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
    2 days ago
















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
    2 days ago














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 2 days ago









Davide Giraudo

125k16150260




125k16150260










asked 2 days ago









Michael Maier

488




488








  • 2




    First prove the case where $f$ is a step function. Can you do that?
    – Severin Schraven
    2 days ago














  • 2




    First prove the case where $f$ is a step function. Can you do that?
    – Severin Schraven
    2 days ago








2




2




First prove the case where $f$ is a step function. Can you do that?
– Severin Schraven
2 days ago




First prove the case where $f$ is a step function. Can you do that?
– Severin Schraven
2 days ago










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
    16 hours ago








  • 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
    16 hours ago










  • 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
    16 hours ago












  • 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
    16 hours ago












  • 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
    15 hours ago











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1 Answer
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1 Answer
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active

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active

oldest

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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
    16 hours ago








  • 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
    16 hours ago










  • 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
    16 hours ago












  • 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
    16 hours ago












  • 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
    15 hours ago
















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
    16 hours ago








  • 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
    16 hours ago










  • 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
    16 hours ago












  • 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
    16 hours ago












  • 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
    15 hours ago














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 15 hours ago

























answered 2 days ago









Davide Giraudo

125k16150260




125k16150260












  • 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
    16 hours ago








  • 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
    16 hours ago










  • 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
    16 hours ago












  • 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
    16 hours ago












  • 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
    15 hours ago


















  • 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
    16 hours ago








  • 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
    16 hours ago










  • 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
    16 hours ago












  • 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
    16 hours ago












  • 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
    15 hours ago
















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
16 hours ago






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
16 hours ago






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
16 hours ago




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
16 hours ago












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
16 hours ago






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
16 hours ago














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
16 hours ago






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
16 hours ago














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
15 hours ago




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
15 hours ago


















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0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
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Understanding the size os this class of aleatory events

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0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
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