Mixing
Exchange limit on bounds of Lebesgue integral
$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be a sequence of measurable sets such that $lim_{n to infty} E_n =E$ for some measurable set $E$. When does it hold that
$$ lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$$
What if we replace $lim$ with $lim sup$ or $lim inf$?
measure-theory lebesgue-integral
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
$endgroup$
|
show 2 more comments
$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be a sequence of measurable sets such that $lim_{n to infty} E_n =E$ for some measurable set $E$. When does it hold that
$$ lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$$
What if we replace $lim$ with $lim sup$ or $lim inf$?
measure-theory lebesgue-integral
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
$endgroup$
$begingroup$
Does it help if you write $displaystyleint_{E_n} f,dmu = displaystyleint_E fchi_{E_n},dmu$?
$endgroup$
– Ted Shifrin
Jan 16 at 19:48
$begingroup$
@TedShifrin Yes I tried that. At least for positive $f$ wouldn't that imply that this is always true? The limit and the integral can be exchanged because $fchi_{E_n}$ is non-decreasing.
$endgroup$
– Jannik Pitt
Jan 16 at 19:50
$begingroup$
What kind of function is $f$? What do you mean by $lim_{n to infty} E_n =E$?
$endgroup$
– BigbearZzz
Jan 16 at 19:56
$begingroup$
@BigbearZzz See en.m.wikipedia.org/wiki/Set-theoretic_limit
$endgroup$
– Jannik Pitt
Jan 16 at 20:14
$begingroup$
@JannikPitt Is $f$ integrable? Locally integrable?
$endgroup$
– BigbearZzz
Jan 16 at 20:15
|
show 2 more comments
$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be a sequence of measurable sets such that $lim_{n to infty} E_n =E$ for some measurable set $E$. When does it hold that
$$ lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$$
What if we replace $lim$ with $lim sup$ or $lim inf$?
measure-theory lebesgue-integral
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
$endgroup$
Let $(E_n)_{n in mathbb{N}}$ be a sequence of measurable sets such that $lim_{n to infty} E_n =E$ for some measurable set $E$. When does it hold that
$$ lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$$
What if we replace $lim$ with $lim sup$ or $lim inf$?
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
asked Jan 16 at 19:45
Jannik PittJannik Pitt
638517
638517
$begingroup$
Does it help if you write $displaystyleint_{E_n} f,dmu = displaystyleint_E fchi_{E_n},dmu$?
$endgroup$
– Ted Shifrin
Jan 16 at 19:48
$begingroup$
@TedShifrin Yes I tried that. At least for positive $f$ wouldn't that imply that this is always true? The limit and the integral can be exchanged because $fchi_{E_n}$ is non-decreasing.
$endgroup$
– Jannik Pitt
Jan 16 at 19:50
$begingroup$
What kind of function is $f$? What do you mean by $lim_{n to infty} E_n =E$?
$endgroup$
– BigbearZzz
Jan 16 at 19:56
$begingroup$
@BigbearZzz See en.m.wikipedia.org/wiki/Set-theoretic_limit
$endgroup$
– Jannik Pitt
Jan 16 at 20:14
$begingroup$
@JannikPitt Is $f$ integrable? Locally integrable?
$endgroup$
– BigbearZzz
Jan 16 at 20:15
|
show 2 more comments
$begingroup$
Does it help if you write $displaystyleint_{E_n} f,dmu = displaystyleint_E fchi_{E_n},dmu$?
$endgroup$
– Ted Shifrin
Jan 16 at 19:48
$begingroup$
@TedShifrin Yes I tried that. At least for positive $f$ wouldn't that imply that this is always true? The limit and the integral can be exchanged because $fchi_{E_n}$ is non-decreasing.
$endgroup$
– Jannik Pitt
Jan 16 at 19:50
$begingroup$
What kind of function is $f$? What do you mean by $lim_{n to infty} E_n =E$?
$endgroup$
– BigbearZzz
Jan 16 at 19:56
$begingroup$
@BigbearZzz See en.m.wikipedia.org/wiki/Set-theoretic_limit
$endgroup$
– Jannik Pitt
Jan 16 at 20:14
$begingroup$
@JannikPitt Is $f$ integrable? Locally integrable?
$endgroup$
– BigbearZzz
Jan 16 at 20:15
$begingroup$
Does it help if you write $displaystyleint_{E_n} f,dmu = displaystyleint_E fchi_{E_n},dmu$?
$endgroup$
– Ted Shifrin
Jan 16 at 19:48
$begingroup$
Does it help if you write $displaystyleint_{E_n} f,dmu = displaystyleint_E fchi_{E_n},dmu$?
$endgroup$
– Ted Shifrin
Jan 16 at 19:48
$begingroup$
@TedShifrin Yes I tried that. At least for positive $f$ wouldn't that imply that this is always true? The limit and the integral can be exchanged because $fchi_{E_n}$ is non-decreasing.
$endgroup$
– Jannik Pitt
Jan 16 at 19:50
$begingroup$
@TedShifrin Yes I tried that. At least for positive $f$ wouldn't that imply that this is always true? The limit and the integral can be exchanged because $fchi_{E_n}$ is non-decreasing.
$endgroup$
– Jannik Pitt
Jan 16 at 19:50
$begingroup$
What kind of function is $f$? What do you mean by $lim_{n to infty} E_n =E$?
$endgroup$
– BigbearZzz
Jan 16 at 19:56
$begingroup$
What kind of function is $f$? What do you mean by $lim_{n to infty} E_n =E$?
$endgroup$
– BigbearZzz
Jan 16 at 19:56
$begingroup$
@BigbearZzz See en.m.wikipedia.org/wiki/Set-theoretic_limit
$endgroup$
– Jannik Pitt
Jan 16 at 20:14
$begingroup$
@BigbearZzz See en.m.wikipedia.org/wiki/Set-theoretic_limit
$endgroup$
– Jannik Pitt
Jan 16 at 20:14
$begingroup$
@JannikPitt Is $f$ integrable? Locally integrable?
$endgroup$
– BigbearZzz
Jan 16 at 20:15
$begingroup$
@JannikPitt Is $f$ integrable? Locally integrable?
$endgroup$
– BigbearZzz
Jan 16 at 20:15
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $f$ be an integrable function. i.e. $fin L^1(Bbb R)$. Then we have $|fchi_{E_n}|le |f|$ for all $ninBbb N$, hence by Dominated Convergence Theorem we have
$$begin{align}
int_Ef,dmu &= int fchi_E,dmu \
&= lim_{ntoinfty} int fchi_{E_n} ,dmu \
&= lim_{ntoinfty} int_{E_n} f,dmu
end{align}$$
provided that $chi_{E_n}tochi_E$ almost everywhere.
The pointwise convergence a.e. of $chi_{E_n}$ can be a result of letting $$E=bigcup_n E_n, $$
for example, or even a weaker hypothesis like $E=liminf_{ntoinfty} E_n$, i.e.
$$
E = bigcup_{n=1}^infty bigcap_{kge n} E_k.
$$
If we only assume that $fin L^1_{text{loc}}(Bbb R)$ then the previous convergent result still holds, provided that we assume further that $E,E_n$ are contained in some bounded set $Bsubset Bbb R$ (because we can apply the above argument by viewing $fin L^1(B)$).
If $f$ is not even locally integrable, then we can find counterexamples so that $int_Ef,dmu ne lim_{ntoinfty} int_{E_n} f,dmu$.
Alternatively, we can consider the pseudo-distance function
$$
d(A,B) := mu(ADelta B)
$$
defined on $mathcal A$, the family of Lebesgue measurable sets of $Bbb R$. Define the equivalent relation $Asim B$ iff $d(A,B)=0$, then the set of equivalent classes $mathcal A/sim$ forms a (complete) metric space. We can now interpret $E = lim_{ntoinfty} E_n$ as
$$
d(E,E_n) to 0
$$
in $mathcal A/sim$.
Now, any $fin L^1(Bbb R)$ induces an absolutely continuous signed measure on $Bbb R$. Hence for any $varepsilon>0$ we can find a number $delta>0$ such that $mu(A)<delta$ implies
$$
int_A |f|,dmu < varepsilon.
$$
Thus for sufficiently large $n$, we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu < varepsilon
$$
since $mu(EDelta E_n)to 0$. This shows that $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$.
Similarly, if we instead assume $fin L^infty (Bbb R)$ then for sufficiently large $n$ we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu le mu(EDelta E_n) ||f||_infty to 0
$$
as $d(E,E_n)to 0$, which implies $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$ also.
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
$endgroup$
1
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
add a comment |
$begingroup$
Take the Lebesgue measured space $(mathbb R, mu)$, $E_n= (-infty, 1/n)$ and
$$f(x)=begin{cases}
0 & x le 0\
1/x & x>0
end{cases}$$
Then $limlimits_{n to infty}E_n =(-infty , 0]$. And
$$infty = lim_{n to infty}int_{E_n}f d mu neq int_{E}f dmu =0$$
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
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oldest
votes
$begingroup$
Let $f$ be an integrable function. i.e. $fin L^1(Bbb R)$. Then we have $|fchi_{E_n}|le |f|$ for all $ninBbb N$, hence by Dominated Convergence Theorem we have
$$begin{align}
int_Ef,dmu &= int fchi_E,dmu \
&= lim_{ntoinfty} int fchi_{E_n} ,dmu \
&= lim_{ntoinfty} int_{E_n} f,dmu
end{align}$$
provided that $chi_{E_n}tochi_E$ almost everywhere.
The pointwise convergence a.e. of $chi_{E_n}$ can be a result of letting $$E=bigcup_n E_n, $$
for example, or even a weaker hypothesis like $E=liminf_{ntoinfty} E_n$, i.e.
$$
E = bigcup_{n=1}^infty bigcap_{kge n} E_k.
$$
If we only assume that $fin L^1_{text{loc}}(Bbb R)$ then the previous convergent result still holds, provided that we assume further that $E,E_n$ are contained in some bounded set $Bsubset Bbb R$ (because we can apply the above argument by viewing $fin L^1(B)$).
If $f$ is not even locally integrable, then we can find counterexamples so that $int_Ef,dmu ne lim_{ntoinfty} int_{E_n} f,dmu$.
Alternatively, we can consider the pseudo-distance function
$$
d(A,B) := mu(ADelta B)
$$
defined on $mathcal A$, the family of Lebesgue measurable sets of $Bbb R$. Define the equivalent relation $Asim B$ iff $d(A,B)=0$, then the set of equivalent classes $mathcal A/sim$ forms a (complete) metric space. We can now interpret $E = lim_{ntoinfty} E_n$ as
$$
d(E,E_n) to 0
$$
in $mathcal A/sim$.
Now, any $fin L^1(Bbb R)$ induces an absolutely continuous signed measure on $Bbb R$. Hence for any $varepsilon>0$ we can find a number $delta>0$ such that $mu(A)<delta$ implies
$$
int_A |f|,dmu < varepsilon.
$$
Thus for sufficiently large $n$, we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu < varepsilon
$$
since $mu(EDelta E_n)to 0$. This shows that $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$.
Similarly, if we instead assume $fin L^infty (Bbb R)$ then for sufficiently large $n$ we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu le mu(EDelta E_n) ||f||_infty to 0
$$
as $d(E,E_n)to 0$, which implies $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$ also.
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
$endgroup$
1
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
add a comment |
$begingroup$
Let $f$ be an integrable function. i.e. $fin L^1(Bbb R)$. Then we have $|fchi_{E_n}|le |f|$ for all $ninBbb N$, hence by Dominated Convergence Theorem we have
$$begin{align}
int_Ef,dmu &= int fchi_E,dmu \
&= lim_{ntoinfty} int fchi_{E_n} ,dmu \
&= lim_{ntoinfty} int_{E_n} f,dmu
end{align}$$
provided that $chi_{E_n}tochi_E$ almost everywhere.
The pointwise convergence a.e. of $chi_{E_n}$ can be a result of letting $$E=bigcup_n E_n, $$
for example, or even a weaker hypothesis like $E=liminf_{ntoinfty} E_n$, i.e.
$$
E = bigcup_{n=1}^infty bigcap_{kge n} E_k.
$$
If we only assume that $fin L^1_{text{loc}}(Bbb R)$ then the previous convergent result still holds, provided that we assume further that $E,E_n$ are contained in some bounded set $Bsubset Bbb R$ (because we can apply the above argument by viewing $fin L^1(B)$).
If $f$ is not even locally integrable, then we can find counterexamples so that $int_Ef,dmu ne lim_{ntoinfty} int_{E_n} f,dmu$.
Alternatively, we can consider the pseudo-distance function
$$
d(A,B) := mu(ADelta B)
$$
defined on $mathcal A$, the family of Lebesgue measurable sets of $Bbb R$. Define the equivalent relation $Asim B$ iff $d(A,B)=0$, then the set of equivalent classes $mathcal A/sim$ forms a (complete) metric space. We can now interpret $E = lim_{ntoinfty} E_n$ as
$$
d(E,E_n) to 0
$$
in $mathcal A/sim$.
Now, any $fin L^1(Bbb R)$ induces an absolutely continuous signed measure on $Bbb R$. Hence for any $varepsilon>0$ we can find a number $delta>0$ such that $mu(A)<delta$ implies
$$
int_A |f|,dmu < varepsilon.
$$
Thus for sufficiently large $n$, we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu < varepsilon
$$
since $mu(EDelta E_n)to 0$. This shows that $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$.
Similarly, if we instead assume $fin L^infty (Bbb R)$ then for sufficiently large $n$ we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu le mu(EDelta E_n) ||f||_infty to 0
$$
as $d(E,E_n)to 0$, which implies $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$ also.
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
$endgroup$
1
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
add a comment |
$begingroup$
Let $f$ be an integrable function. i.e. $fin L^1(Bbb R)$. Then we have $|fchi_{E_n}|le |f|$ for all $ninBbb N$, hence by Dominated Convergence Theorem we have
$$begin{align}
int_Ef,dmu &= int fchi_E,dmu \
&= lim_{ntoinfty} int fchi_{E_n} ,dmu \
&= lim_{ntoinfty} int_{E_n} f,dmu
end{align}$$
provided that $chi_{E_n}tochi_E$ almost everywhere.
The pointwise convergence a.e. of $chi_{E_n}$ can be a result of letting $$E=bigcup_n E_n, $$
for example, or even a weaker hypothesis like $E=liminf_{ntoinfty} E_n$, i.e.
$$
E = bigcup_{n=1}^infty bigcap_{kge n} E_k.
$$
If we only assume that $fin L^1_{text{loc}}(Bbb R)$ then the previous convergent result still holds, provided that we assume further that $E,E_n$ are contained in some bounded set $Bsubset Bbb R$ (because we can apply the above argument by viewing $fin L^1(B)$).
If $f$ is not even locally integrable, then we can find counterexamples so that $int_Ef,dmu ne lim_{ntoinfty} int_{E_n} f,dmu$.
Alternatively, we can consider the pseudo-distance function
$$
d(A,B) := mu(ADelta B)
$$
defined on $mathcal A$, the family of Lebesgue measurable sets of $Bbb R$. Define the equivalent relation $Asim B$ iff $d(A,B)=0$, then the set of equivalent classes $mathcal A/sim$ forms a (complete) metric space. We can now interpret $E = lim_{ntoinfty} E_n$ as
$$
d(E,E_n) to 0
$$
in $mathcal A/sim$.
Now, any $fin L^1(Bbb R)$ induces an absolutely continuous signed measure on $Bbb R$. Hence for any $varepsilon>0$ we can find a number $delta>0$ such that $mu(A)<delta$ implies
$$
int_A |f|,dmu < varepsilon.
$$
Thus for sufficiently large $n$, we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu < varepsilon
$$
since $mu(EDelta E_n)to 0$. This shows that $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$.
Similarly, if we instead assume $fin L^infty (Bbb R)$ then for sufficiently large $n$ we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu le mu(EDelta E_n) ||f||_infty to 0
$$
as $d(E,E_n)to 0$, which implies $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$ also.
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
$endgroup$
Let $f$ be an integrable function. i.e. $fin L^1(Bbb R)$. Then we have $|fchi_{E_n}|le |f|$ for all $ninBbb N$, hence by Dominated Convergence Theorem we have
$$begin{align}
int_Ef,dmu &= int fchi_E,dmu \
&= lim_{ntoinfty} int fchi_{E_n} ,dmu \
&= lim_{ntoinfty} int_{E_n} f,dmu
end{align}$$
provided that $chi_{E_n}tochi_E$ almost everywhere.
The pointwise convergence a.e. of $chi_{E_n}$ can be a result of letting $$E=bigcup_n E_n, $$
for example, or even a weaker hypothesis like $E=liminf_{ntoinfty} E_n$, i.e.
$$
E = bigcup_{n=1}^infty bigcap_{kge n} E_k.
$$
If we only assume that $fin L^1_{text{loc}}(Bbb R)$ then the previous convergent result still holds, provided that we assume further that $E,E_n$ are contained in some bounded set $Bsubset Bbb R$ (because we can apply the above argument by viewing $fin L^1(B)$).
If $f$ is not even locally integrable, then we can find counterexamples so that $int_Ef,dmu ne lim_{ntoinfty} int_{E_n} f,dmu$.
Alternatively, we can consider the pseudo-distance function
$$
d(A,B) := mu(ADelta B)
$$
defined on $mathcal A$, the family of Lebesgue measurable sets of $Bbb R$. Define the equivalent relation $Asim B$ iff $d(A,B)=0$, then the set of equivalent classes $mathcal A/sim$ forms a (complete) metric space. We can now interpret $E = lim_{ntoinfty} E_n$ as
$$
d(E,E_n) to 0
$$
in $mathcal A/sim$.
Now, any $fin L^1(Bbb R)$ induces an absolutely continuous signed measure on $Bbb R$. Hence for any $varepsilon>0$ we can find a number $delta>0$ such that $mu(A)<delta$ implies
$$
int_A |f|,dmu < varepsilon.
$$
Thus for sufficiently large $n$, we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu < varepsilon
$$
since $mu(EDelta E_n)to 0$. This shows that $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$.
Similarly, if we instead assume $fin L^infty (Bbb R)$ then for sufficiently large $n$ we have
$$
left| int_E f,dmu - int_{E_n} f,dmu right|le int_{EDelta E_n} |f|,dmu le mu(EDelta E_n) ||f||_infty to 0
$$
as $d(E,E_n)to 0$, which implies $lim_{n to infty}int_{E_n}f d mu = int_{E}f dmu$ also.
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
edited Jan 16 at 21:55
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
answered Jan 16 at 20:30
BigbearZzzBigbearZzz
8,75121652
8,75121652
1
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
add a comment |
1
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
1
1
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
$begingroup$
Very nice answer!
$endgroup$
– Jannik Pitt
Jan 19 at 14:46
add a comment |
$begingroup$
Take the Lebesgue measured space $(mathbb R, mu)$, $E_n= (-infty, 1/n)$ and
$$f(x)=begin{cases}
0 & x le 0\
1/x & x>0
end{cases}$$
Then $limlimits_{n to infty}E_n =(-infty , 0]$. And
$$infty = lim_{n to infty}int_{E_n}f d mu neq int_{E}f dmu =0$$
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
$begingroup$
Take the Lebesgue measured space $(mathbb R, mu)$, $E_n= (-infty, 1/n)$ and
$$f(x)=begin{cases}
0 & x le 0\
1/x & x>0
end{cases}$$
Then $limlimits_{n to infty}E_n =(-infty , 0]$. And
$$infty = lim_{n to infty}int_{E_n}f d mu neq int_{E}f dmu =0$$
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
$begingroup$
Take the Lebesgue measured space $(mathbb R, mu)$, $E_n= (-infty, 1/n)$ and
$$f(x)=begin{cases}
0 & x le 0\
1/x & x>0
end{cases}$$
Then $limlimits_{n to infty}E_n =(-infty , 0]$. And
$$infty = lim_{n to infty}int_{E_n}f d mu neq int_{E}f dmu =0$$
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
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Take the Lebesgue measured space $(mathbb R, mu)$, $E_n= (-infty, 1/n)$ and
$$f(x)=begin{cases}
0 & x le 0\
1/x & x>0
end{cases}$$
Then $limlimits_{n to infty}E_n =(-infty , 0]$. And
$$infty = lim_{n to infty}int_{E_n}f d mu neq int_{E}f dmu =0$$
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
edited Jan 16 at 20:29
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 16 at 20:24
mathcounterexamples.netmathcounterexamples.net
26.9k22157
26.9k22157
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add a comment |
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Does it help if you write $displaystyleint_{E_n} f,dmu = displaystyleint_E fchi_{E_n},dmu$?
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– Ted Shifrin
Jan 16 at 19:48
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@TedShifrin Yes I tried that. At least for positive $f$ wouldn't that imply that this is always true? The limit and the integral can be exchanged because $fchi_{E_n}$ is non-decreasing.
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– Jannik Pitt
Jan 16 at 19:50
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What kind of function is $f$? What do you mean by $lim_{n to infty} E_n =E$?
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– BigbearZzz
Jan 16 at 19:56
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@BigbearZzz See en.m.wikipedia.org/wiki/Set-theoretic_limit
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– Jannik Pitt
Jan 16 at 20:14
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@JannikPitt Is $f$ integrable? Locally integrable?
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– BigbearZzz
Jan 16 at 20:15