Preimage of open set is Lebesgue measurable only if the function itself is measurable It is a simple result in my book saying the proof is trivial, but I can not seem to show it. If someone can provide a hint just to help me begin my proof, it would be of assistance.
Assume you know that $g:D\to\mathbb{R}$ is Lebesgue measurable. Then how to go about showing that $g^{-1}(U)$ is also measurable, where $U\subset \mathbb{R}$ is open?
 A: According to your definition of measurability, we know that $g^{-1}((-\infty,a))$ is measurable for all $a\in\mathbb{R}$. This implies that
$$ g^{-1}([a,b))=g^{-1}((-\infty,b))\setminus g^{-1}((-\infty,a))$$
is measurable for all real numbers $a<b$, and by taking unions of such intervals we see that sets of the form $g^{-1}((a,b))$ and $g^{-1}((a,\infty))$ are measurable for all $a,b\in\mathbb{R}$. In other words $g^{-1}(I)$ is measurable for all open intervals $I$.
Finally, since every open subset $U\subset\mathbb{R}$ can be written as a countable union of disjoint open intervals $I_j$, it follows that $g^{-1}(U)=\cup_jg^{-1}(I_j)$ is measurable.
A: For the completeness of @carmichael561 answer:
If $\mathcal{G}:\mathbb{R}\rightarrow \mathbb{R}$ is Borel-measurable, then for any set $U \in \mathcal{B}$ (where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$) i.e.: $\mathcal{G}^{-1}(U) \in \mathcal{B}$ (this is by definition of measurable function).
A $\sigma$-algebra is closed under unions and intersections, so if $I1, I2 \in \mathcal{B}$ then: $I1 \cup I2$ and $I1 \cap I2$ are in $\mathcal{B}$. This is for all set in the $\sigma$-algebra.
