Does $\{A\subset Y\mid f^{-1}(A)\in \mathcal F\}=\{A\in \mathcal W\mid f^{-1}(A)\in \mathcal F\}$?

Question

Let $(X,\mathcal F)$ and $(Y,\mathcal W)$ two measure space and $f:X\to Y$. Does $$\{A\subset Y\mid f^{-1}(A)\in \mathcal F\}=\{A\in \mathcal W\mid f^{-1}(A)\in \mathcal F\}\ \ ?$$

The inclusion $$\{A\subset Y\mid f^{-1}(A)\in \mathcal F\}\supset\{A\in \mathcal W\mid f^{-1}(A)\in \mathcal F\},$$ is clear. But I have problem with the converse inclusion. I tried to prove that $\{A\in \mathcal W\mid f^{-1}(A)\in \mathcal F\}$ contain all $A\subset Y$ s.t. $f^{-1}(A)\in \mathcal F$, and since $\{A\in \mathcal W\mid f^{-1}(A)\in \mathcal F\}$ is a $\sigma -$algebra, we would have the result, but it failed.

Answer 1

That's false.

If $f$ is a constant function, then the pre-image of every set is either everything or nothing and both of these are measurable.

In other words the first set becomes $P(Y)$.

Answer 2

No it doesn't hold. For example if $\mathcal W=\{\emptyset, Y\}$, take $\mathcal F=\mathcal B(\mathbb R)$ the borel set and $f(x)=x$. Then $f^{-1}(\{1\})\in \mathcal F$ but $\{1\}\notin \mathcal W$.

Answer 3

Its false. You take in mind that in $W $ there are less set than in all $P(Y) $. In $\mathbb{R}^n $ with Lebesgue measure exists set that not are measurables (Vitali sets).

If you want to prove that

$\Omega = \{f^{-1}(A) : A \in W\} $

Is a $\sigma $-algebra you must use only the definition. For example

$\cup_n f^{-1}(A_n) = f^{-1}(\cup_n An) $

And the numerable union of $A_n $ are in $W $ because $W $ is a $\sigma $-algebra then inversa image of this union are in $\Omega $...