$f$ is measurable $\iff$ $f^{-1}(G)$ is measurable for open $G$. $E$ : Lebesgue measurable set
$f$ : $E \to \mathbb{R}$
I want to prove
$f$ is Lebesgue-measurable function
$\iff$ $f^{-1}(G)$ is Lebesgue measurable for all open set $G$.
($\Leftarrow$) is O.K. For all $a\in \mathbb{R}$, $\{ f>a \} = \{x\in E | f(x)>a \} = f^{-1}((a, \infty)).$
Since $(a, \infty)$ is open, $f^{-1}((a, \infty))$ is Lebesugue-measurable and so is $\{ f>a \}$.
But I couldn't prove $(\Rightarrow)$.
My attempt is as following.
Let $G$ open. Since $G$ is open, I can write $G=\cup_{n=1}^{\infty} O_n$, where each $O_i$ is open interval and $O_i \cap O_j=\phi (i \neq j).$ $f^{-1}(G)= f^{-1}( \cup_{n=1}^{\infty} O_n)=\cup_{n=1}^{\infty} f^{-1}(O_n)$.
I cannot proceed from here.
I would like you to give me some ideas.
 A: If by a Lebesgue set you mean a set in the closure of the Borel $\sigma$-algebra $\mathscr{B}$ with respect the Lebesgue Measure, and by Lebesgue measurable you mean that $f^{-1}(L)$ is a measurable subset of $E$ for all Lebesgue sets $L\subset \mathbb{R}$, then the  statement in your problem in one direction is not true in general.

The statement $f^{-1}(G)$ is a measurable subset of $E$ for any open set $G\subset\mathbb{R}$ is equivalent to the statement that $f$ is (Borel) measurable.  This is because the collection $$\mathcal{C}=\{U\subset\mathbb{R}: f^{-1}(U)\,\text{is a measurable subset of }\,E\}$$
is a $\sigma$-algebra. Since $\mathcal{C}$ contains all the open subsets of $\mathbb{R}$, then it also contains all the Borel sets of $\mathbb{R}$ (on account that $\mathscr{B}$ is the minimal $\sigma$-algebra containing the open sets). This shows that $f$ is (Borel) measurable.
Certainly, if $f$ is Lebesgue measurable, then it is also Borel measurable since the Lebesgue $\sigma$-algebra contains the Borel $\sigma$-algebra. The converse however is not true in general.
Here is a counter example. Consider the Cantor function $f$. The function
$$g(y)=\inf\{x:f(x)=y\}$$
Satisfies $f(g(y))=y$ and so $g$ is injective, $g$ takes values on the $1/3$ Cantor set $C$, and $g$ is monotone nondecreasing. Equipe $[0,1]$ with the Lebesgue $\sigma$-algebra $\mathscr{M}_\lambda([0,1])$ in $[0,1]$.  The fact that $g$ is monotone implies that $g$ is (Borel) measurable, that is $g^{-1}(U)\in\mathscr{B}([0,1])\subset\mathscr{M}_\mu([0,1])$ for all $U\in\mathscr{B}(\mathbb{R})$. Let $A\subset [0,1]$ be a non Lebesgue measurable set (a Vitali set for example). As  $B=g(A)\subset C$, and $\lambda(C)=0$, we have that $B$ is Lebesgue measurable (Lebesgue Measure is complete). The punchline is that $g^{-1}(B)=A$ is not Lebesgue measurable.
