Changing value of a Riemann integrable function on a Lebesgue measure 0 set implies the new function has the...
up vote
0
down vote
favorite
Suppose $[a,b]$ is a compact interval of $mathbb{R}$ and $f:[a,b]tomathbb{R}$ be integrable in the Riemann sense. Then, by Lebesgue's criterion, $f$ is bounded on $[a,b]$ and it's set of discontinuities has Lebesgue measure zero.
Now suppose $tilde{f}:[a,b]tomathbb{R}$ is a new function built by changing the value of $f$ in a Lebesgue measure zero subset of $[a,b]$. Since $tilde{f}$ remains bounded and it's set of discontinuities has still Lebesgue measure zero, we know that $tilde{f}$ is Riemann integrable.
Is it true that $$int_{a}^{b}f=int_{a}^{b}tilde{f}qquad ?$$
real-analysis integration definite-integrals riemann-integration
asked Nov 25 at 11:36
eleguitar
113114
add a comment |
up vote
0
down vote
favorite
Suppose $[a,b]$ is a compact interval of $mathbb{R}$ and $f:[a,b]tomathbb{R}$ be integrable in the Riemann sense. Then, by Lebesgue's criterion, $f$ is bounded on $[a,b]$ and it's set of discontinuities has Lebesgue measure zero.
Now suppose $tilde{f}:[a,b]tomathbb{R}$ is a new function built by changing the value of $f$ in a Lebesgue measure zero subset of $[a,b]$. Since $tilde{f}$ remains bounded and it's set of discontinuities has still Lebesgue measure zero, we know that $tilde{f}$ is Riemann integrable.
Is it true that $$int_{a}^{b}f=int_{a}^{b}tilde{f}qquad ?$$
real-analysis integration definite-integrals riemann-integration
asked Nov 25 at 11:36
eleguitar
113114
1
One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere.
– Lord Shark the Unknown
Nov 25 at 11:42
So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:45
If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence
– Hagen von Eitzen
Nov 25 at 11:46
You could say that if $int f$ and $int tilde f$ exist and $tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal.
– Hagen von Eitzen
Nov 25 at 11:48
There's no reason to think $tilde f$ is bounded.
– zhw.
Nov 25 at 19:19
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $[a,b]$ is a compact interval of $mathbb{R}$ and $f:[a,b]tomathbb{R}$ be integrable in the Riemann sense. Then, by Lebesgue's criterion, $f$ is bounded on $[a,b]$ and it's set of discontinuities has Lebesgue measure zero.
Now suppose $tilde{f}:[a,b]tomathbb{R}$ is a new function built by changing the value of $f$ in a Lebesgue measure zero subset of $[a,b]$. Since $tilde{f}$ remains bounded and it's set of discontinuities has still Lebesgue measure zero, we know that $tilde{f}$ is Riemann integrable.
Is it true that $$int_{a}^{b}f=int_{a}^{b}tilde{f}qquad ?$$
real-analysis integration definite-integrals riemann-integration
asked Nov 25 at 11:36
eleguitar
113114
Suppose $[a,b]$ is a compact interval of $mathbb{R}$ and $f:[a,b]tomathbb{R}$ be integrable in the Riemann sense. Then, by Lebesgue's criterion, $f$ is bounded on $[a,b]$ and it's set of discontinuities has Lebesgue measure zero.
Now suppose $tilde{f}:[a,b]tomathbb{R}$ is a new function built by changing the value of $f$ in a Lebesgue measure zero subset of $[a,b]$. Since $tilde{f}$ remains bounded and it's set of discontinuities has still Lebesgue measure zero, we know that $tilde{f}$ is Riemann integrable.
Is it true that $$int_{a}^{b}f=int_{a}^{b}tilde{f}qquad ?$$
real-analysis integration definite-integrals riemann-integration
real-analysis integration definite-integrals riemann-integration
asked Nov 25 at 11:36
eleguitar
113114
asked Nov 25 at 11:36
eleguitar
113114
asked Nov 25 at 11:36
eleguitar
113114
asked Nov 25 at 11:36
eleguitar
113114
asked Nov 25 at 11:36
eleguitar
113114
113114
1
One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere.
– Lord Shark the Unknown
Nov 25 at 11:42
So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:45
If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence
– Hagen von Eitzen
Nov 25 at 11:46
You could say that if $int f$ and $int tilde f$ exist and $tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal.
– Hagen von Eitzen
Nov 25 at 11:48
There's no reason to think $tilde f$ is bounded.
– zhw.
Nov 25 at 19:19
add a comment |
1
One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere.
– Lord Shark the Unknown
Nov 25 at 11:42
So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:45
If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence
– Hagen von Eitzen
Nov 25 at 11:46
You could say that if $int f$ and $int tilde f$ exist and $tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal.
– Hagen von Eitzen
Nov 25 at 11:48
There's no reason to think $tilde f$ is bounded.
– zhw.
Nov 25 at 19:19
1
1
One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere.
– Lord Shark the Unknown
Nov 25 at 11:42
One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere.
– Lord Shark the Unknown
Nov 25 at 11:42
So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:45
So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:45
If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence
– Hagen von Eitzen
Nov 25 at 11:46
If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence
– Hagen von Eitzen
Nov 25 at 11:46
You could say that if $int f$ and $int tilde f$ exist and $tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal.
– Hagen von Eitzen
Nov 25 at 11:48
You could say that if $int f$ and $int tilde f$ exist and $tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal.
– Hagen von Eitzen
Nov 25 at 11:48
There's no reason to think $tilde f$ is bounded.
– zhw.
Nov 25 at 19:19
There's no reason to think $tilde f$ is bounded.
– zhw.
Nov 25 at 19:19
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $f=0$ and $g(x)=0$ for $x$ irrational, $1$ for $x$ rational. Then $f$ is Riemann integrable, $f=g$ almost everywhere but $g$ is discontinuous everywhere, is it is not Riemann integrable.
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
1
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $f=0$ and $g(x)=0$ for $x$ irrational, $1$ for $x$ rational. Then $f$ is Riemann integrable, $f=g$ almost everywhere but $g$ is discontinuous everywhere, is it is not Riemann integrable.
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
1
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
add a comment |
up vote
1
down vote
accepted
Let $f=0$ and $g(x)=0$ for $x$ irrational, $1$ for $x$ rational. Then $f$ is Riemann integrable, $f=g$ almost everywhere but $g$ is discontinuous everywhere, is it is not Riemann integrable.
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
1
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f=0$ and $g(x)=0$ for $x$ irrational, $1$ for $x$ rational. Then $f$ is Riemann integrable, $f=g$ almost everywhere but $g$ is discontinuous everywhere, is it is not Riemann integrable.
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
Let $f=0$ and $g(x)=0$ for $x$ irrational, $1$ for $x$ rational. Then $f$ is Riemann integrable, $f=g$ almost everywhere but $g$ is discontinuous everywhere, is it is not Riemann integrable.
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
answered Nov 25 at 11:46
Kavi Rama Murthy
45.8k31853
45.8k31853
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
1
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
add a comment |
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
1
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:55
1
1
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points.
– Kavi Rama Murthy
Nov 25 at 11:59
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/…
– eleguitar
Nov 25 at 12:06
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012728%2fchanging-value-of-a-riemann-integrable-function-on-a-lebesgue-measure-0-set-impl%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere.
– Lord Shark the Unknown
Nov 25 at 11:42
So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right?
– eleguitar
Nov 25 at 11:45
If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence
– Hagen von Eitzen
Nov 25 at 11:46
You could say that if $int f$ and $int tilde f$ exist and $tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal.
– Hagen von Eitzen
Nov 25 at 11:48
There's no reason to think $tilde f$ is bounded.
– zhw.
Nov 25 at 19:19