Product of a Lebesgue measurable function with an indicator function is measurable Let $f:X\rightarrow \overline{\mathbb{R}}$ be $\mathcal{A}$-measurable and let $B\in \mathcal{A}$. I would like to show that $\chi_{B}f $ is $\mathcal{A}$-measurable. 
I want to find the set $[\chi_{B}f > a ]$ for all $a \in \mathbb{R}$ , how to do that? 
 A: I'll assume that $f$ takes values in $\mathbb{R}$, and leave it to you to adjust in the case where $f$ actually takes values in the extended reals.
If $a\geq 0$, then $\chi_Bf(x)\gt a$ if and only if $f(x)\gt a$ and $\chi_B(x)=1$; that is,
$$[\chi_Bf\gt a] = [f\gt a]\cap B\text{ if }a\geq 0.$$
Since $f$ is measurable, and $B$ is measurable, the intersection is measurable.
If $a\lt 0$, then $\chi_B f(x)\gt a$ if and only if $x\in B$ and $f(x)\gt a$, or $\chi_B(x)=0$ (since then $\chi_Bf(x)=0\gt a$). So 
$$[\chi_Bf\gt a] = ([f\gt a]\cap B) \cup (\mathbb{R}-B)\text{ if }a\lt 0.$$
This is a union of two measurable sets, hence measurable.
A: Indeed. Suppose that $f$ is a measurable function. Let $D$ the domain of $f$. Then $$\{x:f(x)^2\gt a\}=\{x:f(x)\gt \sqrt{a}\}\cup \{x:f(x)\lt -\sqrt{a}\}$$
if $a\geq 0$ and $$\{x:f(x)^2\gt a\}=D$$ if $a\lt 0$. So, $f^2$ is measurable.
If $f$ and $g$ are measurable functions defined in $D$, since $$2fg=(f+g)^2-f^2-g^2,$$ $fg$ is measurable.
In your particular case, if $B$ is measurable $\chi_B$ is measurable and $f\chi_B$ is measurable. 
I'm assuming that you know that the sum of measurable functions is a measurable function. However, it is not difficult to prove.
