Counterexample for the monotone convergence theorem












5












$begingroup$


Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20
















5












$begingroup$


Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20














5












5








5





$begingroup$


Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$




Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure




real-analysis integration measure-theory lebesgue-integral lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 14:13









YuiTo Cheng

2,1362837




2,1362837










asked Mar 16 at 12:55









Marine GalantinMarine Galantin

940419




940419












  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20


















  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20
















$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
Mar 16 at 13:01




$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
Mar 16 at 13:01












$begingroup$
I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
$endgroup$
– Henry
Mar 16 at 13:18




$begingroup$
I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
$endgroup$
– Henry
Mar 16 at 13:18












$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
Mar 16 at 13:20




$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
Mar 16 at 13:20










1 Answer
1






active

oldest

votes


















4












$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18











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%2f3150362%2fcounterexample-for-the-monotone-convergence-theorem%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









4












$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18
















4












$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18














4












4








4





$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$



Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 16 at 13:05

























answered Mar 16 at 13:00









PierrePierre

19511




19511












  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18


















  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18
















$begingroup$
there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
$endgroup$
– Marine Galantin
Mar 16 at 13:05




$begingroup$
there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
$endgroup$
– Marine Galantin
Mar 16 at 13:05




1




1




$begingroup$
The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
Mar 16 at 13:09






$begingroup$
The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
Mar 16 at 13:09














$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
Mar 16 at 13:18




$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
Mar 16 at 13:18


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150362%2fcounterexample-for-the-monotone-convergence-theorem%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

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more