Let $(\Bbb R,\mathcal A_{\Bbb R}^*,\bar\lambda)$ be the complete Lebesgue measure space. Let $f:\Bbb R\to\Bbb R$ be a function, so the theorem says that:
$$f\;\text{is Lebesgue measurable}\Leftrightarrow\forall\epsilon>0\;\exists\ E\;\text{open}\;\text{with}\;\bar\lambda(E)<\epsilon\;\text{s.t.}\;\;f\mid_{E^c}:\Bbb R\setminus E\to\Bbb R\;\text{is continuous}\;\;$$
I have proved the $(\mid\Rightarrow)$ part, so I'm trying to proved the $(\Leftarrow\mid)$ part. And what I got so far is this:
Proof:
Let $A\in\mathcal A_{\Bbb R}^*$, so I want to prove that:
$$f^{-1}(A)\in\mathcal A_{\Bbb R}^*$$
By hypothesis, we know that $\forall i\in\Bbb N\;\exists\ E_i$ open with $\bar\lambda(E_i)<\frac{1}{i}$ s.t. $f_i=f\mid_{E_i^c}$ is continuous. So let $E=\bigcap_{i=1}^\infty E_i$ thus $\bar\lambda(E)=0$ and $\;\forall U\subset\Bbb R$ open we get that:
$$f_i^{-1}(U)=E_i^c\cap f^{-1}(U)=f^{-1}(U)\setminus E_i\;\;\forall i\in\Bbb N\\ \Rightarrow\ \bigcup_{i=1}^\infty f_i^{-1}(U)=f^{-1}(U)\setminus E$$
thus $f^{-1}(U)\setminus E$ is open $\;\forall U\subset\Bbb R$ open. And stuck here since I don't how to get to the point that $f$ should be Lebesgue measurable. The only two things I've came up with is: (1) Try to show that $f$ is continuous, but can't see how from here and the other one is (2) For given $\epsilon>0$ try to find some open set $G$ s.t. $f^{-1}(A)\subset G$ and $\bar\lambda(G\setminus f^{-1}(A))<\epsilon$ but I don't get find it clearly.
Any help would be appreciated.
I think I got it. So continuiung with a third option, we know that:
$$f^{-1}(A)=\big(f^{-1}(A)\setminus E\big)\;\cup\; \big(f^{-1}(A)\cap E\big)$$
where we know that, since $f_i$ is continuous $\forall i\in\Bbb N$, $f_i$ is Lebesgue measurable $\forall i\in\Bbb N$ and thus $f_i^{-1}(A)\in\mathcal A_{\Bbb R}^*\;\;\forall i\in\Bbb N$, but:
$$f^{-1}(A)\setminus E=f^{-1}(A)\setminus\bigcap_{i\in\Bbb N}E_i=\bigcup_{i\in\Bbb N}\big(f^{-1}(A)\setminus E_i\big)=\bigcup_{i\in\Bbb N}\big(f\mid_{E_i^c}^{-1}(A)\big)=\bigcup_{i=1}^\infty f_i^{-1}(A)\in\mathcal A_{\Bbb R}^*$$
thus $f^{-1}(A)\setminus E\in\mathcal A_{\Bbb R}^*$.
And since $f^{-1}(A)\cap E\subset E\Rightarrow\lambda^*\big(f^{-1}(A)\cap E\big)\le \lambda^*(E)=\bar\lambda(E)=0$ thus $f^{-1}(A)\cap E\in\mathcal A_{\Bbb R}^*$.
Thereby $f^{-1}(A)\in\mathcal A_{\Bbb R}^*$ and hence $f$ is Lebesgue measurable.