Every locally finite measure space is regular











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Prove that every locally finite measure space is regular.




I am not sure how to go about this problem. Certainly, this can be shown to be true for any ball of finite measure that is contained in a larger ball by local finiteness and continuity of finite measure from above and below, but how would I go about proving this for an arbitrary set $A$?






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    Prove that every locally finite measure space is regular.




    I am not sure how to go about this problem. Certainly, this can be shown to be true for any ball of finite measure that is contained in a larger ball by local finiteness and continuity of finite measure from above and below, but how would I go about proving this for an arbitrary set $A$?






    measure-theory







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Prove that every locally finite measure space is regular.




      I am not sure how to go about this problem. Certainly, this can be shown to be true for any ball of finite measure that is contained in a larger ball by local finiteness and continuity of finite measure from above and below, but how would I go about proving this for an arbitrary set $A$?






      measure-theory







      share|cite|improve this question
















      Prove that every locally finite measure space is regular.




      I am not sure how to go about this problem. Certainly, this can be shown to be true for any ball of finite measure that is contained in a larger ball by local finiteness and continuity of finite measure from above and below, but how would I go about proving this for an arbitrary set $A$?






      measure-theory




      measure-theory






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      edited Nov 26 at 11:40









      John Hughes

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      asked Nov 26 at 10:31









      Cute Brownie

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          I don't know where you got this problem from, but it seems to be false, see for example the third example from this section:



          https://en.wikipedia.org/wiki/Regular_measure#Inner_regular_measures_that_are_not_outer_regular



          You have a locally finite measure there that is inner-regular but not outer-regular(and hence not regular).






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            1 Answer
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            I don't know where you got this problem from, but it seems to be false, see for example the third example from this section:



            https://en.wikipedia.org/wiki/Regular_measure#Inner_regular_measures_that_are_not_outer_regular



            You have a locally finite measure there that is inner-regular but not outer-regular(and hence not regular).






            share|cite|improve this answer

























              up vote
              0
              down vote













              I don't know where you got this problem from, but it seems to be false, see for example the third example from this section:



              https://en.wikipedia.org/wiki/Regular_measure#Inner_regular_measures_that_are_not_outer_regular



              You have a locally finite measure there that is inner-regular but not outer-regular(and hence not regular).






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                I don't know where you got this problem from, but it seems to be false, see for example the third example from this section:



                https://en.wikipedia.org/wiki/Regular_measure#Inner_regular_measures_that_are_not_outer_regular



                You have a locally finite measure there that is inner-regular but not outer-regular(and hence not regular).






                share|cite|improve this answer












                I don't know where you got this problem from, but it seems to be false, see for example the third example from this section:



                https://en.wikipedia.org/wiki/Regular_measure#Inner_regular_measures_that_are_not_outer_regular



                You have a locally finite measure there that is inner-regular but not outer-regular(and hence not regular).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 14:58









                Sorin Tirc

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