$\sigma-$algebra , inverse function Is a $\sigma-$algebra a set that contains all the subsets of a set? 
In my lecture notes there isthe following: 
$$f(x)=\sin x \\ f^{-1}\left (\left [\frac{1}{2}, 1\right ]\right )=\bigcup_{k=1}^{\infty} \left [2k \pi +\frac{\pi }{6}, 2k \pi+\frac{5 \pi}{6} \right ]$$ 
To calculate this we want to find the values of $x$ so that $f(x)=\frac{1}{2}$ and $f(x)=1$ or not ??? 
But why is at both limits of the interval $\frac{\pi}{6}$ ??? 
Shouldn't it also be $\frac{\pi}{2}$ ??? 
Edit: 
$$\sin x \leq 1 \Rightarrow x \leq 2k \pi+\frac{\pi}{2} \text{ AND } x \leq 2k \pi -\frac{\pi}{2}=2k\pi +\frac{3\pi}{2}$$ 
$$\sin x \geq \frac{1}{2} \Rightarrow x \geq 2k \pi+\frac{\pi}{6} \text{ AND } x \geq 2k\pi -\frac{\pi}{6}=2k \pi+\frac{11\pi}{6}$$ 
Therefore, we get the following: 
 
So how do we get this interval???
 A: A $\sigma$-algebra $\mathcal{F}$
  of $\Omega$
  is a part of $\mathcal{P}\left(\Omega\right)$
  such that $\emptyset\in\mathcal{F}$
 , $\mathcal{F}$
  is stable by finite intersections and also stable by countable unions.
Now,$$\sin^{-1}\left[\frac{1}{2},1\right]=\bigcup_{k=0}^{+\infty}\left[\frac{\pi}{6}+k2\pi,\frac{5\pi}{6}+k2\pi\right]$$
 because for any $k\geq0$
 , if $$x\in\left[\frac{\pi}{6}+k2\pi,\frac{5\pi}{6}+k2\pi\right],$$
 then $$\sin\left(x\right)\in\left[\frac{1}{2},1\right]
 .$$
Let us consider the case $k=0$.
When $\frac{\pi}{6}\leq x<\frac{\pi}{2},$ we have $\frac{1}{2}\leq\sin\left(x\right)<1$
  and $\sin$
  is stricly increasing on this interval (since $\cos>0$
  here). After, we have clearly $\sin\left(\frac{\pi}{2}\right)=1\in\left[\frac{1}{2},1\right].$
But now, consider $\frac{\pi}{2}<x\leq\frac{5\pi}{6}$
  : here $\sin$
  is stricly decreasing (since $\cos<0$
  here), and ranges from $\sin\left(\frac{\pi}{2}\right)=1$
  to $\sin\left(\frac{5\pi}{6}\right)=\frac{1}{2}.$ We must not forget these possible values for $x$.
 
