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Lebesgue integration: integral of continuous function tends to infinity
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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 yesterday
jffi
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up vote
0
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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 yesterday
jffi
83
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
yesterday
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
yesterday
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 yesterday
jffi
83
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 yesterday
jffi
83
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
jffi
83
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
jffi
83
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
jffi
83
asked yesterday
jffi
83
83
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
jffi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
yesterday
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
yesterday
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
yesterday
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
yesterday
2
2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
yesterday
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
yesterday
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
yesterday
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
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
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 11 hours ago
Gâteau-Gallois
35819
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 11 hours ago
Gâteau-Gallois
35819
add a comment |
up vote
0
down vote
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 11 hours ago
Gâteau-Gallois
35819
add a comment |
up vote
0
down vote
up vote
0
down vote
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 11 hours ago
Gâteau-Gallois
35819
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 11 hours ago
Gâteau-Gallois
35819
answered 11 hours ago
Gâteau-Gallois
35819
answered 11 hours ago
Gâteau-Gallois
35819
answered 11 hours ago
Gâteau-Gallois
35819
35819
add a comment |
add a comment |
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2
I assume you mean for all $c in mathbb{R}$? I believe the Intermediate Value Theorem might help you.
– Gâteau-Gallois
yesterday
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
yesterday