Mixing
An exercise from Stein's Real Analysis about Lebesgue integral
I would be glad if anyone check my solution for the following question.
Suppose $f$ is integrable on $(-pi,pi]$ and extended to $Bbb R$ by making it periodic of period $2pi$. Show that $int_{-pi}^pi f(x)dx=int_I f(x)dx$ where $I$ is any interval in $Bbb R$ of length $2pi$.
Attempt: Note that $I=(-pi,pi]+a=(-pi+a,pi+a]$ for some $ain Bbb R$. So, it is enough to show that for any $ain Bbb R$,
$$int_{-pi}^pi f(x)=int_{-pi+a}^{pi+a}f(x)dx.$$
Here we go:
begin{align}
int_{-pi+a}^{pi+a}f(x)dx &=int_{-pi+a}^{pi}f(x)dx+int_pi^{pi+a}f(x)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x+2pi)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x)dx, quad text{since the period of $f$ is $2pi$ }
\&=int_{-pi}^pi f(x)dx
end{align}
Do we need to prove the equality
$int_pi^{pi+a}f(x)dx=int_{-pi}^{-pi+a}f(x+2pi)dx$ ? If the answer is yes, how?
Thanks!
real-analysis measure-theory proof-verification lebesgue-integral
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
add a comment |
I would be glad if anyone check my solution for the following question.
Suppose $f$ is integrable on $(-pi,pi]$ and extended to $Bbb R$ by making it periodic of period $2pi$. Show that $int_{-pi}^pi f(x)dx=int_I f(x)dx$ where $I$ is any interval in $Bbb R$ of length $2pi$.
Attempt: Note that $I=(-pi,pi]+a=(-pi+a,pi+a]$ for some $ain Bbb R$. So, it is enough to show that for any $ain Bbb R$,
$$int_{-pi}^pi f(x)=int_{-pi+a}^{pi+a}f(x)dx.$$
Here we go:
begin{align}
int_{-pi+a}^{pi+a}f(x)dx &=int_{-pi+a}^{pi}f(x)dx+int_pi^{pi+a}f(x)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x+2pi)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x)dx, quad text{since the period of $f$ is $2pi$ }
\&=int_{-pi}^pi f(x)dx
end{align}
Do we need to prove the equality
$int_pi^{pi+a}f(x)dx=int_{-pi}^{-pi+a}f(x+2pi)dx$ ? If the answer is yes, how?
Thanks!
real-analysis measure-theory proof-verification lebesgue-integral
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
Your first claim about the form of $I$ isn't true. It is true that $I = (-pi,pi] + a$ for some $ainBbb R$, not necessarily an $ain(0,2pi]$, however.
– AOrtiz
Nov 21 '18 at 23:11
Thanks, I updated!
– Ergin Suer
Nov 21 '18 at 23:38
add a comment |
I would be glad if anyone check my solution for the following question.
Suppose $f$ is integrable on $(-pi,pi]$ and extended to $Bbb R$ by making it periodic of period $2pi$. Show that $int_{-pi}^pi f(x)dx=int_I f(x)dx$ where $I$ is any interval in $Bbb R$ of length $2pi$.
Attempt: Note that $I=(-pi,pi]+a=(-pi+a,pi+a]$ for some $ain Bbb R$. So, it is enough to show that for any $ain Bbb R$,
$$int_{-pi}^pi f(x)=int_{-pi+a}^{pi+a}f(x)dx.$$
Here we go:
begin{align}
int_{-pi+a}^{pi+a}f(x)dx &=int_{-pi+a}^{pi}f(x)dx+int_pi^{pi+a}f(x)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x+2pi)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x)dx, quad text{since the period of $f$ is $2pi$ }
\&=int_{-pi}^pi f(x)dx
end{align}
Do we need to prove the equality
$int_pi^{pi+a}f(x)dx=int_{-pi}^{-pi+a}f(x+2pi)dx$ ? If the answer is yes, how?
Thanks!
real-analysis measure-theory proof-verification lebesgue-integral
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
I would be glad if anyone check my solution for the following question.
Suppose $f$ is integrable on $(-pi,pi]$ and extended to $Bbb R$ by making it periodic of period $2pi$. Show that $int_{-pi}^pi f(x)dx=int_I f(x)dx$ where $I$ is any interval in $Bbb R$ of length $2pi$.
Attempt: Note that $I=(-pi,pi]+a=(-pi+a,pi+a]$ for some $ain Bbb R$. So, it is enough to show that for any $ain Bbb R$,
$$int_{-pi}^pi f(x)=int_{-pi+a}^{pi+a}f(x)dx.$$
Here we go:
begin{align}
int_{-pi+a}^{pi+a}f(x)dx &=int_{-pi+a}^{pi}f(x)dx+int_pi^{pi+a}f(x)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x+2pi)dx
\&=int_{-pi+a}^{pi}f(x)dx+int_{-pi}^{-pi+a}f(x)dx, quad text{since the period of $f$ is $2pi$ }
\&=int_{-pi}^pi f(x)dx
end{align}
Do we need to prove the equality
$int_pi^{pi+a}f(x)dx=int_{-pi}^{-pi+a}f(x+2pi)dx$ ? If the answer is yes, how?
Thanks!
real-analysis measure-theory proof-verification lebesgue-integral
real-analysis measure-theory proof-verification lebesgue-integral
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
edited Nov 21 '18 at 23:37
Ergin Suer
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
asked Nov 21 '18 at 23:00
Ergin SuerErgin Suer
1,409921
1,409921
Your first claim about the form of $I$ isn't true. It is true that $I = (-pi,pi] + a$ for some $ainBbb R$, not necessarily an $ain(0,2pi]$, however.
– AOrtiz
Nov 21 '18 at 23:11
Thanks, I updated!
– Ergin Suer
Nov 21 '18 at 23:38
add a comment |
Your first claim about the form of $I$ isn't true. It is true that $I = (-pi,pi] + a$ for some $ainBbb R$, not necessarily an $ain(0,2pi]$, however.
– AOrtiz
Nov 21 '18 at 23:11
Thanks, I updated!
– Ergin Suer
Nov 21 '18 at 23:38
Your first claim about the form of $I$ isn't true. It is true that $I = (-pi,pi] + a$ for some $ainBbb R$, not necessarily an $ain(0,2pi]$, however.
– AOrtiz
Nov 21 '18 at 23:11
Your first claim about the form of $I$ isn't true. It is true that $I = (-pi,pi] + a$ for some $ainBbb R$, not necessarily an $ain(0,2pi]$, however.
– AOrtiz
Nov 21 '18 at 23:11
Thanks, I updated!
– Ergin Suer
Nov 21 '18 at 23:38
Thanks, I updated!
– Ergin Suer
Nov 21 '18 at 23:38
add a comment |
3 Answers
3
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oldest
votes
$a$ need not be in $(0,2pi]$ but this doesn't affect the proof. Yes, you have to justify your second step. Just make the substitution $x=y+2pi$ for the justification.
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
add a comment |
There is a more general result, that follows from the translation invariance of Lebesgue measure, and makes precise the statement "$y=x−2π$ implies $dy=dx$ by differentiation":
Set $g:mathbb Rto mathbb R$ by $g(x)=x+a.$ Then, the image measure $mg^{-1}$ satisfies $mg^{-1}(E)=m(E)$ for an arbitrary measurable set $E$, and so
on the one hand,
$int_Ef(x)d(mg^{-1})=int_Ef(x)dm$
and on the other,
$int_Ef(x)d(mg^{-1})=int_{mathbb R}chi_E(x)f(x)d(mg^{-1})=int_{mathbb R}g(f(x)chi_E(x))dm=int_{mathbb R}f(x+a)chi_E(x+a))dm.$
Therefore,
$int_Ef(x)dm=int_{mathbb R}f(x+a)chi_E(x+a))dm$.
To recover your result, set $a=2pi, E=(-pi,pi]$ and note that $f(x+2pi)=f(x)$
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
add a comment |
begin{align}
int_pi^{pi+a}f(x)dx &=int_{Bbb R}chi_E(x)f(x)dx=int_{Bbb R}chi_E(x+2pi)f(x+2pi)dx \&=int_{Bbb R}chi_{E-2pi}(x)f(x)dx=int_{-pi}^{-pi+a}f(x)dx
end{align}
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$a$ need not be in $(0,2pi]$ but this doesn't affect the proof. Yes, you have to justify your second step. Just make the substitution $x=y+2pi$ for the justification.
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
add a comment |
$a$ need not be in $(0,2pi]$ but this doesn't affect the proof. Yes, you have to justify your second step. Just make the substitution $x=y+2pi$ for the justification.
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
add a comment |
$a$ need not be in $(0,2pi]$ but this doesn't affect the proof. Yes, you have to justify your second step. Just make the substitution $x=y+2pi$ for the justification.
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
$a$ need not be in $(0,2pi]$ but this doesn't affect the proof. Yes, you have to justify your second step. Just make the substitution $x=y+2pi$ for the justification.
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 21 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
add a comment |
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
In the substitution, y varies from $-pi$ to $-pi+a$ as $x$ varies from $pi$ to $pi+a$. Then $y=x-2pi$ implies $dy=dx$ by differentiation. So, we replace $dx$ by $dy$. Is it true?
– Ergin Suer
Nov 21 '18 at 23:35
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
Yes, that is precisely the argument.
– Kavi Rama Murthy
Nov 21 '18 at 23:37
add a comment |
There is a more general result, that follows from the translation invariance of Lebesgue measure, and makes precise the statement "$y=x−2π$ implies $dy=dx$ by differentiation":
Set $g:mathbb Rto mathbb R$ by $g(x)=x+a.$ Then, the image measure $mg^{-1}$ satisfies $mg^{-1}(E)=m(E)$ for an arbitrary measurable set $E$, and so
on the one hand,
$int_Ef(x)d(mg^{-1})=int_Ef(x)dm$
and on the other,
$int_Ef(x)d(mg^{-1})=int_{mathbb R}chi_E(x)f(x)d(mg^{-1})=int_{mathbb R}g(f(x)chi_E(x))dm=int_{mathbb R}f(x+a)chi_E(x+a))dm.$
Therefore,
$int_Ef(x)dm=int_{mathbb R}f(x+a)chi_E(x+a))dm$.
To recover your result, set $a=2pi, E=(-pi,pi]$ and note that $f(x+2pi)=f(x)$
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
add a comment |
There is a more general result, that follows from the translation invariance of Lebesgue measure, and makes precise the statement "$y=x−2π$ implies $dy=dx$ by differentiation":
Set $g:mathbb Rto mathbb R$ by $g(x)=x+a.$ Then, the image measure $mg^{-1}$ satisfies $mg^{-1}(E)=m(E)$ for an arbitrary measurable set $E$, and so
on the one hand,
$int_Ef(x)d(mg^{-1})=int_Ef(x)dm$
and on the other,
$int_Ef(x)d(mg^{-1})=int_{mathbb R}chi_E(x)f(x)d(mg^{-1})=int_{mathbb R}g(f(x)chi_E(x))dm=int_{mathbb R}f(x+a)chi_E(x+a))dm.$
Therefore,
$int_Ef(x)dm=int_{mathbb R}f(x+a)chi_E(x+a))dm$.
To recover your result, set $a=2pi, E=(-pi,pi]$ and note that $f(x+2pi)=f(x)$
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
add a comment |
There is a more general result, that follows from the translation invariance of Lebesgue measure, and makes precise the statement "$y=x−2π$ implies $dy=dx$ by differentiation":
Set $g:mathbb Rto mathbb R$ by $g(x)=x+a.$ Then, the image measure $mg^{-1}$ satisfies $mg^{-1}(E)=m(E)$ for an arbitrary measurable set $E$, and so
on the one hand,
$int_Ef(x)d(mg^{-1})=int_Ef(x)dm$
and on the other,
$int_Ef(x)d(mg^{-1})=int_{mathbb R}chi_E(x)f(x)d(mg^{-1})=int_{mathbb R}g(f(x)chi_E(x))dm=int_{mathbb R}f(x+a)chi_E(x+a))dm.$
Therefore,
$int_Ef(x)dm=int_{mathbb R}f(x+a)chi_E(x+a))dm$.
To recover your result, set $a=2pi, E=(-pi,pi]$ and note that $f(x+2pi)=f(x)$
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
There is a more general result, that follows from the translation invariance of Lebesgue measure, and makes precise the statement "$y=x−2π$ implies $dy=dx$ by differentiation":
Set $g:mathbb Rto mathbb R$ by $g(x)=x+a.$ Then, the image measure $mg^{-1}$ satisfies $mg^{-1}(E)=m(E)$ for an arbitrary measurable set $E$, and so
on the one hand,
$int_Ef(x)d(mg^{-1})=int_Ef(x)dm$
and on the other,
$int_Ef(x)d(mg^{-1})=int_{mathbb R}chi_E(x)f(x)d(mg^{-1})=int_{mathbb R}g(f(x)chi_E(x))dm=int_{mathbb R}f(x+a)chi_E(x+a))dm.$
Therefore,
$int_Ef(x)dm=int_{mathbb R}f(x+a)chi_E(x+a))dm$.
To recover your result, set $a=2pi, E=(-pi,pi]$ and note that $f(x+2pi)=f(x)$
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
answered Nov 21 '18 at 23:59
MatematletaMatematleta
10.1k2918
10.1k2918
add a comment |
add a comment |
begin{align}
int_pi^{pi+a}f(x)dx &=int_{Bbb R}chi_E(x)f(x)dx=int_{Bbb R}chi_E(x+2pi)f(x+2pi)dx \&=int_{Bbb R}chi_{E-2pi}(x)f(x)dx=int_{-pi}^{-pi+a}f(x)dx
end{align}
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
add a comment |
begin{align}
int_pi^{pi+a}f(x)dx &=int_{Bbb R}chi_E(x)f(x)dx=int_{Bbb R}chi_E(x+2pi)f(x+2pi)dx \&=int_{Bbb R}chi_{E-2pi}(x)f(x)dx=int_{-pi}^{-pi+a}f(x)dx
end{align}
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
add a comment |
begin{align}
int_pi^{pi+a}f(x)dx &=int_{Bbb R}chi_E(x)f(x)dx=int_{Bbb R}chi_E(x+2pi)f(x+2pi)dx \&=int_{Bbb R}chi_{E-2pi}(x)f(x)dx=int_{-pi}^{-pi+a}f(x)dx
end{align}
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
begin{align}
int_pi^{pi+a}f(x)dx &=int_{Bbb R}chi_E(x)f(x)dx=int_{Bbb R}chi_E(x+2pi)f(x+2pi)dx \&=int_{Bbb R}chi_{E-2pi}(x)f(x)dx=int_{-pi}^{-pi+a}f(x)dx
end{align}
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
answered Nov 27 '18 at 5:22
Ergin SuerErgin Suer
1,409921
1,409921
add a comment |
add a comment |
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Your first claim about the form of $I$ isn't true. It is true that $I = (-pi,pi] + a$ for some $ainBbb R$, not necessarily an $ain(0,2pi]$, however.
– AOrtiz
Nov 21 '18 at 23:11
Thanks, I updated!
– Ergin Suer
Nov 21 '18 at 23:38