Lebesgue integration: integral of continuous function tends to infinity
up vote
0
down vote
favorite
I'm studying measure theory and Lebesgue integration and I've run into this problem:
let $ f:R rightarrow R$ be a continuous function such that:
- $int f^+dlambda_1 = int f^-dlambda_1 =+infty $
show that $c in R$, then there is $Ain mathscr{B}(R)$ such that $int_A fdlambda_1 = c$.
I know that because of the continuity of $f$, there must be an $A$ such that $0 <int_A fdlambda_1 <infty $. I was thinking of a proof along the lines of the proof of the Rienman series theorem, but I am not sure how to proceed. Some help would be highly appreciated.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 19 at 14:42
jffi
103
add a comment |
up vote
0
down vote
favorite
I'm studying measure theory and Lebesgue integration and I've run into this problem:
let $ f:R rightarrow R$ be a continuous function such that:
- $int f^+dlambda_1 = int f^-dlambda_1 =+infty $
show that $c in R$, then there is $Ain mathscr{B}(R)$ such that $int_A fdlambda_1 = c$.
I know that because of the continuity of $f$, there must be an $A$ such that $0 <int_A fdlambda_1 <infty $. I was thinking of a proof along the lines of the proof of the Rienman series theorem, but I am not sure how to proceed. Some help would be highly appreciated.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 19 at 14:42
jffi
103
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
Nov 19 at 14:45
Yes I meant for all $c in mathbb{R}$. Could you perhaps elaborate a bit more on your proof idea ? I don’t get how you can reach this result from the theorem.
– jffi
Nov 19 at 15:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm studying measure theory and Lebesgue integration and I've run into this problem:
let $ f:R rightarrow R$ be a continuous function such that:
- $int f^+dlambda_1 = int f^-dlambda_1 =+infty $
show that $c in R$, then there is $Ain mathscr{B}(R)$ such that $int_A fdlambda_1 = c$.
I know that because of the continuity of $f$, there must be an $A$ such that $0 <int_A fdlambda_1 <infty $. I was thinking of a proof along the lines of the proof of the Rienman series theorem, but I am not sure how to proceed. Some help would be highly appreciated.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 19 at 14:42
jffi
103
I'm studying measure theory and Lebesgue integration and I've run into this problem:
let $ f:R rightarrow R$ be a continuous function such that:
- $int f^+dlambda_1 = int f^-dlambda_1 =+infty $
show that $c in R$, then there is $Ain mathscr{B}(R)$ such that $int_A fdlambda_1 = c$.
I know that because of the continuity of $f$, there must be an $A$ such that $0 <int_A fdlambda_1 <infty $. I was thinking of a proof along the lines of the proof of the Rienman series theorem, but I am not sure how to proceed. Some help would be highly appreciated.
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Nov 19 at 14:42
jffi
103
asked Nov 19 at 14:42
jffi
103
asked Nov 19 at 14:42
jffi
103
asked Nov 19 at 14:42
jffi
103
asked Nov 19 at 14:42
jffi
103
103
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
Nov 19 at 14:45
Yes I meant for all $c in mathbb{R}$. Could you perhaps elaborate a bit more on your proof idea ? I don’t get how you can reach this result from the theorem.
– jffi
Nov 19 at 15:51
add a comment |
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
Nov 19 at 14:45
Yes I meant for all $c in mathbb{R}$. Could you perhaps elaborate a bit more on your proof idea ? I don’t get how you can reach this result from the theorem.
– jffi
Nov 19 at 15:51
2
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
Nov 19 at 14:45
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
Nov 19 at 14:45
Yes I meant for all $c in mathbb{R}$. Could you perhaps elaborate a bit more on your proof idea ? I don’t get how you can reach this result from the theorem.
– jffi
Nov 19 at 15:51
Yes I meant for all $c in mathbb{R}$. Could you perhaps elaborate a bit more on your proof idea ? I don’t get how you can reach this result from the theorem.
– jffi
Nov 19 at 15:51
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
I'll give it a shot.
Consider $A = {x in mathbb{R}, f(x) geq 0}$ which is well defined since $f$ is continuous.
Consider now $g(x) = int_{A cap (0,x)} f(y) dy + int_{A cap (-x,0)} f(y) dy = int_{-x}^x f(y) mathbf{1}_A(y) dy$.
Clearly $g(0) = 0$. Moreover, $lim limits_{x to infty} g(x) = infty$ by your condition on $f_+$. It only remains to show that $g$ is continuous to use the intermediate value theorem.
Usually, to get $g$ continuous, (see for instance this http://mathonline.wikidot.com/continuity-of-functions-defined-by-lebesgue-integrals), you would need $f_+$ to be bounded by an integrable function, but in fact since $f$ is continuous, on any compact set you have that $f_+$ is bounded (Weierstrass theorem). This is enough for your purpose: for any $c > 0$, it exists $x_c > 0$ such that $g(x_c) > c$ (otherwise the limit of $g$ can not be $+ infty$). Then on the compact $[-x_c - 1, x_c + 1]$, your function $g$ is clearly continuous (since then $f_+$ is bounded) by applying Theorem 1 in the link above.
So, $g(0) = 0, g(x_c) > c$ and g is continuous, therefore it exists $alpha$ such that $g(alpha) = c$ (intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem).
You can do the exact same thing for $c < 0$ by changing the definition of $A$.
answered Nov 20 at 12:07
Gâteau-Gallois
347111
add a comment |
up vote
0
down vote
Consider $A:={x:f^+(x)>0}$ and $B:={x:f^-(x)>0}$. If $c>0$ then define a function $g:[0,infty)rightarrow [0,infty)$ such that $g(r)=int_{[-r,r]}f^+(x)dx=int_{Acap[-r,r]}f(x)dx.$ Notice that $g(0)=0$ and $g(r)rightarrowinfty$ as $rrightarrow infty.$ And $|g(r)-g(s)|=int_{[-r,-s)}f^+(x)dx+int_{(s,r]}f^+(x)dxrightarrow 0$ as $suparrow r.$ So, $g$ is a continuous function. By mean value theorem $exists b>0$ such that $g(b)=c.$ Take $A_0=[-b,b]cap A.$ Then $int_{A_0}f(x)dx=g(b)=c.$ For $c<0$ start with $B$, and for $c=0$ take $A_0=phi.$
answered Nov 23 at 13:48
John_Wick
1,119111
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I'll give it a shot.
Consider $A = {x in mathbb{R}, f(x) geq 0}$ which is well defined since $f$ is continuous.
Consider now $g(x) = int_{A cap (0,x)} f(y) dy + int_{A cap (-x,0)} f(y) dy = int_{-x}^x f(y) mathbf{1}_A(y) dy$.
Clearly $g(0) = 0$. Moreover, $lim limits_{x to infty} g(x) = infty$ by your condition on $f_+$. It only remains to show that $g$ is continuous to use the intermediate value theorem.
Usually, to get $g$ continuous, (see for instance this http://mathonline.wikidot.com/continuity-of-functions-defined-by-lebesgue-integrals), you would need $f_+$ to be bounded by an integrable function, but in fact since $f$ is continuous, on any compact set you have that $f_+$ is bounded (Weierstrass theorem). This is enough for your purpose: for any $c > 0$, it exists $x_c > 0$ such that $g(x_c) > c$ (otherwise the limit of $g$ can not be $+ infty$). Then on the compact $[-x_c - 1, x_c + 1]$, your function $g$ is clearly continuous (since then $f_+$ is bounded) by applying Theorem 1 in the link above.
So, $g(0) = 0, g(x_c) > c$ and g is continuous, therefore it exists $alpha$ such that $g(alpha) = c$ (intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem).
You can do the exact same thing for $c < 0$ by changing the definition of $A$.
answered Nov 20 at 12:07
Gâteau-Gallois
347111
add a comment |
up vote
0
down vote
accepted
I'll give it a shot.
Consider $A = {x in mathbb{R}, f(x) geq 0}$ which is well defined since $f$ is continuous.
Consider now $g(x) = int_{A cap (0,x)} f(y) dy + int_{A cap (-x,0)} f(y) dy = int_{-x}^x f(y) mathbf{1}_A(y) dy$.
Clearly $g(0) = 0$. Moreover, $lim limits_{x to infty} g(x) = infty$ by your condition on $f_+$. It only remains to show that $g$ is continuous to use the intermediate value theorem.
Usually, to get $g$ continuous, (see for instance this http://mathonline.wikidot.com/continuity-of-functions-defined-by-lebesgue-integrals), you would need $f_+$ to be bounded by an integrable function, but in fact since $f$ is continuous, on any compact set you have that $f_+$ is bounded (Weierstrass theorem). This is enough for your purpose: for any $c > 0$, it exists $x_c > 0$ such that $g(x_c) > c$ (otherwise the limit of $g$ can not be $+ infty$). Then on the compact $[-x_c - 1, x_c + 1]$, your function $g$ is clearly continuous (since then $f_+$ is bounded) by applying Theorem 1 in the link above.
So, $g(0) = 0, g(x_c) > c$ and g is continuous, therefore it exists $alpha$ such that $g(alpha) = c$ (intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem).
You can do the exact same thing for $c < 0$ by changing the definition of $A$.
answered Nov 20 at 12:07
Gâteau-Gallois
347111
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I'll give it a shot.
Consider $A = {x in mathbb{R}, f(x) geq 0}$ which is well defined since $f$ is continuous.
Consider now $g(x) = int_{A cap (0,x)} f(y) dy + int_{A cap (-x,0)} f(y) dy = int_{-x}^x f(y) mathbf{1}_A(y) dy$.
Clearly $g(0) = 0$. Moreover, $lim limits_{x to infty} g(x) = infty$ by your condition on $f_+$. It only remains to show that $g$ is continuous to use the intermediate value theorem.
Usually, to get $g$ continuous, (see for instance this http://mathonline.wikidot.com/continuity-of-functions-defined-by-lebesgue-integrals), you would need $f_+$ to be bounded by an integrable function, but in fact since $f$ is continuous, on any compact set you have that $f_+$ is bounded (Weierstrass theorem). This is enough for your purpose: for any $c > 0$, it exists $x_c > 0$ such that $g(x_c) > c$ (otherwise the limit of $g$ can not be $+ infty$). Then on the compact $[-x_c - 1, x_c + 1]$, your function $g$ is clearly continuous (since then $f_+$ is bounded) by applying Theorem 1 in the link above.
So, $g(0) = 0, g(x_c) > c$ and g is continuous, therefore it exists $alpha$ such that $g(alpha) = c$ (intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem).
You can do the exact same thing for $c < 0$ by changing the definition of $A$.
answered Nov 20 at 12:07
Gâteau-Gallois
347111
I'll give it a shot.
Consider $A = {x in mathbb{R}, f(x) geq 0}$ which is well defined since $f$ is continuous.
Consider now $g(x) = int_{A cap (0,x)} f(y) dy + int_{A cap (-x,0)} f(y) dy = int_{-x}^x f(y) mathbf{1}_A(y) dy$.
Clearly $g(0) = 0$. Moreover, $lim limits_{x to infty} g(x) = infty$ by your condition on $f_+$. It only remains to show that $g$ is continuous to use the intermediate value theorem.
Usually, to get $g$ continuous, (see for instance this http://mathonline.wikidot.com/continuity-of-functions-defined-by-lebesgue-integrals), you would need $f_+$ to be bounded by an integrable function, but in fact since $f$ is continuous, on any compact set you have that $f_+$ is bounded (Weierstrass theorem). This is enough for your purpose: for any $c > 0$, it exists $x_c > 0$ such that $g(x_c) > c$ (otherwise the limit of $g$ can not be $+ infty$). Then on the compact $[-x_c - 1, x_c + 1]$, your function $g$ is clearly continuous (since then $f_+$ is bounded) by applying Theorem 1 in the link above.
So, $g(0) = 0, g(x_c) > c$ and g is continuous, therefore it exists $alpha$ such that $g(alpha) = c$ (intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem).
You can do the exact same thing for $c < 0$ by changing the definition of $A$.
answered Nov 20 at 12:07
Gâteau-Gallois
347111
answered Nov 20 at 12:07
Gâteau-Gallois
347111
answered Nov 20 at 12:07
Gâteau-Gallois
347111
answered Nov 20 at 12:07
Gâteau-Gallois
347111
347111
add a comment |
add a comment |
up vote
0
down vote
Consider $A:={x:f^+(x)>0}$ and $B:={x:f^-(x)>0}$. If $c>0$ then define a function $g:[0,infty)rightarrow [0,infty)$ such that $g(r)=int_{[-r,r]}f^+(x)dx=int_{Acap[-r,r]}f(x)dx.$ Notice that $g(0)=0$ and $g(r)rightarrowinfty$ as $rrightarrow infty.$ And $|g(r)-g(s)|=int_{[-r,-s)}f^+(x)dx+int_{(s,r]}f^+(x)dxrightarrow 0$ as $suparrow r.$ So, $g$ is a continuous function. By mean value theorem $exists b>0$ such that $g(b)=c.$ Take $A_0=[-b,b]cap A.$ Then $int_{A_0}f(x)dx=g(b)=c.$ For $c<0$ start with $B$, and for $c=0$ take $A_0=phi.$
answered Nov 23 at 13:48
John_Wick
1,119111
add a comment |
up vote
0
down vote
Consider $A:={x:f^+(x)>0}$ and $B:={x:f^-(x)>0}$. If $c>0$ then define a function $g:[0,infty)rightarrow [0,infty)$ such that $g(r)=int_{[-r,r]}f^+(x)dx=int_{Acap[-r,r]}f(x)dx.$ Notice that $g(0)=0$ and $g(r)rightarrowinfty$ as $rrightarrow infty.$ And $|g(r)-g(s)|=int_{[-r,-s)}f^+(x)dx+int_{(s,r]}f^+(x)dxrightarrow 0$ as $suparrow r.$ So, $g$ is a continuous function. By mean value theorem $exists b>0$ such that $g(b)=c.$ Take $A_0=[-b,b]cap A.$ Then $int_{A_0}f(x)dx=g(b)=c.$ For $c<0$ start with $B$, and for $c=0$ take $A_0=phi.$
answered Nov 23 at 13:48
John_Wick
1,119111
add a comment |
up vote
0
down vote
up vote
0
down vote
Consider $A:={x:f^+(x)>0}$ and $B:={x:f^-(x)>0}$. If $c>0$ then define a function $g:[0,infty)rightarrow [0,infty)$ such that $g(r)=int_{[-r,r]}f^+(x)dx=int_{Acap[-r,r]}f(x)dx.$ Notice that $g(0)=0$ and $g(r)rightarrowinfty$ as $rrightarrow infty.$ And $|g(r)-g(s)|=int_{[-r,-s)}f^+(x)dx+int_{(s,r]}f^+(x)dxrightarrow 0$ as $suparrow r.$ So, $g$ is a continuous function. By mean value theorem $exists b>0$ such that $g(b)=c.$ Take $A_0=[-b,b]cap A.$ Then $int_{A_0}f(x)dx=g(b)=c.$ For $c<0$ start with $B$, and for $c=0$ take $A_0=phi.$
answered Nov 23 at 13:48
John_Wick
1,119111
Consider $A:={x:f^+(x)>0}$ and $B:={x:f^-(x)>0}$. If $c>0$ then define a function $g:[0,infty)rightarrow [0,infty)$ such that $g(r)=int_{[-r,r]}f^+(x)dx=int_{Acap[-r,r]}f(x)dx.$ Notice that $g(0)=0$ and $g(r)rightarrowinfty$ as $rrightarrow infty.$ And $|g(r)-g(s)|=int_{[-r,-s)}f^+(x)dx+int_{(s,r]}f^+(x)dxrightarrow 0$ as $suparrow r.$ So, $g$ is a continuous function. By mean value theorem $exists b>0$ such that $g(b)=c.$ Take $A_0=[-b,b]cap A.$ Then $int_{A_0}f(x)dx=g(b)=c.$ For $c<0$ start with $B$, and for $c=0$ take $A_0=phi.$
answered Nov 23 at 13:48
John_Wick
1,119111
answered Nov 23 at 13:48
John_Wick
1,119111
answered Nov 23 at 13:48
John_Wick
1,119111
answered Nov 23 at 13:48
John_Wick
1,119111
1,119111
add a comment |
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%2f3005015%2flebesgue-integration-integral-of-continuous-function-tends-to-infinity%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
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
Nov 19 at 14:45
Yes I meant for all $c in mathbb{R}$. Could you perhaps elaborate a bit more on your proof idea ? I don’t get how you can reach this result from the theorem.
– jffi
Nov 19 at 15:51