Composition of continuous and nonmeasurable function is measurable.

Question

I'm stuck on part (b) of the question below. This is another question from a practice preliminary exam. Thanks in advance!

Problem

a) Let $g$ be a monotone function on $\mathbb{R}$. Prove that for every measurable function $f$ on a measurable set $E$ the composition $g \circ f$ is measurable.

b) Show that for every continuous not strictly monotone function $g$ on $\mathbb{R}$ there exists a non-measurable function $f$ such that $g \circ f$ is measurable.

My question relates to part (b). I've solved part (a). I'm not sure if they're trying to say, "every continuous [qualifiers removed] function," or if "not strictly monotone" is trying to state that $g$ is monotone but perhaps not strictly so. Just wondering if anyone can solve and/or provide comment or corrections to the second part of the problem above.

Edited part (b)

Show that for every continuous function $g$ on $\mathbb{R}$ which is not strictly monotone there exists a nonmeasurable function $f$ such that $g \circ f$ is measurable.

Answer 1

(b) If $g$ is not strictly monotone, then there are $a,b\in \mathbb R,$ $a\ne b,$ such that $g(a)=g(b).$ Let $E\subset\mathbb R$ be nonmeasurable. Set $f = a \chi_E +b\chi_{\mathbb R \setminus E}.$ Then $f$ is not measurable. However, on $E,$ $g\circ f = g(a),$ and on $\mathbb R\setminus E,$ $g\circ f = g(b).$ Since $g(a)=g(b),$ $g\circ f$ is constant and hence measurable.

Answer 2

It means $g$ is monotone but not one-to-one. The condition is needed because otherwise the statement is never true: If $g$ is one-to-one then $g^{-1}$ exists and it's also monotone, so by part (a), $f$ would be measurable because $f = g^{-1}\circ g \circ f$