A case where Lebesgue integrable implies Riemann integrable
$begingroup$
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
edited Jan 15 at 20:42
Bernard
124k741118
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
$endgroup$
add a comment |
$begingroup$
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
edited Jan 15 at 20:42
Bernard
124k741118
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 15 at 20:42
Bernard
124k741118
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
$endgroup$
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
edited Jan 15 at 20:42
Bernard
124k741118
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
edited Jan 15 at 20:42
Bernard
124k741118
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
edited Jan 15 at 20:42
Bernard
124k741118
edited Jan 15 at 20:42
Bernard
124k741118
edited Jan 15 at 20:42
Bernard
124k741118
124k741118
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
asked Jan 15 at 20:39
Anas BOUALIIAnas BOUALII
1388
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 15 at 20:45
BermudesBermudes
307113
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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$begingroup$
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.
answered Jan 15 at 20:45
BermudesBermudes
307113
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Jan 15 at 20:45
BermudesBermudes
307113
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Jan 15 at 20:45
BermudesBermudes
307113
$endgroup$
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.
answered Jan 15 at 20:45
BermudesBermudes
307113
edited Jan 15 at 21:00
answered Jan 15 at 20:45
BermudesBermudes
307113
answered Jan 15 at 20:45
BermudesBermudes
307113
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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