nondecreasing rearrangement is equimeasurable












4












$begingroup$


Two functions $f(x)$ and $g(x)$ are called equi-measurable if $m({x:f(x)>t})=m({x:g(x)>t})$.
Nondecreasing rearrangement of a function $f(x)$ is defined as $$f^*(tau)=inf{t>0:m({x:f(x)>t}leqtau}.$$
Prove that $f^*(tau)$ and $f(x)$ are equimeasurable.






real-analysis measure-theory decreasing-rearrangements







share|cite|improve this question











$endgroup$








  • 10




    $begingroup$
    What are your thoughts on the matter? Do you see a point where to start?
    $endgroup$
    – t.b.
    Dec 30 '11 at 21:40










  • $begingroup$
    This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12)
    $endgroup$
    – Carucel
    Feb 25 '16 at 10:51
















4












$begingroup$


Two functions $f(x)$ and $g(x)$ are called equi-measurable if $m({x:f(x)>t})=m({x:g(x)>t})$.
Nondecreasing rearrangement of a function $f(x)$ is defined as $$f^*(tau)=inf{t>0:m({x:f(x)>t}leqtau}.$$
Prove that $f^*(tau)$ and $f(x)$ are equimeasurable.






real-analysis measure-theory decreasing-rearrangements







share|cite|improve this question











$endgroup$








  • 10




    $begingroup$
    What are your thoughts on the matter? Do you see a point where to start?
    $endgroup$
    – t.b.
    Dec 30 '11 at 21:40










  • $begingroup$
    This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12)
    $endgroup$
    – Carucel
    Feb 25 '16 at 10:51














4












4








4


4



$begingroup$


Two functions $f(x)$ and $g(x)$ are called equi-measurable if $m({x:f(x)>t})=m({x:g(x)>t})$.
Nondecreasing rearrangement of a function $f(x)$ is defined as $$f^*(tau)=inf{t>0:m({x:f(x)>t}leqtau}.$$
Prove that $f^*(tau)$ and $f(x)$ are equimeasurable.






real-analysis measure-theory decreasing-rearrangements







share|cite|improve this question











$endgroup$




Two functions $f(x)$ and $g(x)$ are called equi-measurable if $m({x:f(x)>t})=m({x:g(x)>t})$.
Nondecreasing rearrangement of a function $f(x)$ is defined as $$f^*(tau)=inf{t>0:m({x:f(x)>t}leqtau}.$$
Prove that $f^*(tau)$ and $f(x)$ are equimeasurable.






real-analysis measure-theory decreasing-rearrangements




real-analysis measure-theory decreasing-rearrangements






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '17 at 13:18









Guy Fsone

17.2k43074




17.2k43074










asked Dec 30 '11 at 21:15









TimTim

562




562








  • 10




    $begingroup$
    What are your thoughts on the matter? Do you see a point where to start?
    $endgroup$
    – t.b.
    Dec 30 '11 at 21:40










  • $begingroup$
    This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12)
    $endgroup$
    – Carucel
    Feb 25 '16 at 10:51














  • 10




    $begingroup$
    What are your thoughts on the matter? Do you see a point where to start?
    $endgroup$
    – t.b.
    Dec 30 '11 at 21:40










  • $begingroup$
    This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12)
    $endgroup$
    – Carucel
    Feb 25 '16 at 10:51








10




10




$begingroup$
What are your thoughts on the matter? Do you see a point where to start?
$endgroup$
– t.b.
Dec 30 '11 at 21:40




$begingroup$
What are your thoughts on the matter? Do you see a point where to start?
$endgroup$
– t.b.
Dec 30 '11 at 21:40












$begingroup$
This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12)
$endgroup$
– Carucel
Feb 25 '16 at 10:51




$begingroup$
This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12)
$endgroup$
– Carucel
Feb 25 '16 at 10:51










1 Answer
1






active

oldest

votes


















0












$begingroup$


You want to prove that ${f>t}^* = {f^*>t},,text{?}$




Fix $t>0$ et $yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$. One can check that for every, $0<s< t$ one has $$left{x in mathbb{R}^n: |f(x)| > t right}subset left{x in mathbb{R}^n: |f(x)| > s right}$$ this entails that,
begin{equation}label{eq-inclu t-s}tag{I}
left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{x in mathbb{R}^n: |f(x)| > s right}^{*}~~~textrm{for all $sin ]0,t[$}.
end{equation}
this implies that,$$ mathbf{1}_{left{ | f| > s right}^*}(y) =1 ~~~sin (0,t)$$



Therefore, from definition of $f^{*}$, if $yin {|f|>t}^*$ then we have
$$begin{align*}
f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
&= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
& = int_{0}^{t} ds+int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds \ &>t.
end{align*}$$



Whence, $$left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{ x in mathbb{R}^n:f^{*}(x)> t right}.$$
On the other hand, if we suppose, $ynotin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$ then for all $s>0$ such that $ yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}$ one has $0<sleq t$.



Indeed, $t>s $ then from eqref{eq-inclu t-s} $$yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,



$$supleft{s>0 : yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}right}leq t. $$
We then deduce that,
$$begin{align*}
f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
&= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ underbrace{int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds}_{=0}leq=t
end{align*}$$
that is $f^*(y)leq t$ or that $ynotin left{x in mathbb{R}^n: f^*(x) > t right}$. We've just prove that,



begin{equation}label{eq}tag{II}
Bbb R^nsetminus left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset Bbb R^nsetminusleft{x in mathbb{R}^n: f^*(x) > s right}~~~textrm{for all $sin ]0,t[$}.
end{equation}



Which end the prove by taking the complementary.






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%2f95277%2fnondecreasing-rearrangement-is-equimeasurable%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









    0












    $begingroup$


    You want to prove that ${f>t}^* = {f^*>t},,text{?}$




    Fix $t>0$ et $yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$. One can check that for every, $0<s< t$ one has $$left{x in mathbb{R}^n: |f(x)| > t right}subset left{x in mathbb{R}^n: |f(x)| > s right}$$ this entails that,
    begin{equation}label{eq-inclu t-s}tag{I}
    left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{x in mathbb{R}^n: |f(x)| > s right}^{*}~~~textrm{for all $sin ]0,t[$}.
    end{equation}
    this implies that,$$ mathbf{1}_{left{ | f| > s right}^*}(y) =1 ~~~sin (0,t)$$



    Therefore, from definition of $f^{*}$, if $yin {|f|>t}^*$ then we have
    $$begin{align*}
    f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
    &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
    & = int_{0}^{t} ds+int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds \ &>t.
    end{align*}$$



    Whence, $$left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{ x in mathbb{R}^n:f^{*}(x)> t right}.$$
    On the other hand, if we suppose, $ynotin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$ then for all $s>0$ such that $ yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}$ one has $0<sleq t$.



    Indeed, $t>s $ then from eqref{eq-inclu t-s} $$yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,



    $$supleft{s>0 : yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}right}leq t. $$
    We then deduce that,
    $$begin{align*}
    f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
    &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ underbrace{int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds}_{=0}leq=t
    end{align*}$$
    that is $f^*(y)leq t$ or that $ynotin left{x in mathbb{R}^n: f^*(x) > t right}$. We've just prove that,



    begin{equation}label{eq}tag{II}
    Bbb R^nsetminus left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset Bbb R^nsetminusleft{x in mathbb{R}^n: f^*(x) > s right}~~~textrm{for all $sin ]0,t[$}.
    end{equation}



    Which end the prove by taking the complementary.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$


      You want to prove that ${f>t}^* = {f^*>t},,text{?}$




      Fix $t>0$ et $yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$. One can check that for every, $0<s< t$ one has $$left{x in mathbb{R}^n: |f(x)| > t right}subset left{x in mathbb{R}^n: |f(x)| > s right}$$ this entails that,
      begin{equation}label{eq-inclu t-s}tag{I}
      left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{x in mathbb{R}^n: |f(x)| > s right}^{*}~~~textrm{for all $sin ]0,t[$}.
      end{equation}
      this implies that,$$ mathbf{1}_{left{ | f| > s right}^*}(y) =1 ~~~sin (0,t)$$



      Therefore, from definition of $f^{*}$, if $yin {|f|>t}^*$ then we have
      $$begin{align*}
      f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
      &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
      & = int_{0}^{t} ds+int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds \ &>t.
      end{align*}$$



      Whence, $$left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{ x in mathbb{R}^n:f^{*}(x)> t right}.$$
      On the other hand, if we suppose, $ynotin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$ then for all $s>0$ such that $ yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}$ one has $0<sleq t$.



      Indeed, $t>s $ then from eqref{eq-inclu t-s} $$yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,



      $$supleft{s>0 : yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}right}leq t. $$
      We then deduce that,
      $$begin{align*}
      f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
      &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ underbrace{int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds}_{=0}leq=t
      end{align*}$$
      that is $f^*(y)leq t$ or that $ynotin left{x in mathbb{R}^n: f^*(x) > t right}$. We've just prove that,



      begin{equation}label{eq}tag{II}
      Bbb R^nsetminus left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset Bbb R^nsetminusleft{x in mathbb{R}^n: f^*(x) > s right}~~~textrm{for all $sin ]0,t[$}.
      end{equation}



      Which end the prove by taking the complementary.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$


        You want to prove that ${f>t}^* = {f^*>t},,text{?}$




        Fix $t>0$ et $yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$. One can check that for every, $0<s< t$ one has $$left{x in mathbb{R}^n: |f(x)| > t right}subset left{x in mathbb{R}^n: |f(x)| > s right}$$ this entails that,
        begin{equation}label{eq-inclu t-s}tag{I}
        left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{x in mathbb{R}^n: |f(x)| > s right}^{*}~~~textrm{for all $sin ]0,t[$}.
        end{equation}
        this implies that,$$ mathbf{1}_{left{ | f| > s right}^*}(y) =1 ~~~sin (0,t)$$



        Therefore, from definition of $f^{*}$, if $yin {|f|>t}^*$ then we have
        $$begin{align*}
        f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
        &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
        & = int_{0}^{t} ds+int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds \ &>t.
        end{align*}$$



        Whence, $$left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{ x in mathbb{R}^n:f^{*}(x)> t right}.$$
        On the other hand, if we suppose, $ynotin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$ then for all $s>0$ such that $ yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}$ one has $0<sleq t$.



        Indeed, $t>s $ then from eqref{eq-inclu t-s} $$yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,



        $$supleft{s>0 : yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}right}leq t. $$
        We then deduce that,
        $$begin{align*}
        f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
        &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ underbrace{int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds}_{=0}leq=t
        end{align*}$$
        that is $f^*(y)leq t$ or that $ynotin left{x in mathbb{R}^n: f^*(x) > t right}$. We've just prove that,



        begin{equation}label{eq}tag{II}
        Bbb R^nsetminus left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset Bbb R^nsetminusleft{x in mathbb{R}^n: f^*(x) > s right}~~~textrm{for all $sin ]0,t[$}.
        end{equation}



        Which end the prove by taking the complementary.






        share|cite|improve this answer









        $endgroup$




        You want to prove that ${f>t}^* = {f^*>t},,text{?}$




        Fix $t>0$ et $yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$. One can check that for every, $0<s< t$ one has $$left{x in mathbb{R}^n: |f(x)| > t right}subset left{x in mathbb{R}^n: |f(x)| > s right}$$ this entails that,
        begin{equation}label{eq-inclu t-s}tag{I}
        left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{x in mathbb{R}^n: |f(x)| > s right}^{*}~~~textrm{for all $sin ]0,t[$}.
        end{equation}
        this implies that,$$ mathbf{1}_{left{ | f| > s right}^*}(y) =1 ~~~sin (0,t)$$



        Therefore, from definition of $f^{*}$, if $yin {|f|>t}^*$ then we have
        $$begin{align*}
        f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
        &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
        & = int_{0}^{t} ds+int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds \ &>t.
        end{align*}$$



        Whence, $$left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset left{ x in mathbb{R}^n:f^{*}(x)> t right}.$$
        On the other hand, if we suppose, $ynotin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$ then for all $s>0$ such that $ yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}$ one has $0<sleq t$.



        Indeed, $t>s $ then from eqref{eq-inclu t-s} $$yin left{x in mathbb{R}^n: |f(x)| > t right}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,



        $$supleft{s>0 : yin left{x in mathbb{R}^n: |f(x)| > s right}^{*}right}leq t. $$
        We then deduce that,
        $$begin{align*}
        f^{*}(y) &:= int_{0}^{+ infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds\
        &= int_{0}^{t} mathbf{1}_{left{ | f| > s right}^*}(y) ds+ underbrace{int_{t}^{+infty} mathbf{1}_{left{ | f| > s right}^*}(y) ds}_{=0}leq=t
        end{align*}$$
        that is $f^*(y)leq t$ or that $ynotin left{x in mathbb{R}^n: f^*(x) > t right}$. We've just prove that,



        begin{equation}label{eq}tag{II}
        Bbb R^nsetminus left{x in mathbb{R}^n: |f(x)| > t right}^{*}subset Bbb R^nsetminusleft{x in mathbb{R}^n: f^*(x) > s right}~~~textrm{for all $sin ]0,t[$}.
        end{equation}



        Which end the prove by taking the complementary.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '17 at 20:55









        Guy FsoneGuy Fsone

        17.2k43074




        17.2k43074






























            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%2f95277%2fnondecreasing-rearrangement-is-equimeasurable%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

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            How to fix TextFormField cause rebuild widget in Flutter

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more