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When does the Riemann Integral-like characterization of Lebesgue Integrals fail?
Wheeden and Zygmund's book "Measure and Integral" gives an interesting characterization of the Lebesgue Integral that is reminiscent of the Riemann Integral. If $E$ be a Lebesgue measurable set, then then the Lebesgue integral of $f$ on $E$ is equal to $supSigma_{k=1}^n (inf_{E_K}f)lambda(E_k)$, where the supremum is taken over all finite partitions $E_1,...,E_n$ of $E$ into Lebesgue measurable sets. (This remains true for arbitrary measure spaces.)
But Wheeden and Zygmund say that if you switch the supremum and infimum, this need not hold true. So my question is, what is an example of a Lebesgue integrable function $f$ such that the Lebesgue integral of $f$ on $E$ is not equal to $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)$?
Also, is there a subset of the set of Lebesgue integrable functions for which the switched version does hold true?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples riemann-integration
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
add a comment |
Wheeden and Zygmund's book "Measure and Integral" gives an interesting characterization of the Lebesgue Integral that is reminiscent of the Riemann Integral. If $E$ be a Lebesgue measurable set, then then the Lebesgue integral of $f$ on $E$ is equal to $supSigma_{k=1}^n (inf_{E_K}f)lambda(E_k)$, where the supremum is taken over all finite partitions $E_1,...,E_n$ of $E$ into Lebesgue measurable sets. (This remains true for arbitrary measure spaces.)
But Wheeden and Zygmund say that if you switch the supremum and infimum, this need not hold true. So my question is, what is an example of a Lebesgue integrable function $f$ such that the Lebesgue integral of $f$ on $E$ is not equal to $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)$?
Also, is there a subset of the set of Lebesgue integrable functions for which the switched version does hold true?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples riemann-integration
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
add a comment |
Wheeden and Zygmund's book "Measure and Integral" gives an interesting characterization of the Lebesgue Integral that is reminiscent of the Riemann Integral. If $E$ be a Lebesgue measurable set, then then the Lebesgue integral of $f$ on $E$ is equal to $supSigma_{k=1}^n (inf_{E_K}f)lambda(E_k)$, where the supremum is taken over all finite partitions $E_1,...,E_n$ of $E$ into Lebesgue measurable sets. (This remains true for arbitrary measure spaces.)
But Wheeden and Zygmund say that if you switch the supremum and infimum, this need not hold true. So my question is, what is an example of a Lebesgue integrable function $f$ such that the Lebesgue integral of $f$ on $E$ is not equal to $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)$?
Also, is there a subset of the set of Lebesgue integrable functions for which the switched version does hold true?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples riemann-integration
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
Wheeden and Zygmund's book "Measure and Integral" gives an interesting characterization of the Lebesgue Integral that is reminiscent of the Riemann Integral. If $E$ be a Lebesgue measurable set, then then the Lebesgue integral of $f$ on $E$ is equal to $supSigma_{k=1}^n (inf_{E_K}f)lambda(E_k)$, where the supremum is taken over all finite partitions $E_1,...,E_n$ of $E$ into Lebesgue measurable sets. (This remains true for arbitrary measure spaces.)
But Wheeden and Zygmund say that if you switch the supremum and infimum, this need not hold true. So my question is, what is an example of a Lebesgue integrable function $f$ such that the Lebesgue integral of $f$ on $E$ is not equal to $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)$?
Also, is there a subset of the set of Lebesgue integrable functions for which the switched version does hold true?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples riemann-integration
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples riemann-integration
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 4:48
Keshav Srinivasan
2,09111441
2,09111441
add a comment |
add a comment |
2 Answers
2
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oldest
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Let $E = (0,1)$ and $f(x) = frac{1}{sqrt{x}}$. Then $int f < infty$ but if $E = sqcup_{k=1}^n E_k$ is a finite partition, then for some $k$, we have $sup_{E_k} f = infty$ and $lambda(E_k) > 0$.
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
add a comment |
Let $E=[0,1]$ and let $f$ be the characteristic function of $E cap mathbb Q$.
answered Nov 20 '18 at 6:34
Fred
44.2k1845
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
1
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
add a comment |
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
Let $E = (0,1)$ and $f(x) = frac{1}{sqrt{x}}$. Then $int f < infty$ but if $E = sqcup_{k=1}^n E_k$ is a finite partition, then for some $k$, we have $sup_{E_k} f = infty$ and $lambda(E_k) > 0$.
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
add a comment |
Let $E = (0,1)$ and $f(x) = frac{1}{sqrt{x}}$. Then $int f < infty$ but if $E = sqcup_{k=1}^n E_k$ is a finite partition, then for some $k$, we have $sup_{E_k} f = infty$ and $lambda(E_k) > 0$.
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
add a comment |
Let $E = (0,1)$ and $f(x) = frac{1}{sqrt{x}}$. Then $int f < infty$ but if $E = sqcup_{k=1}^n E_k$ is a finite partition, then for some $k$, we have $sup_{E_k} f = infty$ and $lambda(E_k) > 0$.
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
Let $E = (0,1)$ and $f(x) = frac{1}{sqrt{x}}$. Then $int f < infty$ but if $E = sqcup_{k=1}^n E_k$ is a finite partition, then for some $k$, we have $sup_{E_k} f = infty$ and $lambda(E_k) > 0$.
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
answered Nov 20 '18 at 15:56
mathworker21
8,6371928
8,6371928
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
add a comment |
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
So is there a subset of the set of Lebesgue integrable functions for which it does hold true?
– Keshav Srinivasan
Nov 21 '18 at 3:10
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
@KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true.
– mathworker21
Nov 21 '18 at 7:23
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds?
– Keshav Srinivasan
Nov 21 '18 at 23:01
add a comment |
Let $E=[0,1]$ and let $f$ be the characteristic function of $E cap mathbb Q$.
answered Nov 20 '18 at 6:34
Fred
44.2k1845
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
1
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
add a comment |
Let $E=[0,1]$ and let $f$ be the characteristic function of $E cap mathbb Q$.
answered Nov 20 '18 at 6:34
Fred
44.2k1845
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
1
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
add a comment |
Let $E=[0,1]$ and let $f$ be the characteristic function of $E cap mathbb Q$.
answered Nov 20 '18 at 6:34
Fred
44.2k1845
Let $E=[0,1]$ and let $f$ be the characteristic function of $E cap mathbb Q$.
answered Nov 20 '18 at 6:34
Fred
44.2k1845
answered Nov 20 '18 at 6:34
Fred
44.2k1845
answered Nov 20 '18 at 6:34
Fred
44.2k1845
answered Nov 20 '18 at 6:34
Fred
44.2k1845
44.2k1845
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
1
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
add a comment |
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
1
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
What partition will give you a nonzero value?
– Keshav Srinivasan
Nov 20 '18 at 12:44
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
For an arbitrary partition $E_1,....,E_n$ we have $ sup_{E_k}f=1$, thus $Sigma_{k=1}^n (sup_{E_K}f)lambda(E_k)= Sigma_{k=1}^n lambda(E_k) = lambda(E)=1$, therefore $infSigma_{k=1}^n (sup_{E_K}f)lambda(E_k)=1 ne 0 = int_E f(x) dx.$
– Fred
Nov 20 '18 at 12:53
1
1
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
@Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = Ecap mathbb{Q}$ and $E_2 = Esetminus E_1$? Then $sup_{E_2} f = 0$ and $lambda(E_1) = 0$, so $sum (sup_{E_k} f) lambda (E_k) = 0$
– mathworker21
Nov 20 '18 at 15:52
add a comment |
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