A function that is continuous almost everywhere is Lebesgue measurable












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?






real-analysis analysis measure-theory lebesgue-measure







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$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – T. Bongers
    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


















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?






real-analysis analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – T. Bongers
    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
















0












0








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?






real-analysis analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$




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?






real-analysis analysis measure-theory lebesgue-measure




real-analysis analysis measure-theory lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 1:30









TuringTester69TuringTester69

307213




307213












  • $begingroup$
    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.
    $endgroup$
    – T. Bongers
    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




















  • $begingroup$
    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.
    $endgroup$
    – T. Bongers
    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


















$begingroup$
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.
$endgroup$
– T. Bongers
Dec 13 '18 at 1:41




$begingroup$
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.
$endgroup$
– T. Bongers
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 ${x in E : f(x) < alpha}$ 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






$begingroup$
@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












1 Answer
1






active

oldest

votes


















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, $x in S’$.






share|cite|improve this answer











$endgroup$









  • 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











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1 Answer
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1 Answer
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active

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active

oldest

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active

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votes









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, $x in S’$.






share|cite|improve this answer











$endgroup$









  • 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
















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, $x in S’$.






share|cite|improve this answer











$endgroup$









  • 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














2












2








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, $x in S’$.






share|cite|improve this answer











$endgroup$



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, $x in S’$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 1:47

























answered Dec 13 '18 at 1:36









MindlackMindlack

4,780210




4,780210








  • 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














  • 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








1




1




$begingroup$
wrong that it's open
$endgroup$
– mathworker21
Dec 13 '18 at 1:37






$begingroup$
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




$begingroup$
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




$begingroup$
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




$begingroup$
+1 nice :)......
$endgroup$
– mathworker21
Dec 13 '18 at 1:47


















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