If a set is measurable then its image set is measurable.

Question

Let $$f: (\Omega_1,\mathscr{F_1})\rightarrow(\Omega_2,\mathscr{F_2})$$ be a measurable function. Then we have $f^{-1}(A)\in \mathscr{F_1}$, when $A\in \mathscr{F_2}$.

I am wondering whether the following is true: if $B\in \mathscr{F_1}$ then there exist $A\in \mathscr{F_2}$ such that $B= f^{-1}(A)$.

Answer 1

Not necessarily.

Simple counterexample: $\Omega_1=\Omega=\Omega_2$ with $\mathcal F_1=\wp(\Omega)$ and $\mathcal F_2=\{\varnothing,\Omega\}$ and $\Omega$ is not a singleton.

Let $f$ be the identity function.

Actually it can be shown that the collection $\{f^{-1}(A)\mid A\in\mathcal F_2\}$ is a $\sigma$-algebra. If $f$ is measurable then it is a subcollection of $\mathcal F_1$ but does not necessarily coincide with $\mathcal F_1$ (as shown).

Answer 2

You can consider a constant function on $\mathbb{R}$ to see that this is not always the case.