characteristic function of complement of non-Borel set Let $C$ be the Cantor set and $M \subseteq C$ a subset, $M \notin \mathcal B(\mathbb R)$. Obviously the characteristic function of $M, \chi_M:[0,1] \rightarrow \mathbb R$, is not (Borel-) measurable. My question is why the restriction
$\chi_M|_{[0,1] \backslash M}$ 
is Borel-measurable though?
 A: If you restrict the domain of a function, you must also redefine what “Borel measurable” means; i.e., you need to redefine the $\sigma$-algebra on the restricted domain. In this particular space, you can define a $\sigma$-algebra on $[0,1]\setminus M$ using $\mathscr B(\mathbb R)$ as follows:
$$\{B\cap[0,1]\setminus M\,|\,B\in\mathscr B(\mathbb R)\}.\tag{$\spadesuit$}$$
Note that this is a legit $\sigma$-algebra in spite of the fact that $M$ is not Borel-measurable.
Now observe that the function $\chi_M\big|([0,1]\setminus M)$ identically vanishes, so it is measurable with respect to any $\sigma$-algebra on the domain $([0,1]\setminus M)$; specifically, with respect to the “Borel $\sigma$-algebra” on $([0,1]\setminus M)$—or, more accurately, the $\sigma$-algebra on $([0,1]\setminus M)$ induced by the Borel $\sigma$-algebra on $\mathbb R$—as defined by $(\spadesuit)$.
At any rate, I wouldn't say that $\chi_M\big|([0,1]\setminus M)$ is “Borel measurable,” because the restricted domain is not a Borel set—even though it is possible to define something one would be tempted to call a “Borel $\sigma$-algebra” on it as in $(\spadesuit)$, but this terminology is likely to be confusing.
