Show that a monotone function f: I -> R is borel-measureable.
I know, that this question has been asked and answered multiple times here, but only for f: R->R und not for f: I->R. The first case could be proofed by proofing that {x $\in$ R: f(x) < a} $\in$ B(R) (in case of decreasing) for any a $\in$ R.
First of all, why is that enough? In my opinion, I have to proof that f^(-1) (A) $\in$ B(R) for all A $\in$ J with J as set of any kind of intervals in R (since it creates B(R) = $\sigma$(J) )
Secondly, how can I consider the interval I in that proof? I think, there is a huge difference between proofing f:R->R and f: I->R as measurable functions.
I am not quite satisfied with that answer: Yes, I is an interval and our measure spaces are:
(I, B(I) = $\sigma$ ($O_I$) ) and (R, B(R) = $\sigma$($O_R$) = $\sigma$ (J)) with $O_I$ and $O_R$ as topologies.
It holds: B(I) = $\sigma$ ($O_I$) $\subseteq$ $\sigma$(J) = B(R). Any kind of interval C $\in$ J is automatically in B(R)=$\sigma$(J), since J $\subseteq$ $\sigma$(J). But I can't quite formally see, why C should be in B(I) = $\sigma$ ($O_I$).
Edit: $C$ = f^(-1) ($A$) = f^(-1) ($\inf$,a]