It does not work because
What other possible sigma-algebras of could we choose to keep measurable w.r.t. ?
Any sigma-algebra which contains to the sigma-algebra generated by the family:
Fernando Revilla
I am using the following definition of "measurable." Suppose we have a set with an associated sigma-algebra of subsets. A function is measurable w.r.t. if: for all , the set belongs to .
So I choose (the reals) and let the function be the identity function . I hope that this innocent bijective function is measurable. And for all , we would have: .
But "measurable" is defined in the context of "with respect to" a sigma-algebra, right? So this requires me to choose a sigma-algebra, no?
Would choosing the trivial sigma-algebra work?
What other possible sigma-algebras of could we choose to keep measurable w.r.t. ?
Thanks!
It does not work because
What other possible sigma-algebras of could we choose to keep measurable w.r.t. ?
Any sigma-algebra which contains to the sigma-algebra generated by the family:
Fernando Revilla