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







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














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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059500%2fshow-that-lim-n-to-infty-int-xfd-mu-n-int-xfd-mu%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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
















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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059500%2fshow-that-lim-n-to-infty-int-xfd-mu-n-int-xfd-mu%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

Display a custom attribute below product name in the front-end Magento 1.9.3.8

Image
0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
Read more