# 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?

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Notice that 0 times f is 0, when x is finite, which will happen outside of B; what will happen on points of B? – gary Nov 11 '11 at 0:48

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.

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Thank you leo……. – VVV Nov 12 '11 at 22:48
You're welcome! – leo Nov 14 '11 at 0:26

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.

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Thank you Arturo. – VVV Nov 12 '11 at 22:47