Measurability of the composition of a measurable map with a surjective map satisfying an expansion condition

Solution 1:

The point here is that the assumptions that $g: \mathbb{R} \to \mathbb{R}$ is surjective and that $|g(x)-g(y)| \geq c|x-y|$ for some $c \gt 0$ imply that $g$ is bijective and that its inverse $h: \mathbb{R} \to\mathbb{R}$ is Lipschitz continuous with Lipschitz constant $c^{-1}$.

In particular, $h$ maps Lebesgue measurable sets to Lebesgue measurable sets. In more detail:

1. $h$ maps Borel sets to Borel sets: Every closed set is a countable union of compact sets and thus $h$ maps closed sets to Borel sets because $h(\bigcup C_n) = \bigcup h(C_n)$ where the right hand side is a countable union of compact sets by continuity of $h$, hence it is Borel measurable. This tells us that $g = h^{-1}$ is Borel measurable, so $h$ indeed maps Borel sets to Borel sets.^(*)
2. $h$ maps null sets to null sets: If $N$ is a null set then given any $\varepsilon \gt 0$ we can cover $N$ by countably many intervals whose total length doesn't exceed $\varepsilon$. By Lipschitz continuity, the images of these intervals will again be intervals whose total length doesn't exceed $c^{-1} \varepsilon$, and this shows that $h(N)$ is a null set.
3. Every Lebesgue measurable set $L$ can be written as $L = B \cup N$ with $B$ Borel and $N$ null. Then $h(L) = h(B) \cup h(N)$ is a union of a Borel set and a null set by 1. and 2., so $h(L)$ is Lebesgue measurable.

In conclusion, $g^{-1}(L) = h(L)$ is Lebesgue measurable for every Lebesgue measurable $L$. Thus, $g^{-1}(f^{-1}(B))$ is Lebesgue measurable for every Borel set $B$ because $L = f^{-1}(B)$ is Lebesgue measurable by measurability of $f$, and this proves that $f \circ g$ is measurable.

^(*) It is a general fact due to Lusin and Souslin that a Borel measurable injection between completely metrizable spaces sends Borel sets to Borel sets, but this is much more difficult to prove than 1 above (see e.g. Kechris, Classical descriptive set theory, Theorem 15.1, page 89). Without injectivity this fails, the continuous image of a Borel set is not a Borel set: it is only analytic in general. See this MO answer for the standard story to be told at this point…