A case where Lebesgue integrable implies Riemann integrable












0












$begingroup$


Let $I$ an interval on $mathbb{R}$ such as $I=(a,b)$, with $a$ or $b$ could be equal to infinity.



And we have $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$, then do we have always
$$int_{(a,b)}fdlambda= int_a^bf(x)dx$$and if not ! When we have this equality, knowing that $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$?






integration lebesgue-integral riemann-integration







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  • 1




    $begingroup$
    Are you looking for this counterexample $f=chi_{mathbb Q}in{mathcal L}^1(0,1)setminus {cal R}(0,1)$?
    $endgroup$
    – Tito Eliatron
    Jan 15 at 20:45
















0












$begingroup$


Let $I$ an interval on $mathbb{R}$ such as $I=(a,b)$, with $a$ or $b$ could be equal to infinity.



And we have $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$, then do we have always
$$int_{(a,b)}fdlambda= int_a^bf(x)dx$$and if not ! When we have this equality, knowing that $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$?






integration lebesgue-integral riemann-integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you looking for this counterexample $f=chi_{mathbb Q}in{mathcal L}^1(0,1)setminus {cal R}(0,1)$?
    $endgroup$
    – Tito Eliatron
    Jan 15 at 20:45














0












0








0





$begingroup$


Let $I$ an interval on $mathbb{R}$ such as $I=(a,b)$, with $a$ or $b$ could be equal to infinity.



And we have $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$, then do we have always
$$int_{(a,b)}fdlambda= int_a^bf(x)dx$$and if not ! When we have this equality, knowing that $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$?






integration lebesgue-integral riemann-integration







share|cite|improve this question











$endgroup$




Let $I$ an interval on $mathbb{R}$ such as $I=(a,b)$, with $a$ or $b$ could be equal to infinity.



And we have $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$, then do we have always
$$int_{(a,b)}fdlambda= int_a^bf(x)dx$$and if not ! When we have this equality, knowing that $fin mathcal{L}^1(I,mathcal{B}(I), lambda)$?






integration lebesgue-integral riemann-integration




integration lebesgue-integral riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 20:42









Bernard

124k741118




124k741118










asked Jan 15 at 20:39









Anas BOUALIIAnas BOUALII

1388




1388








  • 1




    $begingroup$
    Are you looking for this counterexample $f=chi_{mathbb Q}in{mathcal L}^1(0,1)setminus {cal R}(0,1)$?
    $endgroup$
    – Tito Eliatron
    Jan 15 at 20:45














  • 1




    $begingroup$
    Are you looking for this counterexample $f=chi_{mathbb Q}in{mathcal L}^1(0,1)setminus {cal R}(0,1)$?
    $endgroup$
    – Tito Eliatron
    Jan 15 at 20:45








1




1




$begingroup$
Are you looking for this counterexample $f=chi_{mathbb Q}in{mathcal L}^1(0,1)setminus {cal R}(0,1)$?
$endgroup$
– Tito Eliatron
Jan 15 at 20:45




$begingroup$
Are you looking for this counterexample $f=chi_{mathbb Q}in{mathcal L}^1(0,1)setminus {cal R}(0,1)$?
$endgroup$
– Tito Eliatron
Jan 15 at 20:45










1 Answer
1






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oldest

votes


















1












$begingroup$

Take simply the characteristic function of the rationals $f = chi_{mathbb Q}$. $f$ is Lebesgue-integrable (and $int fdlambda$ = 0) but $f$ is not Riemann-integrable.



In the case of bounded functions over closed intervals, a Lebesgue-integrable function $f$ is Riemann-integrable iff the set of discontinuities of $f$ is a null set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:50








  • 3




    $begingroup$
    No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
    $endgroup$
    – Bermudes
    Jan 15 at 20:52












  • $begingroup$
    I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:56










  • $begingroup$
    I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
    $endgroup$
    – Bermudes
    Jan 15 at 21:01






  • 1




    $begingroup$
    Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
    $endgroup$
    – Adrian Keister
    Jan 15 at 21:11












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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1












$begingroup$

Take simply the characteristic function of the rationals $f = chi_{mathbb Q}$. $f$ is Lebesgue-integrable (and $int fdlambda$ = 0) but $f$ is not Riemann-integrable.



In the case of bounded functions over closed intervals, a Lebesgue-integrable function $f$ is Riemann-integrable iff the set of discontinuities of $f$ is a null set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:50








  • 3




    $begingroup$
    No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
    $endgroup$
    – Bermudes
    Jan 15 at 20:52












  • $begingroup$
    I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:56










  • $begingroup$
    I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
    $endgroup$
    – Bermudes
    Jan 15 at 21:01






  • 1




    $begingroup$
    Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
    $endgroup$
    – Adrian Keister
    Jan 15 at 21:11
















1












$begingroup$

Take simply the characteristic function of the rationals $f = chi_{mathbb Q}$. $f$ is Lebesgue-integrable (and $int fdlambda$ = 0) but $f$ is not Riemann-integrable.



In the case of bounded functions over closed intervals, a Lebesgue-integrable function $f$ is Riemann-integrable iff the set of discontinuities of $f$ is a null set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:50








  • 3




    $begingroup$
    No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
    $endgroup$
    – Bermudes
    Jan 15 at 20:52












  • $begingroup$
    I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:56










  • $begingroup$
    I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
    $endgroup$
    – Bermudes
    Jan 15 at 21:01






  • 1




    $begingroup$
    Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
    $endgroup$
    – Adrian Keister
    Jan 15 at 21:11














1












1








1





$begingroup$

Take simply the characteristic function of the rationals $f = chi_{mathbb Q}$. $f$ is Lebesgue-integrable (and $int fdlambda$ = 0) but $f$ is not Riemann-integrable.



In the case of bounded functions over closed intervals, a Lebesgue-integrable function $f$ is Riemann-integrable iff the set of discontinuities of $f$ is a null set.






share|cite|improve this answer











$endgroup$



Take simply the characteristic function of the rationals $f = chi_{mathbb Q}$. $f$ is Lebesgue-integrable (and $int fdlambda$ = 0) but $f$ is not Riemann-integrable.



In the case of bounded functions over closed intervals, a Lebesgue-integrable function $f$ is Riemann-integrable iff the set of discontinuities of $f$ is a null set.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 15 at 21:00

























answered Jan 15 at 20:45









BermudesBermudes

307113




307113












  • $begingroup$
    But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:50








  • 3




    $begingroup$
    No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
    $endgroup$
    – Bermudes
    Jan 15 at 20:52












  • $begingroup$
    I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:56










  • $begingroup$
    I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
    $endgroup$
    – Bermudes
    Jan 15 at 21:01






  • 1




    $begingroup$
    Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
    $endgroup$
    – Adrian Keister
    Jan 15 at 21:11


















  • $begingroup$
    But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:50








  • 3




    $begingroup$
    No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
    $endgroup$
    – Bermudes
    Jan 15 at 20:52












  • $begingroup$
    I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
    $endgroup$
    – Anas BOUALII
    Jan 15 at 20:56










  • $begingroup$
    I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
    $endgroup$
    – Bermudes
    Jan 15 at 21:01






  • 1




    $begingroup$
    Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
    $endgroup$
    – Adrian Keister
    Jan 15 at 21:11
















$begingroup$
But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
$endgroup$
– Anas BOUALII
Jan 15 at 20:50






$begingroup$
But in this example isn't the measure of the set of discontinuity is $0$ since it's $mathbb{Q}$?
$endgroup$
– Anas BOUALII
Jan 15 at 20:50






3




3




$begingroup$
No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
$endgroup$
– Bermudes
Jan 15 at 20:52






$begingroup$
No: $f$ is nowhere continuous. Take any $x in mathbb R$. For every $delta > 0$, you have $sup_{y in B_delta(x)}f(y) = 1$ and $inf_{y in B_delta(x)}f(y) = 0$.
$endgroup$
– Bermudes
Jan 15 at 20:52














$begingroup$
I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
$endgroup$
– Anas BOUALII
Jan 15 at 20:56




$begingroup$
I understand, so under the condition condition of ''continue a.e" we have every lebesgue integrable function is Riemann integrable function.
$endgroup$
– Anas BOUALII
Jan 15 at 20:56












$begingroup$
I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
$endgroup$
– Bermudes
Jan 15 at 21:01




$begingroup$
I edited my answer. This is true for bounded functions over closed intervals. I don't think this is true in general.
$endgroup$
– Bermudes
Jan 15 at 21:01




1




1




$begingroup$
Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
$endgroup$
– Adrian Keister
Jan 15 at 21:11




$begingroup$
Could be wrong, but a function being Lebesgue integrable is theoretically irrelevant as to whether it's Riemann integrable. A function is Riemann integrable iff its set of discontinuities has measure zero. Being Lebesgue integrable doesn't gain you anything, so far as I can make out.
$endgroup$
– Adrian Keister
Jan 15 at 21:11


















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