Q:

How to prove that preimage of measurable space is measurable?

Given a function mapping between two sample spaces $S_1$ and $S_2$, if $S_2$,$F_2$ is measurable how do I show that preimage of $S_2$,$F_2$ in $S_1$ is measurable set?

A:

Strictly speaking, this is trivial: the preimage of $(S_2, F_2)$ is all of $S_1$ (by definition), which is measurable (by definition).
Perhaps you want to conclude that the preimage of any measurable subset of $S_2$ is measurable: that is a nice property of a function.  However, this conclusion is not true in general, either.  For example, let $(S_1, F_1)$ contain a nonmeasurable set and let $(S_2, F_2)$ contain two disjoint measurable atoms.  Map every element of the nonmeasurable set to one of the atoms and map every element of the complement of the nonmeasurable set to the other atom.  The preimage of the first atom is not measurable, whence this map is not measurable.

