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Let $u: \mathbb{R} \to \mathbb{R}, \ x \to \sin(x)$ and $\lambda$ the lebesgue measure on $\mathbb{R}$

Determine the $\sigma (u)$ on $\mathbb{R}$

I know that $\sigma (u)$ is the smallst $\sigma$-algebra s.t u is $\sigma (u) / B(\mathbb{R})$ measureable. We have that $\sigma (u) = \sigma ( \{u^{-1} (B) : B \in B(\mathbb{R} \})$, and that the inverse of $\sin (x)$ is $\arcsin (x)$.

If we study $u(x)$ on the intervals $x \in (- \pi/2 + k\pi, \pi/2 + k\pi]$ we can write the inverses as $u^{-1} = (-1)^k \arcsin(x)$. Then every preimage can be written as $(-1)^k \arcsin([-1,1]) + k\pi$ and thus $\sigma (u) = \sigma ( \{ \cup_{k \in \mathbb{Z}} ((-1)^{k} \arcsin (B \cap [-1,1]) + k\pi ) : B \in B(\mathbb{R}) \})$

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    Would you min explaining your train of thought? It toke me some time to understand this, and to hear how someone else thinks about this, might be helpfull. @drhab – 2017-02-16

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In general if $f:X\to Y$ is a function then a subset $A$ of $X$ is by definition $f$-saturated if $A=f^{-1}(B)$ for some $B\subseteq Y$.

If that is the case then there can be several choices for $B$ and one of them is $f(A)$. So we could also say that $A$ is $f$-saturated if $A=f^{-1}(f(A))$.

Let $\mathcal B$ denote the Borel-$\sigma$ algebra on $\mathbb R$ and note that the function $u:\mathbb R\to\mathbb R$ prescribed by $x\mapsto\sin x$ has the following properties:$$\sigma(u):=u^{-1}(\mathcal B)\subseteq\mathcal B\tag1$$and: $$u(\mathcal B)\subseteq\mathcal B\tag2$$$(1)$ because $u$ is Borel-measurable, and $(2)$ actually needs a proof, but I will leave that out here.

Now define: $$\mathcal B_u:=\{A\in\mathcal B\mid A\text{ is }u\text{-saturated}\}$$ We claim that: $$\sigma(u)=\mathcal B_u$$

If $A\in\sigma(u)$ then $(1)$ tells us that $A=u^{-1}(B)$ for some $B\in\mathcal B$, and consequently $A\in\mathcal B_u$.

If $A\in\mathcal B_u\subseteq\mathcal B$ and $(2)$ tells us that $u(A)\in\mathcal B$, and $(2)$ assures that $A=u^{-1}(u(A))\in\sigma(u)$.

You could wonder: what do $u$-saturated sets looks like?

If $A$ is $u$-saturated then for any $r\in\mathbb R$ you can define $A_r:=A\cap[r,r+2\pi)$, and write: $$A=\bigcup_{n\in\mathbb Z}(2n\pi+A_r)$$Of course it is handsome here to choose for e.g. $r=0$. Also note that $A\in\mathcal B\iff A_r\in\mathcal B$.