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?



$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$







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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.
    – 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?
    – 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.
    – copper.hat
    Aug 27 at 17:54














up vote
2
down vote

favorite












Generalizing this question, when can we integrate in a similar way for a countable sum of measures?



$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$







share|cite|improve this question
















  • 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.
    – 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?
    – 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.
    – copper.hat
    Aug 27 at 17:54












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?



$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$







share|cite|improve this question












Generalizing this question, when can we integrate in a similar way for a countable sum of measures?



$$int f,dm=sum_n in mathbbN a_n int f,dm_n$$









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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.
    – 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?
    – 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.
    – copper.hat
    Aug 27 at 17:54












  • 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.
    – 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?
    – 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.
    – 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.
– 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.
– 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?
– 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?
– 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.
– 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.
– copper.hat
Aug 27 at 17:54










2 Answers
2






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 $k$






share|cite|improve this answer




















  • thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
    – An old man in the sea.
    Aug 27 at 17:21










  • @Anoldmaninthesea. Do you mean measure theory or sum of measures?
    – 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...
    – 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.
    – 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... ;)
    – An old man in the sea.
    Aug 27 at 19:31

















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






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

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    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 $k$






    share|cite|improve this answer




















    • thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
      – An old man in the sea.
      Aug 27 at 17:21










    • @Anoldmaninthesea. Do you mean measure theory or sum of measures?
      – 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...
      – 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.
      – 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... ;)
      – An old man in the sea.
      Aug 27 at 19:31














    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 $k$






    share|cite|improve this answer




















    • thanks for your answer. I'll read it soon. By the way, do you have some reference for this specific question?
      – An old man in the sea.
      Aug 27 at 17:21










    • @Anoldmaninthesea. Do you mean measure theory or sum of measures?
      – 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...
      – 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.
      – 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... ;)
      – An old man in the sea.
      Aug 27 at 19:31












    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 $k$






    share|cite|improve this answer












    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 $k$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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?
      – An old man in the sea.
      Aug 27 at 17:21










    • @Anoldmaninthesea. Do you mean measure theory or sum of measures?
      – 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...
      – 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.
      – 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... ;)
      – 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?
      – An old man in the sea.
      Aug 27 at 17:21










    • @Anoldmaninthesea. Do you mean measure theory or sum of measures?
      – 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...
      – 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.
      – 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... ;)
      – 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?
    – 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?
    – An old man in the sea.
    Aug 27 at 17:21












    @Anoldmaninthesea. Do you mean measure theory or sum of measures?
    – i try
    Aug 27 at 19:21




    @Anoldmaninthesea. Do you mean measure theory or sum of measures?
    – 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...
    – 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...
    – 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.
    – i try
    Aug 27 at 19:23




    It is problem $4.6$ in the book: Measures, Integrals and Martingales by Rene Schilling.
    – 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... ;)
    – 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... ;)
    – An old man in the sea.
    Aug 27 at 19:31










    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






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      up vote
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      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






      share|cite|improve this answer






















        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






        share|cite|improve this answer












        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







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Aug 28 at 21:31









        Daniele Tampieri

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