Prove that $int_{E}f = int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx$ using Fubini's Theorem
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Problem. Let $E$ be a domain in the plane bounded by the continuous curves $y = phi(x)$ and $y = psi(x)$ for $x in I = [a,b]$, where $phi(x), psi(x)$. Prove that if $f$ is a Borel measurable, integrable function defined on $E$, then
$$int_{E}f = int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx.$$
My attempt. Use the Fubini's Theorem that says
$$int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx = int_{I times [phi(x),psi(x)]}f(x,y)d(y times x)$$
where $x times y$ represents the product measure. But I could not relate $int_{E}f$ with $int_{I times [phi(x),psi(x)]}f$. I believe that this is the main point of the question. Can someone help me?
integration measure-theory
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
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add a comment |
$begingroup$
Problem. Let $E$ be a domain in the plane bounded by the continuous curves $y = phi(x)$ and $y = psi(x)$ for $x in I = [a,b]$, where $phi(x), psi(x)$. Prove that if $f$ is a Borel measurable, integrable function defined on $E$, then
$$int_{E}f = int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx.$$
My attempt. Use the Fubini's Theorem that says
$$int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx = int_{I times [phi(x),psi(x)]}f(x,y)d(y times x)$$
where $x times y$ represents the product measure. But I could not relate $int_{E}f$ with $int_{I times [phi(x),psi(x)]}f$. I believe that this is the main point of the question. Can someone help me?
integration measure-theory
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
$endgroup$
$begingroup$
Your equation at the end doesn't make sense as stated: In $Itimes [phi(x),psi(x)]$, you haven't said what $x$ is.
$endgroup$
– user25959
Dec 8 '18 at 21:03
add a comment |
$begingroup$
Problem. Let $E$ be a domain in the plane bounded by the continuous curves $y = phi(x)$ and $y = psi(x)$ for $x in I = [a,b]$, where $phi(x), psi(x)$. Prove that if $f$ is a Borel measurable, integrable function defined on $E$, then
$$int_{E}f = int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx.$$
My attempt. Use the Fubini's Theorem that says
$$int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx = int_{I times [phi(x),psi(x)]}f(x,y)d(y times x)$$
where $x times y$ represents the product measure. But I could not relate $int_{E}f$ with $int_{I times [phi(x),psi(x)]}f$. I believe that this is the main point of the question. Can someone help me?
integration measure-theory
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
$endgroup$
Problem. Let $E$ be a domain in the plane bounded by the continuous curves $y = phi(x)$ and $y = psi(x)$ for $x in I = [a,b]$, where $phi(x), psi(x)$. Prove that if $f$ is a Borel measurable, integrable function defined on $E$, then
$$int_{E}f = int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx.$$
My attempt. Use the Fubini's Theorem that says
$$int_{I}int_{[phi(x),psi(x)]}f(x,y)dydx = int_{I times [phi(x),psi(x)]}f(x,y)d(y times x)$$
where $x times y$ represents the product measure. But I could not relate $int_{E}f$ with $int_{I times [phi(x),psi(x)]}f$. I believe that this is the main point of the question. Can someone help me?
integration measure-theory
integration measure-theory
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
asked Dec 8 '18 at 20:52
Lucas CorrêaLucas Corrêa
1,5951421
1,5951421
$begingroup$
Your equation at the end doesn't make sense as stated: In $Itimes [phi(x),psi(x)]$, you haven't said what $x$ is.
$endgroup$
– user25959
Dec 8 '18 at 21:03
add a comment |
$begingroup$
Your equation at the end doesn't make sense as stated: In $Itimes [phi(x),psi(x)]$, you haven't said what $x$ is.
$endgroup$
– user25959
Dec 8 '18 at 21:03
$begingroup$
Your equation at the end doesn't make sense as stated: In $Itimes [phi(x),psi(x)]$, you haven't said what $x$ is.
$endgroup$
– user25959
Dec 8 '18 at 21:03
$begingroup$
Your equation at the end doesn't make sense as stated: In $Itimes [phi(x),psi(x)]$, you haven't said what $x$ is.
$endgroup$
– user25959
Dec 8 '18 at 21:03
add a comment |
1 Answer
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$begingroup$
The idea is to apply Fubini's theorem in $mathbb R^2.$ Recall that for any $g$ Borel measurable and integrable on $mathbb R^2,$ we have,
$$ int_{mathbb R^2} g = int_{mathbb R}int_{mathbb R} g(x,y),mathrm{d}x,mathrm{d}y. $$
The key trick is to embed the information about $E$ into this function $g.$ So we set,
$$ g = chi_E f. $$
Where $chi_E$ is the characteristic function of $E subset mathbb R^2.$ By our assumptions $g$ is indeed Borel measurable and integrable, such that,
$$ int_{mathbb R^2} chi_Ef = int_E f = int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y. $$
Now $(x,y) in E$ if and only if $x in I$ and $varphi(x) leq y leq psi(x).$ So we conclude that,
$$ int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y = int_I int_{psi(x)}^{varphi(x)} f(x,y),mathrm{d}x,mathrm{d}y, $$
from which the result follows.
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
$endgroup$
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
add a comment |
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$begingroup$
The idea is to apply Fubini's theorem in $mathbb R^2.$ Recall that for any $g$ Borel measurable and integrable on $mathbb R^2,$ we have,
$$ int_{mathbb R^2} g = int_{mathbb R}int_{mathbb R} g(x,y),mathrm{d}x,mathrm{d}y. $$
The key trick is to embed the information about $E$ into this function $g.$ So we set,
$$ g = chi_E f. $$
Where $chi_E$ is the characteristic function of $E subset mathbb R^2.$ By our assumptions $g$ is indeed Borel measurable and integrable, such that,
$$ int_{mathbb R^2} chi_Ef = int_E f = int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y. $$
Now $(x,y) in E$ if and only if $x in I$ and $varphi(x) leq y leq psi(x).$ So we conclude that,
$$ int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y = int_I int_{psi(x)}^{varphi(x)} f(x,y),mathrm{d}x,mathrm{d}y, $$
from which the result follows.
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
$endgroup$
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
add a comment |
$begingroup$
The idea is to apply Fubini's theorem in $mathbb R^2.$ Recall that for any $g$ Borel measurable and integrable on $mathbb R^2,$ we have,
$$ int_{mathbb R^2} g = int_{mathbb R}int_{mathbb R} g(x,y),mathrm{d}x,mathrm{d}y. $$
The key trick is to embed the information about $E$ into this function $g.$ So we set,
$$ g = chi_E f. $$
Where $chi_E$ is the characteristic function of $E subset mathbb R^2.$ By our assumptions $g$ is indeed Borel measurable and integrable, such that,
$$ int_{mathbb R^2} chi_Ef = int_E f = int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y. $$
Now $(x,y) in E$ if and only if $x in I$ and $varphi(x) leq y leq psi(x).$ So we conclude that,
$$ int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y = int_I int_{psi(x)}^{varphi(x)} f(x,y),mathrm{d}x,mathrm{d}y, $$
from which the result follows.
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
$endgroup$
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
add a comment |
$begingroup$
The idea is to apply Fubini's theorem in $mathbb R^2.$ Recall that for any $g$ Borel measurable and integrable on $mathbb R^2,$ we have,
$$ int_{mathbb R^2} g = int_{mathbb R}int_{mathbb R} g(x,y),mathrm{d}x,mathrm{d}y. $$
The key trick is to embed the information about $E$ into this function $g.$ So we set,
$$ g = chi_E f. $$
Where $chi_E$ is the characteristic function of $E subset mathbb R^2.$ By our assumptions $g$ is indeed Borel measurable and integrable, such that,
$$ int_{mathbb R^2} chi_Ef = int_E f = int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y. $$
Now $(x,y) in E$ if and only if $x in I$ and $varphi(x) leq y leq psi(x).$ So we conclude that,
$$ int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y = int_I int_{psi(x)}^{varphi(x)} f(x,y),mathrm{d}x,mathrm{d}y, $$
from which the result follows.
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
$endgroup$
The idea is to apply Fubini's theorem in $mathbb R^2.$ Recall that for any $g$ Borel measurable and integrable on $mathbb R^2,$ we have,
$$ int_{mathbb R^2} g = int_{mathbb R}int_{mathbb R} g(x,y),mathrm{d}x,mathrm{d}y. $$
The key trick is to embed the information about $E$ into this function $g.$ So we set,
$$ g = chi_E f. $$
Where $chi_E$ is the characteristic function of $E subset mathbb R^2.$ By our assumptions $g$ is indeed Borel measurable and integrable, such that,
$$ int_{mathbb R^2} chi_Ef = int_E f = int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y. $$
Now $(x,y) in E$ if and only if $x in I$ and $varphi(x) leq y leq psi(x).$ So we conclude that,
$$ int_{mathbb R}int_{mathbb R} chi_E(x,y)f(x,y),mathrm{d}x,mathrm{d}y = int_I int_{psi(x)}^{varphi(x)} f(x,y),mathrm{d}x,mathrm{d}y, $$
from which the result follows.
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
answered Dec 8 '18 at 21:03
ktoiktoi
2,4161618
2,4161618
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
add a comment |
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
$begingroup$
Nice, I got it! Thank you!
$endgroup$
– Lucas Corrêa
Dec 8 '18 at 21:11
add a comment |
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$begingroup$
Your equation at the end doesn't make sense as stated: In $Itimes [phi(x),psi(x)]$, you haven't said what $x$ is.
$endgroup$
– user25959
Dec 8 '18 at 21:03