How to integrate for a countable sum of measures?

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Generalizing this question, when can we integrate in a similar way for a countable sum of measures?
$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$
probability-theory measure-theory
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up vote
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down vote
favorite
Generalizing this question, when can we integrate in a similar way for a countable sum of measures?
$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$
probability-theory measure-theory
1
You need to add some conditions on $a_n, m_n$. For example, with $a_n=1$ and $m_n $ the Lebesgue measure, the only integrable functions are those that are zero ae.
â copper.hat
Aug 27 at 14:45
@copper.hat thanks for your comment. I don't quite understand the example you gave. What would be the problem of adding countable number of equal terms?
â An old man in the sea.
Aug 27 at 17:30
For a finite linear combination, if $f$ is integrable with respect to each constituent measure, then it is integrable with respect to the linear combination. For a countable combination, this is not necessarily the case. To some extent, this is more about summations that measures. If the $a_n$ are (absolutely) summable, then $f$ would be integrable assuming that it is integrable with respect to each constituent measure.
â copper.hat
Aug 27 at 17:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Generalizing this question, when can we integrate in a similar way for a countable sum of measures?
$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$
probability-theory measure-theory
Generalizing this question, when can we integrate in a similar way for a countable sum of measures?
$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$
probability-theory measure-theory
asked Aug 27 at 13:50
An old man in the sea.
1,2901929
1,2901929
1
You need to add some conditions on $a_n, m_n$. For example, with $a_n=1$ and $m_n $ the Lebesgue measure, the only integrable functions are those that are zero ae.
â copper.hat
Aug 27 at 14:45
@copper.hat thanks for your comment. I don't quite understand the example you gave. What would be the problem of adding countable number of equal terms?
â An old man in the sea.
Aug 27 at 17:30
For a finite linear combination, if $f$ is integrable with respect to each constituent measure, then it is integrable with respect to the linear combination. For a countable combination, this is not necessarily the case. To some extent, this is more about summations that measures. If the $a_n$ are (absolutely) summable, then $f$ would be integrable assuming that it is integrable with respect to each constituent measure.
â copper.hat
Aug 27 at 17:54
add a comment |Â
1
You need to add some conditions on $a_n, m_n$. For example, with $a_n=1$ and $m_n $ the Lebesgue measure, the only integrable functions are those that are zero ae.
â copper.hat
Aug 27 at 14:45
@copper.hat thanks for your comment. I don't quite understand the example you gave. What would be the problem of adding countable number of equal terms?
â An old man in the sea.
Aug 27 at 17:30
For a finite linear combination, if $f$ is integrable with respect to each constituent measure, then it is integrable with respect to the linear combination. For a countable combination, this is not necessarily the case. To some extent, this is more about summations that measures. If the $a_n$ are (absolutely) summable, then $f$ would be integrable assuming that it is integrable with respect to each constituent measure.
â copper.hat
Aug 27 at 17:54
1
1
You need to add some conditions on $a_n, m_n$. For example, with $a_n=1$ and $m_n $ the Lebesgue measure, the only integrable functions are those that are zero ae.
â copper.hat
Aug 27 at 14:45
You need to add some conditions on $a_n, m_n$. For example, with $a_n=1$ and $m_n $ the Lebesgue measure, the only integrable functions are those that are zero ae.
â copper.hat
Aug 27 at 14:45
@copper.hat thanks for your comment. I don't quite understand the example you gave. What would be the problem of adding countable number of equal terms?
â An old man in the sea.
Aug 27 at 17:30
@copper.hat thanks for your comment. I don't quite understand the example you gave. What would be the problem of adding countable number of equal terms?
â An old man in the sea.
Aug 27 at 17:30
For a finite linear combination, if $f$ is integrable with respect to each constituent measure, then it is integrable with respect to the linear combination. For a countable combination, this is not necessarily the case. To some extent, this is more about summations that measures. If the $a_n$ are (absolutely) summable, then $f$ would be integrable assuming that it is integrable with respect to each constituent measure.
â copper.hat
Aug 27 at 17:54
For a finite linear combination, if $f$ is integrable with respect to each constituent measure, then it is integrable with respect to the linear combination. For a countable combination, this is not necessarily the case. To some extent, this is more about summations that measures. If the $a_n$ are (absolutely) summable, then $f$ would be integrable assuming that it is integrable with respect to each constituent measure.
â copper.hat
Aug 27 at 17:54
add a comment |Â
2 Answers
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oldest
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up vote
1
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accepted
It works as long as the $a_n$ are non-negativ. Let us denote with $mathcalA_n$ the $sigma$-Algebra of $m_n$-measurable sets. Then
$$mathcalA=bigcap_n mathcalA_n$$
is another $sigma$-Algebra on which we need to work to define $m$:
For $Ain mathcalA$ we set
$$m(A):=sum_na_nm_n(A).$$
Now $m$ becomes a measure because if you take pairwise disjoint and countable many $A_kinmathcalA$ you get
$$m(cup_k A_k)= sum_na_nm_n(cup A_k) = sum_n a_nsum_k m_n(A_k) = sum_ksum_n a_n m_n(A_k)=sum_k m(A_k),$$
since $A_kin mathcalA_n$ and by $a_ngeq0$ the corresponding sums converge absolutely.
Now $m$ is well defined and we can proceed to the integrals. First for simple $f(x)=sum_k=1^N b_k 1_B_k(x)$, $B_kinmathcalA$ and $b_kgeq 0$
$$int fdm= sum_k=1^Nb_km(B_k) = sum_k=1^N b_k sum_n a_nm_n(B_k) = sum_na_nsum_k=1^Nb_km_n(B_k)\ =sum_n a_nint fdm_n.$$
Now approximate a $mathcalA$-measurable $fgeq0$ pointwise by a non-decreasing sequence of simple functions $f_k$. Beppo-Levi's theorem yields
$$int fd_m = lim_krightarrowinftyint f_kdm = lim_krightarrowinftysum_na_nint f_kdm_n=sum_n a_nlim_krightarrowinftyint f_kdm_n = sum_na_nint fdm_n.$$
Please note that the exchange of $sum$ and $lim$ is also due to Beppo-Levi applied to the counting measure and the fact that $kmapsto a_nint f_kdm_n$ is non-decreasing in $k$
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
 |Â
show 2 more comments
up vote
0
down vote
There is a more precise answer to the question, as a matter of fact a necessary and sufficient condition, as I pointed out a few months ago answering another question on the MathOverflow. Since
$$
m=sum_ninmathbbNa_n m_ntriangleqlim_Ntoinfty sum_leq Na_n m_ntag1label1
$$
the OP question is equivalent to asking when it is possible the passage to the limit under the integral symbol, i.e. when
$$
int fmathrmdmtriangleqint fmathrmdleft(sum_ninmathbbNa_n m_nright)=lim_Ntoinfty sum_leq Na_nint f mathrmdm_ntriangleqsum_ninmathbbNa_n int f mathrmdm_n; ?tag2label2
$$
The answer requires the two following definitions
Definition 1. Let $(E,mathcalE)$ be a measure space and $phi:mathcalEtooverlinemathbbR$ a numerical set function:
$phi$ is called exhaustive if
$$
lim_nphi(A_n)=0
$$
for all families $A_n$ of pairwise disjoint sets in $mathcalE$.
Definition 2. Let $(E,mathcalE)$ be a measure space and $H$ a set (and thus possibly a family) of numerical set functions defined on $mathcalE$: $H$ is called uniformly exhaustive if the numerical set function
$$
Amapstosup_phiin H vertphi(A)vert;text is exhaustive.
$$
By Cafiero's theorem (on the passage to the limit under the integral), formula eqref2 holds if and only if the limit eqref2 exist pointwise and
$$
bigg(fcdotsum_leq Na_n m_nbigg)_Ngeq 1=bigg(sum_leq Nfcdot a_n m_nbigg)_Ngeq 1=bigg(sum_leq N a_n, fcdot m_nbigg)_Ngeq 1text is uniformly exaustive.
$$
A minimal bibliography on the theorem is available in my MathOverflow post cited above
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It works as long as the $a_n$ are non-negativ. Let us denote with $mathcalA_n$ the $sigma$-Algebra of $m_n$-measurable sets. Then
$$mathcalA=bigcap_n mathcalA_n$$
is another $sigma$-Algebra on which we need to work to define $m$:
For $Ain mathcalA$ we set
$$m(A):=sum_na_nm_n(A).$$
Now $m$ becomes a measure because if you take pairwise disjoint and countable many $A_kinmathcalA$ you get
$$m(cup_k A_k)= sum_na_nm_n(cup A_k) = sum_n a_nsum_k m_n(A_k) = sum_ksum_n a_n m_n(A_k)=sum_k m(A_k),$$
since $A_kin mathcalA_n$ and by $a_ngeq0$ the corresponding sums converge absolutely.
Now $m$ is well defined and we can proceed to the integrals. First for simple $f(x)=sum_k=1^N b_k 1_B_k(x)$, $B_kinmathcalA$ and $b_kgeq 0$
$$int fdm= sum_k=1^Nb_km(B_k) = sum_k=1^N b_k sum_n a_nm_n(B_k) = sum_na_nsum_k=1^Nb_km_n(B_k)\ =sum_n a_nint fdm_n.$$
Now approximate a $mathcalA$-measurable $fgeq0$ pointwise by a non-decreasing sequence of simple functions $f_k$. Beppo-Levi's theorem yields
$$int fd_m = lim_krightarrowinftyint f_kdm = lim_krightarrowinftysum_na_nint f_kdm_n=sum_n a_nlim_krightarrowinftyint f_kdm_n = sum_na_nint fdm_n.$$
Please note that the exchange of $sum$ and $lim$ is also due to Beppo-Levi applied to the counting measure and the fact that $kmapsto a_nint f_kdm_n$ is non-decreasing in $k$
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
 |Â
show 2 more comments
up vote
1
down vote
accepted
It works as long as the $a_n$ are non-negativ. Let us denote with $mathcalA_n$ the $sigma$-Algebra of $m_n$-measurable sets. Then
$$mathcalA=bigcap_n mathcalA_n$$
is another $sigma$-Algebra on which we need to work to define $m$:
For $Ain mathcalA$ we set
$$m(A):=sum_na_nm_n(A).$$
Now $m$ becomes a measure because if you take pairwise disjoint and countable many $A_kinmathcalA$ you get
$$m(cup_k A_k)= sum_na_nm_n(cup A_k) = sum_n a_nsum_k m_n(A_k) = sum_ksum_n a_n m_n(A_k)=sum_k m(A_k),$$
since $A_kin mathcalA_n$ and by $a_ngeq0$ the corresponding sums converge absolutely.
Now $m$ is well defined and we can proceed to the integrals. First for simple $f(x)=sum_k=1^N b_k 1_B_k(x)$, $B_kinmathcalA$ and $b_kgeq 0$
$$int fdm= sum_k=1^Nb_km(B_k) = sum_k=1^N b_k sum_n a_nm_n(B_k) = sum_na_nsum_k=1^Nb_km_n(B_k)\ =sum_n a_nint fdm_n.$$
Now approximate a $mathcalA$-measurable $fgeq0$ pointwise by a non-decreasing sequence of simple functions $f_k$. Beppo-Levi's theorem yields
$$int fd_m = lim_krightarrowinftyint f_kdm = lim_krightarrowinftysum_na_nint f_kdm_n=sum_n a_nlim_krightarrowinftyint f_kdm_n = sum_na_nint fdm_n.$$
Please note that the exchange of $sum$ and $lim$ is also due to Beppo-Levi applied to the counting measure and the fact that $kmapsto a_nint f_kdm_n$ is non-decreasing in $k$
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
 |Â
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It works as long as the $a_n$ are non-negativ. Let us denote with $mathcalA_n$ the $sigma$-Algebra of $m_n$-measurable sets. Then
$$mathcalA=bigcap_n mathcalA_n$$
is another $sigma$-Algebra on which we need to work to define $m$:
For $Ain mathcalA$ we set
$$m(A):=sum_na_nm_n(A).$$
Now $m$ becomes a measure because if you take pairwise disjoint and countable many $A_kinmathcalA$ you get
$$m(cup_k A_k)= sum_na_nm_n(cup A_k) = sum_n a_nsum_k m_n(A_k) = sum_ksum_n a_n m_n(A_k)=sum_k m(A_k),$$
since $A_kin mathcalA_n$ and by $a_ngeq0$ the corresponding sums converge absolutely.
Now $m$ is well defined and we can proceed to the integrals. First for simple $f(x)=sum_k=1^N b_k 1_B_k(x)$, $B_kinmathcalA$ and $b_kgeq 0$
$$int fdm= sum_k=1^Nb_km(B_k) = sum_k=1^N b_k sum_n a_nm_n(B_k) = sum_na_nsum_k=1^Nb_km_n(B_k)\ =sum_n a_nint fdm_n.$$
Now approximate a $mathcalA$-measurable $fgeq0$ pointwise by a non-decreasing sequence of simple functions $f_k$. Beppo-Levi's theorem yields
$$int fd_m = lim_krightarrowinftyint f_kdm = lim_krightarrowinftysum_na_nint f_kdm_n=sum_n a_nlim_krightarrowinftyint f_kdm_n = sum_na_nint fdm_n.$$
Please note that the exchange of $sum$ and $lim$ is also due to Beppo-Levi applied to the counting measure and the fact that $kmapsto a_nint f_kdm_n$ is non-decreasing in $k$
It works as long as the $a_n$ are non-negativ. Let us denote with $mathcalA_n$ the $sigma$-Algebra of $m_n$-measurable sets. Then
$$mathcalA=bigcap_n mathcalA_n$$
is another $sigma$-Algebra on which we need to work to define $m$:
For $Ain mathcalA$ we set
$$m(A):=sum_na_nm_n(A).$$
Now $m$ becomes a measure because if you take pairwise disjoint and countable many $A_kinmathcalA$ you get
$$m(cup_k A_k)= sum_na_nm_n(cup A_k) = sum_n a_nsum_k m_n(A_k) = sum_ksum_n a_n m_n(A_k)=sum_k m(A_k),$$
since $A_kin mathcalA_n$ and by $a_ngeq0$ the corresponding sums converge absolutely.
Now $m$ is well defined and we can proceed to the integrals. First for simple $f(x)=sum_k=1^N b_k 1_B_k(x)$, $B_kinmathcalA$ and $b_kgeq 0$
$$int fdm= sum_k=1^Nb_km(B_k) = sum_k=1^N b_k sum_n a_nm_n(B_k) = sum_na_nsum_k=1^Nb_km_n(B_k)\ =sum_n a_nint fdm_n.$$
Now approximate a $mathcalA$-measurable $fgeq0$ pointwise by a non-decreasing sequence of simple functions $f_k$. Beppo-Levi's theorem yields
$$int fd_m = lim_krightarrowinftyint f_kdm = lim_krightarrowinftysum_na_nint f_kdm_n=sum_n a_nlim_krightarrowinftyint f_kdm_n = sum_na_nint fdm_n.$$
Please note that the exchange of $sum$ and $lim$ is also due to Beppo-Levi applied to the counting measure and the fact that $kmapsto a_nint f_kdm_n$ is non-decreasing in $k$
answered Aug 27 at 15:09
humanStampedist
2,036213
2,036213
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
 |Â
show 2 more comments
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
â An old man in the sea.
Aug 27 at 17:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@Anoldmaninthesea. Do you mean measure theory or sum of measures?
â i try
Aug 27 at 19:21
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
@itry for sum of measures. I've searched some books, and they don't reflect much on sum of measures...
â An old man in the sea.
Aug 27 at 19:22
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
â i try
Aug 27 at 19:23
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
@itry I know that a countable sum of measures is itself a measure. What I'm interested is in a reference on when is it possible to integrate as I write on the question... ;)
â An old man in the sea.
Aug 27 at 19:31
 |Â
show 2 more comments
up vote
0
down vote
There is a more precise answer to the question, as a matter of fact a necessary and sufficient condition, as I pointed out a few months ago answering another question on the MathOverflow. Since
$$
m=sum_ninmathbbNa_n m_ntriangleqlim_Ntoinfty sum_leq Na_n m_ntag1label1
$$
the OP question is equivalent to asking when it is possible the passage to the limit under the integral symbol, i.e. when
$$
int fmathrmdmtriangleqint fmathrmdleft(sum_ninmathbbNa_n m_nright)=lim_Ntoinfty sum_leq Na_nint f mathrmdm_ntriangleqsum_ninmathbbNa_n int f mathrmdm_n; ?tag2label2
$$
The answer requires the two following definitions
Definition 1. Let $(E,mathcalE)$ be a measure space and $phi:mathcalEtooverlinemathbbR$ a numerical set function:
$phi$ is called exhaustive if
$$
lim_nphi(A_n)=0
$$
for all families $A_n$ of pairwise disjoint sets in $mathcalE$.
Definition 2. Let $(E,mathcalE)$ be a measure space and $H$ a set (and thus possibly a family) of numerical set functions defined on $mathcalE$: $H$ is called uniformly exhaustive if the numerical set function
$$
Amapstosup_phiin H vertphi(A)vert;text is exhaustive.
$$
By Cafiero's theorem (on the passage to the limit under the integral), formula eqref2 holds if and only if the limit eqref2 exist pointwise and
$$
bigg(fcdotsum_leq Na_n m_nbigg)_Ngeq 1=bigg(sum_leq Nfcdot a_n m_nbigg)_Ngeq 1=bigg(sum_leq N a_n, fcdot m_nbigg)_Ngeq 1text is uniformly exaustive.
$$
A minimal bibliography on the theorem is available in my MathOverflow post cited above
add a comment |Â
up vote
0
down vote
There is a more precise answer to the question, as a matter of fact a necessary and sufficient condition, as I pointed out a few months ago answering another question on the MathOverflow. Since
$$
m=sum_ninmathbbNa_n m_ntriangleqlim_Ntoinfty sum_leq Na_n m_ntag1label1
$$
the OP question is equivalent to asking when it is possible the passage to the limit under the integral symbol, i.e. when
$$
int fmathrmdmtriangleqint fmathrmdleft(sum_ninmathbbNa_n m_nright)=lim_Ntoinfty sum_leq Na_nint f mathrmdm_ntriangleqsum_ninmathbbNa_n int f mathrmdm_n; ?tag2label2
$$
The answer requires the two following definitions
Definition 1. Let $(E,mathcalE)$ be a measure space and $phi:mathcalEtooverlinemathbbR$ a numerical set function:
$phi$ is called exhaustive if
$$
lim_nphi(A_n)=0
$$
for all families $A_n$ of pairwise disjoint sets in $mathcalE$.
Definition 2. Let $(E,mathcalE)$ be a measure space and $H$ a set (and thus possibly a family) of numerical set functions defined on $mathcalE$: $H$ is called uniformly exhaustive if the numerical set function
$$
Amapstosup_phiin H vertphi(A)vert;text is exhaustive.
$$
By Cafiero's theorem (on the passage to the limit under the integral), formula eqref2 holds if and only if the limit eqref2 exist pointwise and
$$
bigg(fcdotsum_leq Na_n m_nbigg)_Ngeq 1=bigg(sum_leq Nfcdot a_n m_nbigg)_Ngeq 1=bigg(sum_leq N a_n, fcdot m_nbigg)_Ngeq 1text is uniformly exaustive.
$$
A minimal bibliography on the theorem is available in my MathOverflow post cited above
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is a more precise answer to the question, as a matter of fact a necessary and sufficient condition, as I pointed out a few months ago answering another question on the MathOverflow. Since
$$
m=sum_ninmathbbNa_n m_ntriangleqlim_Ntoinfty sum_leq Na_n m_ntag1label1
$$
the OP question is equivalent to asking when it is possible the passage to the limit under the integral symbol, i.e. when
$$
int fmathrmdmtriangleqint fmathrmdleft(sum_ninmathbbNa_n m_nright)=lim_Ntoinfty sum_leq Na_nint f mathrmdm_ntriangleqsum_ninmathbbNa_n int f mathrmdm_n; ?tag2label2
$$
The answer requires the two following definitions
Definition 1. Let $(E,mathcalE)$ be a measure space and $phi:mathcalEtooverlinemathbbR$ a numerical set function:
$phi$ is called exhaustive if
$$
lim_nphi(A_n)=0
$$
for all families $A_n$ of pairwise disjoint sets in $mathcalE$.
Definition 2. Let $(E,mathcalE)$ be a measure space and $H$ a set (and thus possibly a family) of numerical set functions defined on $mathcalE$: $H$ is called uniformly exhaustive if the numerical set function
$$
Amapstosup_phiin H vertphi(A)vert;text is exhaustive.
$$
By Cafiero's theorem (on the passage to the limit under the integral), formula eqref2 holds if and only if the limit eqref2 exist pointwise and
$$
bigg(fcdotsum_leq Na_n m_nbigg)_Ngeq 1=bigg(sum_leq Nfcdot a_n m_nbigg)_Ngeq 1=bigg(sum_leq N a_n, fcdot m_nbigg)_Ngeq 1text is uniformly exaustive.
$$
A minimal bibliography on the theorem is available in my MathOverflow post cited above
There is a more precise answer to the question, as a matter of fact a necessary and sufficient condition, as I pointed out a few months ago answering another question on the MathOverflow. Since
$$
m=sum_ninmathbbNa_n m_ntriangleqlim_Ntoinfty sum_leq Na_n m_ntag1label1
$$
the OP question is equivalent to asking when it is possible the passage to the limit under the integral symbol, i.e. when
$$
int fmathrmdmtriangleqint fmathrmdleft(sum_ninmathbbNa_n m_nright)=lim_Ntoinfty sum_leq Na_nint f mathrmdm_ntriangleqsum_ninmathbbNa_n int f mathrmdm_n; ?tag2label2
$$
The answer requires the two following definitions
Definition 1. Let $(E,mathcalE)$ be a measure space and $phi:mathcalEtooverlinemathbbR$ a numerical set function:
$phi$ is called exhaustive if
$$
lim_nphi(A_n)=0
$$
for all families $A_n$ of pairwise disjoint sets in $mathcalE$.
Definition 2. Let $(E,mathcalE)$ be a measure space and $H$ a set (and thus possibly a family) of numerical set functions defined on $mathcalE$: $H$ is called uniformly exhaustive if the numerical set function
$$
Amapstosup_phiin H vertphi(A)vert;text is exhaustive.
$$
By Cafiero's theorem (on the passage to the limit under the integral), formula eqref2 holds if and only if the limit eqref2 exist pointwise and
$$
bigg(fcdotsum_leq Na_n m_nbigg)_Ngeq 1=bigg(sum_leq Nfcdot a_n m_nbigg)_Ngeq 1=bigg(sum_leq N a_n, fcdot m_nbigg)_Ngeq 1text is uniformly exaustive.
$$
A minimal bibliography on the theorem is available in my MathOverflow post cited above
answered Aug 28 at 21:31
Daniele Tampieri
7851314
7851314
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1
You need to add some conditions on $a_n, m_n$. For example, with $a_n=1$ and $m_n $ the Lebesgue measure, the only integrable functions are those that are zero ae.
â copper.hat
Aug 27 at 14:45
@copper.hat thanks for your comment. I don't quite understand the example you gave. What would be the problem of adding countable number of equal terms?
â An old man in the sea.
Aug 27 at 17:30
For a finite linear combination, if $f$ is integrable with respect to each constituent measure, then it is integrable with respect to the linear combination. For a countable combination, this is not necessarily the case. To some extent, this is more about summations that measures. If the $a_n$ are (absolutely) summable, then $f$ would be integrable assuming that it is integrable with respect to each constituent measure.
â copper.hat
Aug 27 at 17:54