The inverse image of a measurable set under a measurable function is measurable?

Question

I have a confusion with measurable functions. I just saw that the statement "The inverse image of a measurable set under a measurable function is measurable" is false with counter-example the function on Cantor set but I know the definition of f measurable is:

Let $f:(X,O_X)\to (Y,O_Y)$ with $O_X$ $\sigma$-algebra of $X$ and $O_Y$ $\sigma$-algebra of $Y$. $f$ said $(O_X-O_Y)$-measurable function if for all $B\in O_Y\ f^{-1}(B)\in O_X$.

But, What is the difference with "The inverse image of a measurable set under a measurable function is measurable? "

Answer 1

It depends what $\sigma$ algebra you are considering on the target space.

When everyone talks about measureable functions on $\mathbb R$, they mean that $\mathcal O_Y$ is the the $\sigma$-algebra of Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue measurable set may not be measurable.