I am using the following definition of "measurable." Suppose we have a set $\displaystyle \Omega$ with an associated sigma-algebra $\displaystyle F$ of subsets. A function $\displaystyle f: \Omega \rightarrow R$ is measurable w.r.t. $\displaystyle F$ if: for all $\displaystyle c \in R$, the set $\displaystyle \{\omega \in \Omega | f(\omega) < c\} = f^{-1}((-\infty,c))$ belongs to $\displaystyle F$.

So I choose $\displaystyle \Omega = R$ (the reals) and let the function $\displaystyle f$ be the identity function $\displaystyle f(x) = x$. I hope that this innocent bijective function is measurable. And for all $\displaystyle c$, we would have: $\displaystyle f^{-1}((-\infty,c)) = (-\infty,c)$.

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 $\displaystyle F= \{\Omega,\emptyset\} = \{R,\emptyset\}$ work?

What other possible sigma-algebras $\displaystyle F$ of $\displaystyle \Omega$ could we choose to keep $\displaystyle f$ measurable w.r.t. $\displaystyle F$?

Thanks!