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?

  • $\begingroup$ The first set is of the form $\mathcal{O} \setminus (D \cap \mathcal{O})$ where $\mathcal{O}$ is open, and $D \cap \mathcal{O}$ is a subset of a set of measure zero. $\endgroup$ – user296602 Dec 13 '18 at 1:41
  • $\begingroup$ @T.Bongers $\{x \in E : f(x) < \alpha\}$ is not open necessarily $\endgroup$ – mathworker21 Dec 13 '18 at 1:43
  • $\begingroup$ @T.Bongers there’s no reason why the first one should necessarily be open. If $f$ is the 0 function on $[0,1]$ then it is continuous and for any positive $\alpha$, $\{x \in [0,1] : f(x) < \alpha\} = [0,1]$. $\endgroup$ – TuringTester69 Dec 13 '18 at 1:46

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, $x \in S’$.

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  • 1
    $\begingroup$ wrong that it's open $\endgroup$ – mathworker21 Dec 13 '18 at 1:37
  • $\begingroup$ Is it? Maybe I’m missing something but that doesn’t seem obvious to me. $\endgroup$ – TuringTester69 Dec 13 '18 at 1:37
  • $\begingroup$ You are right, I am editing. $\endgroup$ – Mindlack Dec 13 '18 at 1:39
  • $\begingroup$ +1 nice :)...... $\endgroup$ – mathworker21 Dec 13 '18 at 1:47

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