If $f(x,u(x))$ measuable with conditions












1












$begingroup$


recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?



copy the question as follows:




Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
continuous for each $x$. If $u(t)$ is continuous, how can I show that
the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
measurable?




The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.






real-analysis functional-analysis measure-theory measurable-functions







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?



    copy the question as follows:




    Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
    continuous for each $x$. If $u(t)$ is continuous, how can I show that
    the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
    measurable?




    The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.






    real-analysis functional-analysis measure-theory measurable-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?



      copy the question as follows:




      Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
      continuous for each $x$. If $u(t)$ is continuous, how can I show that
      the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
      measurable?




      The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.






      real-analysis functional-analysis measure-theory measurable-functions







      share|cite|improve this question









      $endgroup$




      recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?



      copy the question as follows:




      Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
      continuous for each $x$. If $u(t)$ is continuous, how can I show that
      the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
      measurable?




      The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.






      real-analysis functional-analysis measure-theory measurable-functions




      real-analysis functional-analysis measure-theory measurable-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 13:11









      Larry EppesLarry Eppes

      471311




      471311






















          1 Answer
          1






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          $begingroup$

          To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.



          The proof usually goes like this: Let's define
          $$
          g(x) := f(x,u(x)),
          $$
          where $u$ is a simple function
          $$
          u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
          $$
          where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
          $$
          {xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
          $$
          Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.



          In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
          $$
          lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
          $$
          for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.



          The case where $u$ is continuous follows immediately.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:39










          • $begingroup$
            @LarryEppes Glad I could help :)
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:40










          • $begingroup$
            I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:43










          • $begingroup$
            @LarryEppes Yes, they're the same conditions as the ones in your question.
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:45










          • $begingroup$
            I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 19:23











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          $begingroup$

          To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.



          The proof usually goes like this: Let's define
          $$
          g(x) := f(x,u(x)),
          $$
          where $u$ is a simple function
          $$
          u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
          $$
          where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
          $$
          {xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
          $$
          Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.



          In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
          $$
          lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
          $$
          for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.



          The case where $u$ is continuous follows immediately.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:39










          • $begingroup$
            @LarryEppes Glad I could help :)
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:40










          • $begingroup$
            I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:43










          • $begingroup$
            @LarryEppes Yes, they're the same conditions as the ones in your question.
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:45










          • $begingroup$
            I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 19:23
















          1












          $begingroup$

          To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.



          The proof usually goes like this: Let's define
          $$
          g(x) := f(x,u(x)),
          $$
          where $u$ is a simple function
          $$
          u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
          $$
          where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
          $$
          {xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
          $$
          Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.



          In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
          $$
          lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
          $$
          for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.



          The case where $u$ is continuous follows immediately.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:39










          • $begingroup$
            @LarryEppes Glad I could help :)
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:40










          • $begingroup$
            I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:43










          • $begingroup$
            @LarryEppes Yes, they're the same conditions as the ones in your question.
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:45










          • $begingroup$
            I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 19:23














          1












          1








          1





          $begingroup$

          To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.



          The proof usually goes like this: Let's define
          $$
          g(x) := f(x,u(x)),
          $$
          where $u$ is a simple function
          $$
          u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
          $$
          where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
          $$
          {xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
          $$
          Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.



          In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
          $$
          lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
          $$
          for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.



          The case where $u$ is continuous follows immediately.






          share|cite|improve this answer











          $endgroup$



          To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.



          The proof usually goes like this: Let's define
          $$
          g(x) := f(x,u(x)),
          $$
          where $u$ is a simple function
          $$
          u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
          $$
          where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
          $$
          {xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
          $$
          Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.



          In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
          $$
          lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
          $$
          for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.



          The case where $u$ is continuous follows immediately.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 2:32

























          answered Dec 13 '18 at 16:45









          BigbearZzzBigbearZzz

          8,67621652




          8,67621652












          • $begingroup$
            +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:39










          • $begingroup$
            @LarryEppes Glad I could help :)
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:40










          • $begingroup$
            I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:43










          • $begingroup$
            @LarryEppes Yes, they're the same conditions as the ones in your question.
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:45










          • $begingroup$
            I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 19:23


















          • $begingroup$
            +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:39










          • $begingroup$
            @LarryEppes Glad I could help :)
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:40










          • $begingroup$
            I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 17:43










          • $begingroup$
            @LarryEppes Yes, they're the same conditions as the ones in your question.
            $endgroup$
            – BigbearZzz
            Dec 13 '18 at 17:45










          • $begingroup$
            I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
            $endgroup$
            – Larry Eppes
            Dec 13 '18 at 19:23
















          $begingroup$
          +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
          $endgroup$
          – Larry Eppes
          Dec 13 '18 at 17:39




          $begingroup$
          +1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
          $endgroup$
          – Larry Eppes
          Dec 13 '18 at 17:39












          $begingroup$
          @LarryEppes Glad I could help :)
          $endgroup$
          – BigbearZzz
          Dec 13 '18 at 17:40




          $begingroup$
          @LarryEppes Glad I could help :)
          $endgroup$
          – BigbearZzz
          Dec 13 '18 at 17:40












          $begingroup$
          I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
          $endgroup$
          – Larry Eppes
          Dec 13 '18 at 17:43




          $begingroup$
          I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
          $endgroup$
          – Larry Eppes
          Dec 13 '18 at 17:43












          $begingroup$
          @LarryEppes Yes, they're the same conditions as the ones in your question.
          $endgroup$
          – BigbearZzz
          Dec 13 '18 at 17:45




          $begingroup$
          @LarryEppes Yes, they're the same conditions as the ones in your question.
          $endgroup$
          – BigbearZzz
          Dec 13 '18 at 17:45












          $begingroup$
          I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
          $endgroup$
          – Larry Eppes
          Dec 13 '18 at 19:23




          $begingroup$
          I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
          $endgroup$
          – Larry Eppes
          Dec 13 '18 at 19:23


















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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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