Lebesgue-measurable almost everywhere equals Borel measurable function [closed]












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Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.



I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.



I would very much appreciate your help.



Best, KingDingeling










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closed as off-topic by Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg Jan 7 at 23:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg

If this question can be reworded to fit the rules in the help center, please edit the question.





















    1












    $begingroup$


    Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.



    I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.



    I would very much appreciate your help.



    Best, KingDingeling










    share|cite|improve this question









    $endgroup$



    closed as off-topic by Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg Jan 7 at 23:19


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      1












      1








      1





      $begingroup$


      Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.



      I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.



      I would very much appreciate your help.



      Best, KingDingeling










      share|cite|improve this question









      $endgroup$




      Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.



      I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.



      I would very much appreciate your help.



      Best, KingDingeling







      measure-theory lebesgue-measure measurable-functions borel-measures






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 12:05









      KingDingelingKingDingeling

      1857




      1857




      closed as off-topic by Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg Jan 7 at 23:19


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg Jan 7 at 23:19


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, Namaste, José Carlos Santos, egreg

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






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          2












          $begingroup$

          I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am sorry, but I don't know how to do the algebraic induction here.
            $endgroup$
            – KingDingeling
            Jan 7 at 12:41



















          1












          $begingroup$

          Sketch (just an elaboration of the other answer):





          • $f$ is the pointwise limit of simple functions $f_n.$

          • Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.

          • Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$

          • Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show


          • $f$ is the pointwise limit of $g_n$.






          share|cite|improve this answer









          $endgroup$




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I am sorry, but I don't know how to do the algebraic induction here.
              $endgroup$
              – KingDingeling
              Jan 7 at 12:41
















            2












            $begingroup$

            I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I am sorry, but I don't know how to do the algebraic induction here.
              $endgroup$
              – KingDingeling
              Jan 7 at 12:41














            2












            2








            2





            $begingroup$

            I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.






            share|cite|improve this answer









            $endgroup$



            I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 12:12









            Kavi Rama MurthyKavi Rama Murthy

            74.6k53270




            74.6k53270












            • $begingroup$
              I am sorry, but I don't know how to do the algebraic induction here.
              $endgroup$
              – KingDingeling
              Jan 7 at 12:41


















            • $begingroup$
              I am sorry, but I don't know how to do the algebraic induction here.
              $endgroup$
              – KingDingeling
              Jan 7 at 12:41
















            $begingroup$
            I am sorry, but I don't know how to do the algebraic induction here.
            $endgroup$
            – KingDingeling
            Jan 7 at 12:41




            $begingroup$
            I am sorry, but I don't know how to do the algebraic induction here.
            $endgroup$
            – KingDingeling
            Jan 7 at 12:41











            1












            $begingroup$

            Sketch (just an elaboration of the other answer):





            • $f$ is the pointwise limit of simple functions $f_n.$

            • Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.

            • Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$

            • Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show


            • $f$ is the pointwise limit of $g_n$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Sketch (just an elaboration of the other answer):





              • $f$ is the pointwise limit of simple functions $f_n.$

              • Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.

              • Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$

              • Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show


              • $f$ is the pointwise limit of $g_n$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Sketch (just an elaboration of the other answer):





                • $f$ is the pointwise limit of simple functions $f_n.$

                • Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.

                • Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$

                • Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show


                • $f$ is the pointwise limit of $g_n$.






                share|cite|improve this answer









                $endgroup$



                Sketch (just an elaboration of the other answer):





                • $f$ is the pointwise limit of simple functions $f_n.$

                • Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.

                • Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$

                • Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show


                • $f$ is the pointwise limit of $g_n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 16:52









                MatematletaMatematleta

                12.2k21020




                12.2k21020















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