Let $f: mathbb{R} rightarrow mathbb{R}$ be a measurable function, is $f circ sin$ measurable?












4












$begingroup$


If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44
















4












$begingroup$


If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44














4












4








4





$begingroup$


If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure




measure-theory lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 4:42







Non

















asked Dec 9 '18 at 2:00









NonNon

538




538








  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44














  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44








3




3




$begingroup$
the composition of measurable functions is measurable.
$endgroup$
– Tsemo Aristide
Dec 9 '18 at 2:01




$begingroup$
the composition of measurable functions is measurable.
$endgroup$
– Tsemo Aristide
Dec 9 '18 at 2:01




2




2




$begingroup$
@TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
$endgroup$
– Will M.
Dec 9 '18 at 2:05






$begingroup$
@TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
$endgroup$
– Will M.
Dec 9 '18 at 2:05






2




2




$begingroup$
Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
$endgroup$
– Shaun
Dec 9 '18 at 2:20






$begingroup$
Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
$endgroup$
– Shaun
Dec 9 '18 at 2:20






5




5




$begingroup$
@Tsemo actually that’s not true in general
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:42




$begingroup$
@Tsemo actually that’s not true in general
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:42




4




4




$begingroup$
@Tsemo see this post for instance
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:44




$begingroup$
@Tsemo see this post for instance
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:44










2 Answers
2






active

oldest

votes


















3












$begingroup$

Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean the pre-image of a lebesque by a absolutely continuous?
    $endgroup$
    – Non
    Dec 17 '18 at 2:13



















0












$begingroup$

Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






share|cite|improve this answer









$endgroup$













    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%2f3031916%2flet-f-mathbbr-rightarrow-mathbbr-be-a-measurable-function-is-f-circ%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13
















    3












    $begingroup$

    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13














    3












    3








    3





    $begingroup$

    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






    share|cite|improve this answer











    $endgroup$



    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 13:23









    jgon

    15.6k32143




    15.6k32143










    answered Dec 9 '18 at 4:07









    zoidbergzoidberg

    1,080113




    1,080113












    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13


















    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13
















    $begingroup$
    You mean the pre-image of a lebesque by a absolutely continuous?
    $endgroup$
    – Non
    Dec 17 '18 at 2:13




    $begingroup$
    You mean the pre-image of a lebesque by a absolutely continuous?
    $endgroup$
    – Non
    Dec 17 '18 at 2:13











    0












    $begingroup$

    Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






        share|cite|improve this answer









        $endgroup$



        Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 4:43









        NonNon

        538




        538






























            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%2f3031916%2flet-f-mathbbr-rightarrow-mathbbr-be-a-measurable-function-is-f-circ%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

            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

            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