I am watching a video lecture on Lebesgue Integral. In the video, the instructor says, let $(\Omega, \Sigma, \mu)$ be a measure space, and let $(\mathbb{R},\mathcal{B})$ be the measurable space. Then a simple function $h:\Omega\rightarrow \mathbb{R}$ is defined as $$h(\omega)=\sum_{i=1}^{n}c_i\mathcal{X}_{A_i}(\omega),$$ where $c_i>0$, $A_i\in\Sigma$, and $\mathcal{X}_{A_i}(\omega)$ denotes the indicator function. Since $h(\omega)$ is measurable and takes only finite values, the function $h(\omega)$ can be written in its canonical form as $$h(\omega)=\sum_{t\in h(\omega)}t\mathcal{X}_{\{\omega\in\Omega\ : h(\omega)=t\}}.$$
Then the instructor says the set $\Omega$ can be partitioned into the two subsets as $\Omega=\tilde{\Omega}\cup\tilde{\Omega}^{c}$ such that $\mu(\tilde{\Omega}^{c})=0$. Then he says the set $\{\omega\in\Omega : h(\omega)=t\}$ can be partitioned as $$\{\omega\in\Omega : h(\omega)=t\}=\{\omega\in\tilde{\Omega} : h(\omega)=t\}\cup \{\omega\in\tilde{\Omega}^{c} : h(\omega)=t\}.$$
I understood the lecture up to this point because it is obvious that the singleton $\{t\}\in\mathcal{B}$ and the preimage of $\{t\}$ is the set $\{\omega\in\Omega : h(\omega)=t\}$, and since $h(w)$ is measurable, we have that $\{\omega\in\Omega : h(\omega)=t\}\in\Sigma$.
But then he goes on to say $$h(\omega)=\sum_{t\in h(\omega)}t\big(\mathcal{X}_{\{\omega\in\tilde{\Omega} : h(\omega)=t\}\cup \{\omega\in\tilde{\Omega}^{c}:h(\omega)=t\}}\big)$$
Therefore, \begin{eqnarray} \int h(\omega)d(\mu)&=&\sum_{t\in h(\omega)}t \mu\big({\{\omega\in\tilde{\Omega} : h(\omega)=t\}}\cup{\{\omega\in\tilde{\Omega}^{c}:h(\omega)=t\}}\big)\\ &=& \sum_{t\in h(\omega)}t \big(\mu({\{\omega\in\tilde{\Omega} : h(\omega)=t\}})+ \mu ({\{\omega\in\tilde{\Omega}^{c}:h(\omega)=t\}})\big)\\ &=& \sum_{t\in h(\omega)}t\mu({\{\omega\in\tilde{\Omega} : h(\omega)=t\}}) \end{eqnarray}
Now, the thing that I don't understand is why the following two sets are measurable:
(1). $\{\omega\in\tilde{\Omega} : h(\omega)=t\}\in\Sigma$ ?
(2). $\{\omega\in\tilde{\Omega}^{c}:h(\omega)=t\}\in\Sigma$ ?
I don't see why these two sets would be in $\Sigma$. Clearly, the preimage of $\{t\}$ is neither $\{\omega\in\tilde{\Omega} : h(\omega)=t\}$ nor $\{\omega\in\tilde{\Omega}^{c}:h(\omega)=t\}$. So these two sets do not necessarily lie in $\Sigma$. But it is obvious that the union of these two sets is the preimage of $\{t\}$, so the union of these two sets would necessarily be in $\Sigma$.
Can somebody shed some light on why these two sets are measurable according to my instructor?