A function that is continuous almost everywhere is Lebesgue measurable If $f: E \to \mathfrak{M}$ (where $\mathfrak{M}$ is the Lebesgue measurable sets) is continuous a.e., is it true that $f$ is Lebesgue measurable? 
I know that continuous functions on $E \in \mathfrak{M}$ are Lebesgue measurable, but I am wondering if this can be extended to functions that are continuous a.e.? 
My intuition is that the answer is yes. 
Let $D = \{x \in E: f(x) \text{ discontinuous}\}$ and $\alpha \in \mathbb{R}$. Then:
$f^{-1}((-\infty, \alpha)) = ((\{x \in E: f(x) < \alpha\} \setminus D) \cup (\{x \in E: f(x) < \alpha\} \cap D))$
The second set is a subset of $D$, which has measure 0, so it is measurable. But is the first set also measurable? Is there any easier way to prove (or disprove) the statement?

The first set is open, hence measurable. 
Edit: indeed, the first set is not open. 
However, let us denote $S_1$ the first set, $S_2$ the second one, $S=S_1 \cup S_2$. Then $S_2$ has null measure and $S_1 \subset S’ \subset S=S_1 \cup S_2$ where $S’$ is the interior of $S$. 
So $S$ has symmetric difference of null measure with its interior, thus is measurable.
Edit2: Let $x \in S_1$. Then $f(x) < \alpha$ and $f$ is continuous at $x$. Thus, there exists an open interval $J$ containing $x$ such that if $y \in J$, $f(y) < \alpha$, hence $x \in J \subset S$, and since $J$ is open, $x \in S’$. 
