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Proving the Riemann integrability of a function [closed]
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Let $f : [a, b] to mathbbR$ be a Riemann integrable function. If $g : [a, b] to mathbbR$ is another function and $S = lbrace x : f(x) neq g(x)rbrace$ contains exactly $n$ points, show that $g$ is also Riemann integrable from any of the equivalent definitions of Riemann integrability. Can anyone give me some hints?
riemann-integration
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
asked Jul 18 at 20:56
Walls
31
closed as off-topic by amWhy, user223391, Trần Thúc Minh TrÃ, Isaac Browne, John Ma Jul 19 at 4:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Trần Thúc Minh TrÃÂ, Isaac Browne, John Ma
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Let $f : [a, b] to mathbbR$ be a Riemann integrable function. If $g : [a, b] to mathbbR$ is another function and $S = lbrace x : f(x) neq g(x)rbrace$ contains exactly $n$ points, show that $g$ is also Riemann integrable from any of the equivalent definitions of Riemann integrability. Can anyone give me some hints?
riemann-integration
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
asked Jul 18 at 20:56
Walls
31
closed as off-topic by amWhy, user223391, Trần Thúc Minh TrÃ, Isaac Browne, John Ma Jul 19 at 4:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Trần Thúc Minh TrÃÂ, Isaac Browne, John Ma
2
One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero.
– Davey
Jul 18 at 21:02
1
The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero.
– user577471
Jul 18 at 21:02
These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps.
– Pi_die_die
Jul 18 at 21:03
One of my concerns is that what dose the set S imply?
– Walls
Jul 18 at 21:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f : [a, b] to mathbbR$ be a Riemann integrable function. If $g : [a, b] to mathbbR$ is another function and $S = lbrace x : f(x) neq g(x)rbrace$ contains exactly $n$ points, show that $g$ is also Riemann integrable from any of the equivalent definitions of Riemann integrability. Can anyone give me some hints?
riemann-integration
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
asked Jul 18 at 20:56
Walls
31
Let $f : [a, b] to mathbbR$ be a Riemann integrable function. If $g : [a, b] to mathbbR$ is another function and $S = lbrace x : f(x) neq g(x)rbrace$ contains exactly $n$ points, show that $g$ is also Riemann integrable from any of the equivalent definitions of Riemann integrability. Can anyone give me some hints?
riemann-integration
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
asked Jul 18 at 20:56
Walls
31
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
edited Jul 18 at 22:15
Theo Bendit
12.1k1843
12.1k1843
asked Jul 18 at 20:56
Walls
31
asked Jul 18 at 20:56
Walls
31
asked Jul 18 at 20:56
Walls
31
31
closed as off-topic by amWhy, user223391, Trần Thúc Minh TrÃ, Isaac Browne, John Ma Jul 19 at 4:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Trần Thúc Minh TrÃÂ, Isaac Browne, John Ma
closed as off-topic by amWhy, user223391, Trần Thúc Minh TrÃ, Isaac Browne, John Ma Jul 19 at 4:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Trần Thúc Minh TrÃÂ, Isaac Browne, John Ma
2
One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero.
– Davey
Jul 18 at 21:02
1
The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero.
– user577471
Jul 18 at 21:02
These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps.
– Pi_die_die
Jul 18 at 21:03
One of my concerns is that what dose the set S imply?
– Walls
Jul 18 at 21:07
add a comment |Â
2
One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero.
– Davey
Jul 18 at 21:02
1
The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero.
– user577471
Jul 18 at 21:02
These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps.
– Pi_die_die
Jul 18 at 21:03
One of my concerns is that what dose the set S imply?
– Walls
Jul 18 at 21:07
2
2
One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero.
– Davey
Jul 18 at 21:02
One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero.
– Davey
Jul 18 at 21:02
1
1
The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero.
– user577471
Jul 18 at 21:02
The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero.
– user577471
Jul 18 at 21:02
These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps.
– Pi_die_die
Jul 18 at 21:03
These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps.
– Pi_die_die
Jul 18 at 21:03
One of my concerns is that what dose the set S imply?
– Walls
Jul 18 at 21:07
One of my concerns is that what dose the set S imply?
– Walls
Jul 18 at 21:07
add a comment |Â
1 Answer
1
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oldest
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up vote
2
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Let $S:=x_1,x_2,cdots,x_n$.
I will suppose below that $a<x_1$ and $x_n<b$.
Clearly, $S$ is isolated.
Thus, we can find $delta>0$ such that $x_k+1-x_k>2delta$.
We may also suppose that $x_1-a>delta$ and $b-x_n>delta$.
Now, consider the following.
beginalign
int_a^bf(u)mathrmdu
=&int_a^x_1-deltaf(u)mathrmdu+int_x_1+delta^x_2-deltaf(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltaf(u)mathrmdu+int_x_n+delta^bf(u)mathrmdu\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu\
=&int_a^x_1-deltag(u)mathrmdu+int_x_1+delta^x_2-deltag(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltag(u)mathrmdu+int_x_n+delta^bg(u)mathrmdutag1labeleq1\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu.tag2labeleq2
endalign
Note that as $deltato0$, each term in eqrefeq2 tends to $0$,
and eqrefeq1 becomes $int_a^bg(u)mathrmdu$.
answered Jul 18 at 21:27
bkarpuz
457210
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $S:=x_1,x_2,cdots,x_n$.
I will suppose below that $a<x_1$ and $x_n<b$.
Clearly, $S$ is isolated.
Thus, we can find $delta>0$ such that $x_k+1-x_k>2delta$.
We may also suppose that $x_1-a>delta$ and $b-x_n>delta$.
Now, consider the following.
beginalign
int_a^bf(u)mathrmdu
=&int_a^x_1-deltaf(u)mathrmdu+int_x_1+delta^x_2-deltaf(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltaf(u)mathrmdu+int_x_n+delta^bf(u)mathrmdu\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu\
=&int_a^x_1-deltag(u)mathrmdu+int_x_1+delta^x_2-deltag(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltag(u)mathrmdu+int_x_n+delta^bg(u)mathrmdutag1labeleq1\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu.tag2labeleq2
endalign
Note that as $deltato0$, each term in eqrefeq2 tends to $0$,
and eqrefeq1 becomes $int_a^bg(u)mathrmdu$.
answered Jul 18 at 21:27
bkarpuz
457210
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
add a comment |Â
up vote
2
down vote
accepted
Let $S:=x_1,x_2,cdots,x_n$.
I will suppose below that $a<x_1$ and $x_n<b$.
Clearly, $S$ is isolated.
Thus, we can find $delta>0$ such that $x_k+1-x_k>2delta$.
We may also suppose that $x_1-a>delta$ and $b-x_n>delta$.
Now, consider the following.
beginalign
int_a^bf(u)mathrmdu
=&int_a^x_1-deltaf(u)mathrmdu+int_x_1+delta^x_2-deltaf(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltaf(u)mathrmdu+int_x_n+delta^bf(u)mathrmdu\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu\
=&int_a^x_1-deltag(u)mathrmdu+int_x_1+delta^x_2-deltag(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltag(u)mathrmdu+int_x_n+delta^bg(u)mathrmdutag1labeleq1\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu.tag2labeleq2
endalign
Note that as $deltato0$, each term in eqrefeq2 tends to $0$,
and eqrefeq1 becomes $int_a^bg(u)mathrmdu$.
answered Jul 18 at 21:27
bkarpuz
457210
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $S:=x_1,x_2,cdots,x_n$.
I will suppose below that $a<x_1$ and $x_n<b$.
Clearly, $S$ is isolated.
Thus, we can find $delta>0$ such that $x_k+1-x_k>2delta$.
We may also suppose that $x_1-a>delta$ and $b-x_n>delta$.
Now, consider the following.
beginalign
int_a^bf(u)mathrmdu
=&int_a^x_1-deltaf(u)mathrmdu+int_x_1+delta^x_2-deltaf(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltaf(u)mathrmdu+int_x_n+delta^bf(u)mathrmdu\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu\
=&int_a^x_1-deltag(u)mathrmdu+int_x_1+delta^x_2-deltag(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltag(u)mathrmdu+int_x_n+delta^bg(u)mathrmdutag1labeleq1\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu.tag2labeleq2
endalign
Note that as $deltato0$, each term in eqrefeq2 tends to $0$,
and eqrefeq1 becomes $int_a^bg(u)mathrmdu$.
answered Jul 18 at 21:27
bkarpuz
457210
Let $S:=x_1,x_2,cdots,x_n$.
I will suppose below that $a<x_1$ and $x_n<b$.
Clearly, $S$ is isolated.
Thus, we can find $delta>0$ such that $x_k+1-x_k>2delta$.
We may also suppose that $x_1-a>delta$ and $b-x_n>delta$.
Now, consider the following.
beginalign
int_a^bf(u)mathrmdu
=&int_a^x_1-deltaf(u)mathrmdu+int_x_1+delta^x_2-deltaf(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltaf(u)mathrmdu+int_x_n+delta^bf(u)mathrmdu\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu\
=&int_a^x_1-deltag(u)mathrmdu+int_x_1+delta^x_2-deltag(u)mathrmdu+cdots+int_x_n-1+delta^x_n-deltag(u)mathrmdu+int_x_n+delta^bg(u)mathrmdutag1labeleq1\
&+int_x_1-delta^x_1+deltaf(u)mathrmdu+cdots+int_x_n-delta^x_n+deltaf(u)mathrmdu.tag2labeleq2
endalign
Note that as $deltato0$, each term in eqrefeq2 tends to $0$,
and eqrefeq1 becomes $int_a^bg(u)mathrmdu$.
answered Jul 18 at 21:27
bkarpuz
457210
edited Jul 22 at 17:23
answered Jul 18 at 21:27
bkarpuz
457210
answered Jul 18 at 21:27
bkarpuz
457210
answered Jul 18 at 21:27
bkarpuz
457210
457210
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
add a comment |Â
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
Minor typo: I think the integrals in the last line of your equation should have $g(u)$ as the integrand.
– hryghr
Jul 18 at 21:57
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
@hryghr Nope, $f$ and $g$ are different on those intervals (but sure the integrals are the same), so keep it as is.
– bkarpuz
Jul 18 at 22:30
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
My bad, misread the integration limits.
– hryghr
Jul 23 at 12:33
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
Actually, a proof by using partitions would be much better. I can say that this is some kind of cheating...
– bkarpuz
Jul 23 at 14:53
add a comment |Â
2
One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero.
– Davey
Jul 18 at 21:02
1
The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero.
– user577471
Jul 18 at 21:02
These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps.
– Pi_die_die
Jul 18 at 21:03
One of my concerns is that what dose the set S imply?
– Walls
Jul 18 at 21:07