Prove that if $E$ is measurable and $A \subset E$ is measurable, $\gamma$ is simple, then $\int_A \gamma = \int_E \gamma \chi_A $

Question

Let $\gamma: E \rightarrow \mathbb{R}$ be a simple function on $E$, with $|E| < \infty$ (where $|*|$ represents the Lebesgue measure). Suppose $A \subset E$. A simple function is a function that has only a finite number of values in it's range.

We define $\int_E \gamma = \Sigma_{k=1}^n c_k |E_k|$

Where $E_k = \gamma^{-1}(c_k)$ and $c_k$ for $1 \leq k \leq n$ are the finite number of different values that $\gamma$ takes on $E$.

Prove that:

$\int_A \gamma = \int_E \gamma \chi_A $

$\text{Proof}$:

$\int_A \gamma = \Sigma_{k=1}^n c_i |A_i|$

while

$\int_E \gamma \chi_A = \Sigma_{l=1}^m b_i |E_i|$

where $\cup A_i = A$ and $\cup E_i = E$ are disjoint unions. and $c_i$ and $b_i$ are the distinct values the simple functions take when partitioned on their respective domains...

I feel like I'm making this way more complicated then it needs to be. Can anyone make this simple for me?? Thanks!!