Let $N$ be a nonmeasurable subset of $(0,1)$, and let $f(x)=x*char_{N}(x)$ where $char_{N}(x) =$ {$1$ if $x \in N$ and $0$ if $ x \notin N$}. Show that $f$ is nonmeasurable, but that each of the sets {$f=\alpha$} = {$x \in D | f(x) = \alpha$} is measurable.
Okay, so i'm definitely going to need help with this one.
$f(x): (0,1) \rightarrow \mathbb{R}$. Even this i'm not sure is right, the question never states the domain of $f$, but since the value is going to be zero on anything not in $N$ which is a subset of $(0,1)$, i suppose having the domain be $(0,1)$ isn't wrong.
First I need to show that {$f=\alpha$} $=$ {$x \in D | f(x) = \alpha$} is measurable $\forall \alpha \in \mathbb{R}$, then I need to show that $f$ itself is NOT measurable, which means that there must be an open set $U \subset \mathbb{R}$ s.t. $f^{-1}(U)$ is not a measurable subset of $(0,1)$. It would be nice if $N$ was open in $(0,1)$, then we could just take the preimage of $N$, since it is nonmeasurable.
Anyway, that's all I've got, can one of you math genius's lend me a hand?? Thanks!
$\{x:x \chi_N (x) \neq 0 \}=N$ which is not measurable. Hence the inverse image of the Borel set $\mathbb R \setminus \{0\}$ is not measurable and this proves that $f$ is not measurable.