$u(x) = \sin(x)$. Determine the $\sigma (u)$ on $\mathbb{R}$ 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 $\sigma (u) = \sigma ( \{ \cup_{k \in \mathbb{Z}} ((-1)^{k} \arcsin (B \cap [-1,1]) + k\pi ) : B \in B(\mathbb{R})  \})$
 A: 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 $A\in\mathcal B\iff A_r\in\mathcal B$.
