# Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function, is $f \circ \sin$ measurable?

If $$f \circ \sin$$ is measurable, then we need to show that $$(f \circ \sin)^{-1}(B) = \sin^{-1}(f^{-1}(B))$$ is measurable for every Borel set $$B$$, it is sufficient to show that $$\sin^{-1}$$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $$\arcsin$$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $$\sin^{-1}(f^{-1}(B)) = \cup_{n \in \mathbb{Z}} \arcsin(f^{-1}(B)) + 2n\pi$$ would measurable.

It could happen that $$f\circ \sin$$ is not measurable, but the task of finding a set $$B$$ such that $$\sin^{-1}(f^{-1}(B))$$ is not measurable seems very difficult (maybe impossible?).

• the composition of measurable functions is measurable. Dec 9, 2018 at 2:01
• @TsemoAristide He is considering $f$ to be measurable $\Lambda/\mathscr{B},$ so he wants to prove $\sin$ is $\Lambda/\Lambda$ measurable, I guess. Dec 9, 2018 at 2:05
• Use $\sin$ for $\sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $\arcsin$ for $\arcsin$, and other trigonometric functions. Dec 9, 2018 at 2:20
• @Tsemo actually that’s not true in general Dec 9, 2018 at 2:42
• @Tsemo see this post for instance Dec 9, 2018 at 2:44

Yes, $$f \circ \sin$$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.
Here is the bit of math that was missing to complete the demonstration that $$f\ \circ \sin$$ is measurable: Image of Lebesgue measurable set by $$C^1$$ function is measurable.