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












1














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







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  • 2




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
















1














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




















  • 2




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














1












1








1







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















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

125k16150261




125k16150261










asked Jan 2 at 13:49









Michael MaierMichael Maier

638




638








  • 2




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














  • 2




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








2




2




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




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










1 Answer
1






active

oldest

votes


















1














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























  • 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?
    – Michael Maier
    Jan 4 at 10:00








  • 1




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










  • 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.
    – Michael Maier
    Jan 4 at 10:16












  • 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...
    – Michael Maier
    Jan 4 at 10:31












  • 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.
    – Davide Giraudo
    Jan 4 at 11:31











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1 Answer
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1 Answer
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active

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1














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























  • 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?
    – Michael Maier
    Jan 4 at 10:00








  • 1




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










  • 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.
    – Michael Maier
    Jan 4 at 10:16












  • 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...
    – Michael Maier
    Jan 4 at 10:31












  • 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.
    – Davide Giraudo
    Jan 4 at 11:31
















1














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























  • 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?
    – Michael Maier
    Jan 4 at 10:00








  • 1




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










  • 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.
    – Michael Maier
    Jan 4 at 10:16












  • 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...
    – Michael Maier
    Jan 4 at 10:31












  • 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.
    – Davide Giraudo
    Jan 4 at 11:31














1












1








1






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














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

125k16150261




125k16150261












  • 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?
    – Michael Maier
    Jan 4 at 10:00








  • 1




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










  • 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.
    – Michael Maier
    Jan 4 at 10:16












  • 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...
    – Michael Maier
    Jan 4 at 10:31












  • 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.
    – Davide Giraudo
    Jan 4 at 11:31


















  • 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?
    – Michael Maier
    Jan 4 at 10:00








  • 1




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










  • 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.
    – Michael Maier
    Jan 4 at 10:16












  • 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...
    – Michael Maier
    Jan 4 at 10:31












  • 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.
    – Davide Giraudo
    Jan 4 at 11:31
















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?
– Michael Maier
Jan 4 at 10:00






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?
– Michael Maier
Jan 4 at 10:00






1




1




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




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












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.
– Michael Maier
Jan 4 at 10:16






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.
– Michael Maier
Jan 4 at 10:16














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...
– Michael Maier
Jan 4 at 10:31






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...
– Michael Maier
Jan 4 at 10:31














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.
– Davide Giraudo
Jan 4 at 11:31




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.
– Davide Giraudo
Jan 4 at 11:31


















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Management

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Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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Investment

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Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
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