Show that $lim_{ntoinfty}int_{X}fdmu_n = int_Xfdmu$












1












$begingroup$


Given is a a $sigma$ - Algebra $mathcal{A}$, a measure space $(X, mathcal{A})$ and a sequence of measures $left(mu_n :mathcal{A} to [0,infty]right)_{nin mathbb{N}}$ with $mu_n(A) leq mu_{n+1}(A):forall nin mathbb{N}$. I have to show that if $f:Xto[0,infty]$ is measureable with respect to $mathcal{A}$, it follows that $lim_{ntoinfty}int_{X}fdmu_n = int_Xfdmu$, where $mu: mathcal{A}to[0,infty]$, $mu(A):=sup{mu_n(A):ninmathbb{N}}$ and $Ainmathcal{A}$. We have defind integrals with step functions, hence $f=sum_{k=1}^Nalpha_k1_{A_k}$ and it follows by definition that $int_Xfdmu=sum_{k=1}^Nalpha_kmu(A_k)$ for an arbitrary measure $mu$. I have literally no idea where to start...






real-analysis measure-theory convergence







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    First prove the case where $f$ is a step function. Can you do that?
    $endgroup$
    – Severin Schraven
    Jan 2 at 14:02
















1












$begingroup$


Given is a a $sigma$ - Algebra $mathcal{A}$, a measure space $(X, mathcal{A})$ and a sequence of measures $left(mu_n :mathcal{A} to [0,infty]right)_{nin mathbb{N}}$ with $mu_n(A) leq mu_{n+1}(A):forall nin mathbb{N}$. I have to show that if $f:Xto[0,infty]$ is measureable with respect to $mathcal{A}$, it follows that $lim_{ntoinfty}int_{X}fdmu_n = int_Xfdmu$, where $mu: mathcal{A}to[0,infty]$, $mu(A):=sup{mu_n(A):ninmathbb{N}}$ and $Ainmathcal{A}$. We have defind integrals with step functions, hence $f=sum_{k=1}^Nalpha_k1_{A_k}$ and it follows by definition that $int_Xfdmu=sum_{k=1}^Nalpha_kmu(A_k)$ for an arbitrary measure $mu$. I have literally no idea where to start...






real-analysis measure-theory convergence







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    First prove the case where $f$ is a step function. Can you do that?
    $endgroup$
    – Severin Schraven
    Jan 2 at 14:02














1












1








1





$begingroup$


Given is a a $sigma$ - Algebra $mathcal{A}$, a measure space $(X, mathcal{A})$ and a sequence of measures $left(mu_n :mathcal{A} to [0,infty]right)_{nin mathbb{N}}$ with $mu_n(A) leq mu_{n+1}(A):forall nin mathbb{N}$. I have to show that if $f:Xto[0,infty]$ is measureable with respect to $mathcal{A}$, it follows that $lim_{ntoinfty}int_{X}fdmu_n = int_Xfdmu$, where $mu: mathcal{A}to[0,infty]$, $mu(A):=sup{mu_n(A):ninmathbb{N}}$ and $Ainmathcal{A}$. We have defind integrals with step functions, hence $f=sum_{k=1}^Nalpha_k1_{A_k}$ and it follows by definition that $int_Xfdmu=sum_{k=1}^Nalpha_kmu(A_k)$ for an arbitrary measure $mu$. I have literally no idea where to start...






real-analysis measure-theory convergence







share|cite|improve this question











$endgroup$




Given is a a $sigma$ - Algebra $mathcal{A}$, a measure space $(X, mathcal{A})$ and a sequence of measures $left(mu_n :mathcal{A} to [0,infty]right)_{nin mathbb{N}}$ with $mu_n(A) leq mu_{n+1}(A):forall nin mathbb{N}$. I have to show that if $f:Xto[0,infty]$ is measureable with respect to $mathcal{A}$, it follows that $lim_{ntoinfty}int_{X}fdmu_n = int_Xfdmu$, where $mu: mathcal{A}to[0,infty]$, $mu(A):=sup{mu_n(A):ninmathbb{N}}$ and $Ainmathcal{A}$. We have defind integrals with step functions, hence $f=sum_{k=1}^Nalpha_k1_{A_k}$ and it follows by definition that $int_Xfdmu=sum_{k=1}^Nalpha_kmu(A_k)$ for an arbitrary measure $mu$. I have literally no idea where to start...






real-analysis measure-theory convergence




real-analysis measure-theory convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 0:00









Davide Giraudo

126k16150261




126k16150261










asked Jan 2 at 13:49









Michael MaierMichael Maier

819




819








  • 2




    $begingroup$
    First prove the case where $f$ is a step function. Can you do that?
    $endgroup$
    – Severin Schraven
    Jan 2 at 14:02














  • 2




    $begingroup$
    First prove the case where $f$ is a step function. Can you do that?
    $endgroup$
    – Severin Schraven
    Jan 2 at 14:02








2




2




$begingroup$
First prove the case where $f$ is a step function. Can you do that?
$endgroup$
– Severin Schraven
Jan 2 at 14:02




$begingroup$
First prove the case where $f$ is a step function. Can you do that?
$endgroup$
– Severin Schraven
Jan 2 at 14:02










1 Answer
1






active

oldest

votes


















1












$begingroup$

Here are some hints.




  1. First it is not clear that $mu$ is indeed a measure.

  2. Show that the sequence $left( int_{X}fdmu_nright)_{ngeqslant 1}$ is non-decreasing.

  3. As a consequence, $lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}int_{X}fdmu_n$.

  4. By definition of the Lebesgue integral, $$int_{X}fdmu_n=sup_{sinmathcal S} int_{X}sdmu_n ,$$
    where $mathcal S=left{smid 0leqslant sleqslant f, .s=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal Aright}$. Switching the two supremum reduces to establish the result when $f=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal A$. Indeed, by 3.
    $$
    lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}sup_{sinmathcal S}int_{X}sdmu_n=sup_{sinmathcal S}sup_{ngeqslant 1}int_{X}sdmu_n.
    $$
    If we manage to show that for each $sinmathcal S$, the equality $sup_{ngeqslant 1}int_{X}sdmu_n=int_{X}sdmu$, then we are done.

  5. By linearity it suffices to do it when $f =mathbf 1_{A},Ainmathcal A$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
    $endgroup$
    – Michael Maier
    Jan 4 at 10:00








  • 1




    $begingroup$
    I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 10:09










  • $begingroup$
    Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
    $endgroup$
    – Michael Maier
    Jan 4 at 10:16












  • $begingroup$
    And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
    $endgroup$
    – Michael Maier
    Jan 4 at 10:31












  • $begingroup$
    Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 11:31











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Here are some hints.




  1. First it is not clear that $mu$ is indeed a measure.

  2. Show that the sequence $left( int_{X}fdmu_nright)_{ngeqslant 1}$ is non-decreasing.

  3. As a consequence, $lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}int_{X}fdmu_n$.

  4. By definition of the Lebesgue integral, $$int_{X}fdmu_n=sup_{sinmathcal S} int_{X}sdmu_n ,$$
    where $mathcal S=left{smid 0leqslant sleqslant f, .s=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal Aright}$. Switching the two supremum reduces to establish the result when $f=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal A$. Indeed, by 3.
    $$
    lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}sup_{sinmathcal S}int_{X}sdmu_n=sup_{sinmathcal S}sup_{ngeqslant 1}int_{X}sdmu_n.
    $$
    If we manage to show that for each $sinmathcal S$, the equality $sup_{ngeqslant 1}int_{X}sdmu_n=int_{X}sdmu$, then we are done.

  5. By linearity it suffices to do it when $f =mathbf 1_{A},Ainmathcal A$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
    $endgroup$
    – Michael Maier
    Jan 4 at 10:00








  • 1




    $begingroup$
    I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 10:09










  • $begingroup$
    Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
    $endgroup$
    – Michael Maier
    Jan 4 at 10:16












  • $begingroup$
    And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
    $endgroup$
    – Michael Maier
    Jan 4 at 10:31












  • $begingroup$
    Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 11:31
















1












$begingroup$

Here are some hints.




  1. First it is not clear that $mu$ is indeed a measure.

  2. Show that the sequence $left( int_{X}fdmu_nright)_{ngeqslant 1}$ is non-decreasing.

  3. As a consequence, $lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}int_{X}fdmu_n$.

  4. By definition of the Lebesgue integral, $$int_{X}fdmu_n=sup_{sinmathcal S} int_{X}sdmu_n ,$$
    where $mathcal S=left{smid 0leqslant sleqslant f, .s=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal Aright}$. Switching the two supremum reduces to establish the result when $f=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal A$. Indeed, by 3.
    $$
    lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}sup_{sinmathcal S}int_{X}sdmu_n=sup_{sinmathcal S}sup_{ngeqslant 1}int_{X}sdmu_n.
    $$
    If we manage to show that for each $sinmathcal S$, the equality $sup_{ngeqslant 1}int_{X}sdmu_n=int_{X}sdmu$, then we are done.

  5. By linearity it suffices to do it when $f =mathbf 1_{A},Ainmathcal A$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
    $endgroup$
    – Michael Maier
    Jan 4 at 10:00








  • 1




    $begingroup$
    I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 10:09










  • $begingroup$
    Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
    $endgroup$
    – Michael Maier
    Jan 4 at 10:16












  • $begingroup$
    And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
    $endgroup$
    – Michael Maier
    Jan 4 at 10:31












  • $begingroup$
    Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 11:31














1












1








1





$begingroup$

Here are some hints.




  1. First it is not clear that $mu$ is indeed a measure.

  2. Show that the sequence $left( int_{X}fdmu_nright)_{ngeqslant 1}$ is non-decreasing.

  3. As a consequence, $lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}int_{X}fdmu_n$.

  4. By definition of the Lebesgue integral, $$int_{X}fdmu_n=sup_{sinmathcal S} int_{X}sdmu_n ,$$
    where $mathcal S=left{smid 0leqslant sleqslant f, .s=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal Aright}$. Switching the two supremum reduces to establish the result when $f=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal A$. Indeed, by 3.
    $$
    lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}sup_{sinmathcal S}int_{X}sdmu_n=sup_{sinmathcal S}sup_{ngeqslant 1}int_{X}sdmu_n.
    $$
    If we manage to show that for each $sinmathcal S$, the equality $sup_{ngeqslant 1}int_{X}sdmu_n=int_{X}sdmu$, then we are done.

  5. By linearity it suffices to do it when $f =mathbf 1_{A},Ainmathcal A$.






share|cite|improve this answer











$endgroup$



Here are some hints.




  1. First it is not clear that $mu$ is indeed a measure.

  2. Show that the sequence $left( int_{X}fdmu_nright)_{ngeqslant 1}$ is non-decreasing.

  3. As a consequence, $lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}int_{X}fdmu_n$.

  4. By definition of the Lebesgue integral, $$int_{X}fdmu_n=sup_{sinmathcal S} int_{X}sdmu_n ,$$
    where $mathcal S=left{smid 0leqslant sleqslant f, .s=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal Aright}$. Switching the two supremum reduces to establish the result when $f=sum_{i=1}^Nc_imathbf 1_{A_i},A_iinmathcal A$. Indeed, by 3.
    $$
    lim_{nto +infty}int_{X}fdmu_n=sup_{ngeqslant 1}sup_{sinmathcal S}int_{X}sdmu_n=sup_{sinmathcal S}sup_{ngeqslant 1}int_{X}sdmu_n.
    $$
    If we manage to show that for each $sinmathcal S$, the equality $sup_{ngeqslant 1}int_{X}sdmu_n=int_{X}sdmu$, then we are done.

  5. By linearity it suffices to do it when $f =mathbf 1_{A},Ainmathcal A$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 11:30

























answered Jan 2 at 14:15









Davide GiraudoDavide Giraudo

126k16150261




126k16150261












  • $begingroup$
    Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
    $endgroup$
    – Michael Maier
    Jan 4 at 10:00








  • 1




    $begingroup$
    I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 10:09










  • $begingroup$
    Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
    $endgroup$
    – Michael Maier
    Jan 4 at 10:16












  • $begingroup$
    And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
    $endgroup$
    – Michael Maier
    Jan 4 at 10:31












  • $begingroup$
    Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 11:31


















  • $begingroup$
    Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
    $endgroup$
    – Michael Maier
    Jan 4 at 10:00








  • 1




    $begingroup$
    I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 10:09










  • $begingroup$
    Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
    $endgroup$
    – Michael Maier
    Jan 4 at 10:16












  • $begingroup$
    And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
    $endgroup$
    – Michael Maier
    Jan 4 at 10:31












  • $begingroup$
    Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
    $endgroup$
    – Davide Giraudo
    Jan 4 at 11:31
















$begingroup$
Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
$endgroup$
– Michael Maier
Jan 4 at 10:00






$begingroup$
Thanks! However, we defined Lebesgue integral as $int_xfdmu_n=sup{int_xs_kdmu_n, kinmathbb{N}}$, where $s_k$ denotes a sequence of step funktions as defined by you. What am I missing?
$endgroup$
– Michael Maier
Jan 4 at 10:00






1




1




$begingroup$
I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
$endgroup$
– Davide Giraudo
Jan 4 at 10:09




$begingroup$
I guess that in your definition, we have $s_k(x)uparrow f(x)$ for all $x$. Ten the two definition are equivalent.
$endgroup$
– Davide Giraudo
Jan 4 at 10:09












$begingroup$
Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
$endgroup$
– Michael Maier
Jan 4 at 10:16






$begingroup$
Alright. To show part 2.), I have to show that (using my definition) $sup{int_xs_kdmu_n, kinmathbb{N}} leq sup{int_xs_kdmu_{n+1}, kinmathbb{N}}$. I can now drop the two suprema (right?), and after writing out the integral for a step function $s_k$ i get $sum_{i=1}^inftyalpha_imu_n(A_i)leqsum_{i=1}^inftyalpha_imu_{n+1}(A_i)$, which is true due to my initial conditions.
$endgroup$
– Michael Maier
Jan 4 at 10:16














$begingroup$
And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
$endgroup$
– Michael Maier
Jan 4 at 10:31






$begingroup$
And one more (hopefully) last question? Why does switching the suprema reduce the case to f being a step function? I then have something like $sup_{kinmathbb{N}}left(sup_{ngeq1}{int_Xs_kdmu_n}right)$, and I do not see how that leads me further...
$endgroup$
– Michael Maier
Jan 4 at 10:31














$begingroup$
Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
$endgroup$
– Davide Giraudo
Jan 4 at 11:31




$begingroup$
Regard your first comment, yes. The only thing you have to modify is the index of the sum (the $infty$ should be $N$). For your second comment, see the edit.
$endgroup$
– Davide Giraudo
Jan 4 at 11:31


















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