Find unit decomposition of multiplication operator $(Ax)(t) = \sin(t)\,x(t)$ in $L^2(\mathbb{R})$ space.
I started from spectrum of operator $\sigma(A)=[-1,1] \implies E_\lambda \lambda=E(-\infty,\lambda)= 0$, if $\lambda<-1$ and $E_\lambda \lambda=I$, if $\lambda>1$. $E$ is projection-valued measure.
Then $f$ is continuous function. $f(A)\,x(t) = f(\sin(t))\,x(t)$.
Taking the limit $f(t) \to \chi_{(-\infty,\lambda)}(t)$, from this moment I didn't understand:
We have $$ E_\lambda: x(t) \mapsto \chi_{[-\arcsin(\lambda) + \pi + 2\pi n, \arcsin(\lambda) + 3\pi +2\pi n]}(t) $$ because restriction of $\chi_{(-\infty,\lambda)}(\sin(t))$ to $[-1,1]$ equals to $$ \sum\chi_{[-\arcsin(\lambda) + \pi + 2\pi n, \arcsin(\lambda) + 3\pi + 2\pi n]}(t) $$ and the answer is $$ E_\lambda x(t) = \begin{cases} 0, &\lambda \leq - 1 \\ \sum\chi_{[-\arcsin(\lambda) + \pi + 2\pi n, \arcsin(\lambda) + 3\pi + 2\pi n]}(t), &-1 < \lambda \leq 1 \\x(t), &\lambda > 1 \end{cases} $$
So I didn't understand how Indicator of $\sin(x)$ ($\chi_{(-\infty,\lambda)}$) restricted to $$ \sum\chi_{[-\arcsin(\lambda) + \pi + 2\pi n, \arcsin(\lambda) + 3\pi + 2\pi n]}(t). $$
Why borders changed to $[-\arcsin(\lambda) + \pi + 2\pi n, \arcsin(\lambda) + 3\pi + 2\pi n]$?
P.S. Sorry for my English, I'm not native speaker.