Prove that a Jordan measurable set of measure 0 is Riemann integrable
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My problem is: I have a set $A$ which is Jordan measurable, and I have a function $f:Arightarrowmathbb{R}$. I need to prove that if the measure of A is $0$, then $f$ is Riemann integrable and furthermore $int_{A}f(x)dx = 0$.
Thank you!
measure-theory riemann-integration
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
$endgroup$
add a comment |
$begingroup$
My problem is: I have a set $A$ which is Jordan measurable, and I have a function $f:Arightarrowmathbb{R}$. I need to prove that if the measure of A is $0$, then $f$ is Riemann integrable and furthermore $int_{A}f(x)dx = 0$.
Thank you!
measure-theory riemann-integration
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
$endgroup$
$begingroup$
You need to show some effort in this and all questions asked here. I sketched a proof below. See if you can fill in the gaps and improve your question.
$endgroup$
– RRL
Jan 14 at 16:18
add a comment |
$begingroup$
My problem is: I have a set $A$ which is Jordan measurable, and I have a function $f:Arightarrowmathbb{R}$. I need to prove that if the measure of A is $0$, then $f$ is Riemann integrable and furthermore $int_{A}f(x)dx = 0$.
Thank you!
measure-theory riemann-integration
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
$endgroup$
My problem is: I have a set $A$ which is Jordan measurable, and I have a function $f:Arightarrowmathbb{R}$. I need to prove that if the measure of A is $0$, then $f$ is Riemann integrable and furthermore $int_{A}f(x)dx = 0$.
Thank you!
measure-theory riemann-integration
measure-theory riemann-integration
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
asked Jan 14 at 14:58
Raducu MihaiRaducu Mihai
345210
345210
$begingroup$
You need to show some effort in this and all questions asked here. I sketched a proof below. See if you can fill in the gaps and improve your question.
$endgroup$
– RRL
Jan 14 at 16:18
add a comment |
$begingroup$
You need to show some effort in this and all questions asked here. I sketched a proof below. See if you can fill in the gaps and improve your question.
$endgroup$
– RRL
Jan 14 at 16:18
$begingroup$
You need to show some effort in this and all questions asked here. I sketched a proof below. See if you can fill in the gaps and improve your question.
$endgroup$
– RRL
Jan 14 at 16:18
$begingroup$
You need to show some effort in this and all questions asked here. I sketched a proof below. See if you can fill in the gaps and improve your question.
$endgroup$
– RRL
Jan 14 at 16:18
add a comment |
1 Answer
1
active
oldest
votes
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Since $A$ is Jordan measureable it is bounded. The Riemann integral over $A$ is defined as $displaystyle int_Q f chi_A$ where $Q$ is a rectangle containing $A$.
(1) Prove first that the integral of $fchi_A$ over $Q$ actually exists by showing that the integrand is continuous except on a set of measure $0$.
(2) Take any partition $P$ of $Q$. Every subrectangle $R$ of $P$ has nonzero content and cannot be contained in $A$. Thus, $R$ always contains a point where $f(x)chi_A(x) = 0$ and
$$inf_{x in R} f(x)chi_A(x) leqslant 0 leqslant sup_{x in R} f(x)chi_A(x), $$
This implies that lower and upper Darboux sums satisfy $L(P, fchi_A) leqslant 0 leqslant U(P,f chi_A )$. From here you should be able to prove that the integral (which falls between lower and upper sums) is $0$.
answered Jan 14 at 16:17
RRLRRL
50.6k42573
$endgroup$
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Since $A$ is Jordan measureable it is bounded. The Riemann integral over $A$ is defined as $displaystyle int_Q f chi_A$ where $Q$ is a rectangle containing $A$.
(1) Prove first that the integral of $fchi_A$ over $Q$ actually exists by showing that the integrand is continuous except on a set of measure $0$.
(2) Take any partition $P$ of $Q$. Every subrectangle $R$ of $P$ has nonzero content and cannot be contained in $A$. Thus, $R$ always contains a point where $f(x)chi_A(x) = 0$ and
$$inf_{x in R} f(x)chi_A(x) leqslant 0 leqslant sup_{x in R} f(x)chi_A(x), $$
This implies that lower and upper Darboux sums satisfy $L(P, fchi_A) leqslant 0 leqslant U(P,f chi_A )$. From here you should be able to prove that the integral (which falls between lower and upper sums) is $0$.
answered Jan 14 at 16:17
RRLRRL
50.6k42573
$endgroup$
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
add a comment |
$begingroup$
Since $A$ is Jordan measureable it is bounded. The Riemann integral over $A$ is defined as $displaystyle int_Q f chi_A$ where $Q$ is a rectangle containing $A$.
(1) Prove first that the integral of $fchi_A$ over $Q$ actually exists by showing that the integrand is continuous except on a set of measure $0$.
(2) Take any partition $P$ of $Q$. Every subrectangle $R$ of $P$ has nonzero content and cannot be contained in $A$. Thus, $R$ always contains a point where $f(x)chi_A(x) = 0$ and
$$inf_{x in R} f(x)chi_A(x) leqslant 0 leqslant sup_{x in R} f(x)chi_A(x), $$
This implies that lower and upper Darboux sums satisfy $L(P, fchi_A) leqslant 0 leqslant U(P,f chi_A )$. From here you should be able to prove that the integral (which falls between lower and upper sums) is $0$.
answered Jan 14 at 16:17
RRLRRL
50.6k42573
$endgroup$
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
add a comment |
$begingroup$
Since $A$ is Jordan measureable it is bounded. The Riemann integral over $A$ is defined as $displaystyle int_Q f chi_A$ where $Q$ is a rectangle containing $A$.
(1) Prove first that the integral of $fchi_A$ over $Q$ actually exists by showing that the integrand is continuous except on a set of measure $0$.
(2) Take any partition $P$ of $Q$. Every subrectangle $R$ of $P$ has nonzero content and cannot be contained in $A$. Thus, $R$ always contains a point where $f(x)chi_A(x) = 0$ and
$$inf_{x in R} f(x)chi_A(x) leqslant 0 leqslant sup_{x in R} f(x)chi_A(x), $$
This implies that lower and upper Darboux sums satisfy $L(P, fchi_A) leqslant 0 leqslant U(P,f chi_A )$. From here you should be able to prove that the integral (which falls between lower and upper sums) is $0$.
answered Jan 14 at 16:17
RRLRRL
50.6k42573
$endgroup$
Since $A$ is Jordan measureable it is bounded. The Riemann integral over $A$ is defined as $displaystyle int_Q f chi_A$ where $Q$ is a rectangle containing $A$.
(1) Prove first that the integral of $fchi_A$ over $Q$ actually exists by showing that the integrand is continuous except on a set of measure $0$.
(2) Take any partition $P$ of $Q$. Every subrectangle $R$ of $P$ has nonzero content and cannot be contained in $A$. Thus, $R$ always contains a point where $f(x)chi_A(x) = 0$ and
$$inf_{x in R} f(x)chi_A(x) leqslant 0 leqslant sup_{x in R} f(x)chi_A(x), $$
This implies that lower and upper Darboux sums satisfy $L(P, fchi_A) leqslant 0 leqslant U(P,f chi_A )$. From here you should be able to prove that the integral (which falls between lower and upper sums) is $0$.
answered Jan 14 at 16:17
RRLRRL
50.6k42573
answered Jan 14 at 16:17
RRLRRL
50.6k42573
answered Jan 14 at 16:17
RRLRRL
50.6k42573
answered Jan 14 at 16:17
RRLRRL
50.6k42573
50.6k42573
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
add a comment |
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
By $chi_A$ you denoted the Jordan measure of $A$, right? I am asking you because I usually denote it by $lambda_A $
$endgroup$
– Raducu Mihai
Jan 15 at 11:37
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
@RaducuMihai: No -- that is not correct. This is the indicator function: $chi_A(x) = 1$ if $x in A$ and $chi_A(x) = 0$ if $x notin A$. We are dealing with Riemann integration. In $mathbb{R}^n$ Riemann integrals are defined first over rectangles where the notion of partitions into subrectangles makes sense. For arbitrary sets the integral is defined as I stated above.
$endgroup$
– RRL
Jan 15 at 22:56
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
$begingroup$
Also I want to help you but I'm hesitant to take this further. Questions without personal input tend to be closed and ultimately deleted.
$endgroup$
– RRL
Jan 15 at 22:58
add a comment |
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$begingroup$
You need to show some effort in this and all questions asked here. I sketched a proof below. See if you can fill in the gaps and improve your question.
$endgroup$
– RRL
Jan 14 at 16:18