Solution
= Solution
\b[True in both cases.] For every open $U$, <continuity> of $f$ makes $f^{-1}(U)$ open. Then
$$
(f\circ g)^{-1}(U)=g^{-1}(f^{-1}(U))
$$
is <Lebesgue measurable> by <measurability> of $g$. The value $f(0)$ has no bearing on this <preimage> argument.