Lebesgue outer measure regularity












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In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.



Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.






measure-theory lebesgue-measure outer-measure







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$endgroup$

















    0












    $begingroup$


    In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.



    Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.






    measure-theory lebesgue-measure outer-measure







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.



      Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.






      measure-theory lebesgue-measure outer-measure







      share|cite|improve this question











      $endgroup$




      In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.



      Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.






      measure-theory lebesgue-measure outer-measure




      measure-theory lebesgue-measure outer-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 16:22









      amWhy

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      asked Jan 31 '17 at 15:10









      ArindamArindam

      338213




      338213






















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          $begingroup$

          You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
          $$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
          This follows because the left infimum has a greater range.
          By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.



          Therefore, we have
          $$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
            $$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
            This follows because the left infimum has a greater range.
            By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.



            Therefore, we have
            $$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
              $$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
              This follows because the left infimum has a greater range.
              By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.



              Therefore, we have
              $$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
                $$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
                This follows because the left infimum has a greater range.
                By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.



                Therefore, we have
                $$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$






                share|cite|improve this answer











                $endgroup$



                You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
                $$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
                This follows because the left infimum has a greater range.
                By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.



                Therefore, we have
                $$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 16:06









                amWhy

                1




                1










                answered Jan 31 '17 at 15:14









                Henricus V.Henricus V.

                15.1k22048




                15.1k22048






























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