I know that this question is going to get a lot of downvotes. But pardon me, because I have stuck here and have tried many things with no success. Now I state my problem.
Let $(\Omega, \Sigma, \mu)$ be a measured space and let $(\mathbb{R}, \mathcal{B})$ be a measurable space. Let $h:\Omega\rightarrow\mathbb{R}$ be a simple function of the form: $$h(\omega)=\sum_{i=1}^{n}c_i\mathcal{X}_{A_i}(\omega),$$ where $c_i>0$, $A_i\in\Sigma$ and $\mathcal{X}_{A_i}(\omega)$ denotes the indicator function. Here, I am assuming that the function $h(w)$ is written in a non-canonical form, meaning, for some $p,q\in\{1,2,...,n\} (p\neq q)$, we have that $A_p\cap A_q\neq \emptyset$. Now, it is a fact that a simple function can only take some finite non-negative values. So, let $t_1< t_2< t_3<...<t_m$ be the set of values which the function $h(w)$ takes. So, the function $h(w)$ can be written in its canonical form as follows $$h(\omega)=\sum_{i=1}^{m} t_i \mathcal{X}_{\{\omega\in\Omega: h(\omega)=t_i\}}.$$ Note that in the above expression for $h(\omega)$, for $p,q\in\{1,2,3,...,m\}(p\neq q)$, we have that ${\{\omega\in\Omega: h(\omega)=t_p\}}\cap \{\omega\in\Omega: h(\omega)=t_q\}=\emptyset$ (since $h(w)$ is written in canonical form).
Now, I have to prove the following $$\int\big(\sum_{i=1}^{n}c_i\mathcal{X}_{A_i}(\omega)\big)d(\mu)=\int \big(\sum_{i=1}^{m} t_i \mathcal{X}_{\{\omega\in\Omega: h(\omega)=t_i\}}\big)d(\mu),$$ which means I have to prove the following $$\sum_{i=1}^{n}c_i\mu(A_i)=\sum_{i=1}^{m}t_i\mu(\{\omega\in\Omega: h(\omega)=t_i\}).\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
Here is what I have tried to prove the equation number $(1)$:
Let $t_r\in \{t_1,t_2,t_3,...,t_m\}$ be arbitrary such that $t_r\neq 0$. Let $B=\{\omega\in\Omega: h(\omega)=t_r\}$. Define a collection of sets $\mathcal{F}$ as follows $$\mathcal{F}=\{A_1\cap B, A_2\cap B, A_3\cap B,...,A_n\cap B\}.$$
I am able to prove that $$B= (A_1\cap B)\cup(A_2\cap B)\cup(A_3\cap B)\cup...\cup (A_n\cap B).$$ But I am not able to figure out what to do next.
Can somebody help me mathematically derive the equation number $(1)$?
There is a nice and intuitive proof (without using the linearity of the integral) in Theorem 3.13 of Sheldon Axler's Measure, Integration and Real Analysis. Just set $b_k = t_k$, and $B_k = h^{-1}(\{t_k\}) = \{\omega \in \Omega:h(\omega) = t_k\}$. Hope it helps!