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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?).

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    $\begingroup$ the composition of measurable functions is measurable. $\endgroup$ – Tsemo Aristide Dec 9 '18 at 2:01
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    $\begingroup$ @TsemoAristide He is considering $f$ to be measurable $\Lambda/\mathscr{B},$ so he wants to prove $\sin$ is $\Lambda/\Lambda$ measurable, I guess. $\endgroup$ – Will M. Dec 9 '18 at 2:05
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    $\begingroup$ 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. $\endgroup$ – Shaun Dec 9 '18 at 2:20
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    $\begingroup$ @Tsemo actually that’s not true in general $\endgroup$ – Omnomnomnom Dec 9 '18 at 2:42
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    $\begingroup$ @Tsemo see this post for instance $\endgroup$ – Omnomnomnom Dec 9 '18 at 2:44
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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.

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  • $\begingroup$ You mean the pre-image of a lebesque by a absolutely continuous? $\endgroup$ – Non Dec 17 '18 at 2:13
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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.

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