_{j,k}\mu(E_j\cap F_k\cap H_i)$$
Moreover, for all $j$, $k$, if $E_j\cap F_k  \neq \emptyset$, then $a_j\leq b_k$.
\begin{align*}
\int\phi  &= \sum_{j}a_j\mu(E_j)= \sum_{j,k}aj\mu(E_j\cap F_k)= \sum_{\substack{j,k \\ E_j\cap F_k  \neq \emptyset}}a_j\mu(E_j\cap F_k) \leq \\ & \leq \sum_{\substack{j,k \\ E_j\cap F_k  \neq \emptyset}}b_k\mu(E_j\cap F_k)= \sum_{j,k }b_k\mu(E_j\cap F_k)= \\
 &= \sum_k b_k\sum_{j }\mu(E_j\cap F_k )= \sum_k b_k\mu( F_k )= 
 \int \psi
\end{align*}
Proof d.) Since for all $A \in M$, $\int_{A}d\mu=\int\chi_{A}d\mu=\mu(A)$, we have that the map $A\rightarrow \int_{A}d\mu$ is just the map $A \rightarrow \mu(A)$ a measure on $M$.
