How determine $\chi_{E} * S$?

Question

Let $(X, \mathscr{M}, \mu)$ be a measure space, $E \subset X$ such that $E \neq \varnothing \neq E^\mathsf{c}$, and $S:X \rightarrow [0, +\infty)$ a simple measurable function, determine by $S = \sum_{i=1}^{n} \alpha_{i}*\chi_{A_{i}}$ $(*)$.

1. How can I write $\chi_{E} * S$ in terms of (*)?
2. It is obviously simple, so How I can calculate it inverse images $B_{i}$?
3. How a can write it Lebesgue integral in terms of the inverse images (i.e $A_{i}'s$) of S?

Note to 2.: I tried to calculate it inverse images (i.e $B_{i}$), and I get that If $\alpha_{i} \neq 0$, then $B_{i} = A_{i} \cap E$. But, is possible that $0$ doesn't be in the range of $S$ and it always be in the range of $\chi_{E} * S$. So, my problem is with $0$...

Answer 1

$$\chi_E S = \sum\limits_{i=1}^{n} \alpha_i \chi_E \chi_{A_i}$$ $$ =\sum\limits_{i=1}^{n} \alpha_i \chi_{E \cap A_i}.$$ This is a simple function. Simply omit the terms where $E \cap A_i$ is empty. This implies that $\int \chi_E d\mu=\sum\limits_{i=1}^{n} \alpha_i \mu (\chi_{A_i}\cap E)$. It makes no difference whether $0$ is in the range or not.