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

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? "

• Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain. Nov 6, 2018 at 3:13
• @rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions. Nov 6, 2018 at 4:02

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.