0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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, $x \in S’$.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.