Let, m be the lebesgue outer measure. Let, E ⊂ R . m ( E ) ≤ m ( E ∩ ( − ∞ , c ) ) + m ( E ∩ [ c , ∞ ) ) ,is true from definition of outer measure, but I'm having trouble showing the reverse inequality.Where, m ( E ) = inf { ∑ j = 1 ∞ | B j | : E ⊆ ⋃ j = 1 ∞ B j , B j is a open interval }Any help?Thanks.

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Let,   m be the lebesgue outer measure. Let,   E  ⊂      R  .  m  (  E  )  ≤  m  (  E  ∩  (  −  ∞  ,  c  )  )  +  m  (  E  ∩  [  c  ,  ∞  )  )  ,is true from definition of outer measure, but I&#039;m having trouble showing the reverse inequality.Where,  m  (  E  )  =  inf  {      ∑          j      =      1              ∞            |        B    j        |    :  E  ⊆      ⋃          j      =      1              ∞            B    j    ,      B    j         is a open interval    }Any help?Thanks.

odjeljkur58ws · Accepted Answer

It is enough to show,                    m        (        E        )                            ≥        m        (        E        ∩        (        −        ∞        ,        c        )        )        +        m        (        E        ∩        [        c        ,        ∞        )        )            Let,       E    1    =  E  ∩  (  −  ∞  ,  c  ) and       E    2    =  E  ∩  [  c  ,  ∞  )If   m  (  E  )  =  ∞, then nothing to prove.Assume,   m  (  E  )  &amp;lt;  ∞Then given any   ϵ  &amp;gt;  0 , there exists a countable collection of open intervals   (      I    n    ) which covers   E such that      ∑          n      =      1              ∞            ℓ    (          I      n        )    &amp;lt;    m    (    E    )    +    ϵ  Let,       I    n    ′    =      I    n    ∩  (  −  ∞  ,  c  ) and       I    n    ″    =      I    n    ∩  [  c  ,  ∞  )Then,       I    n    ′    ∪      I    n    ″    =      I    n   and       I    n    ′    ∩      I    n    ″    =  ∅Hence,   ℓ  (      I    n    )  =  ℓ  (      I    n    ′    )  +  ℓ  (      I    n    ″    )Now,                              E          1                =        E        ∩        (        −        ∞        ,        c        )                            ⊆                  ∪                      n            =            1                                ∞                                    I          n                ∩        (        −        ∞        ,        c        )                                          =                  ∪                      n            =            1                                ∞                                    (                      I            n                    ∩          (          −          ∞          ,          c          )          )                                                  =                  ∪                      n            =            1                                ∞                                    I          n          ′                    Similarly,       E    2    ⊆      ∪          n      =      1              ∞            I    n    ″  Hence,   m  (      E    1    )  =  m  (      ∪          n      =      1              ∞            I    n    ′    )  ≤      ∑          n      =      1              ∞        m  (      I    n    ′    )Similarly,   m  (      E    2    )  =  m  (      ∪          n      =      1              ∞            I    n    ″    )  ≤      ∑          n      =      1              ∞        m  (      I    n    ″    )Hence,                    m        (                  E          1                )        +        m        (                  E          2                )                            ≤                  ∑                      n            =            1                                ∞                          {        m        (                  I          n          ″                )        +        m        (                  I          n          ″                )        }                            ​                            =                  ∑                      n            =            1                                ∞                          ℓ        (                  I          n          ′                )        +        ℓ        (                  I          n                )                                          =                  ∑                      n            =            1                                ∞                          ℓ        (                  I          n                )                                m        (                  E          1                +                  E          2                )                            ≤                  ∑                      n            =            1                                ∞                          ℓ        (                  I          n                )                                          &amp;lt;        m        (        E        )        +        ϵ            Since,   ϵ  &amp;gt;  0 arbitry, it follows that  m  (      E    1    +      E    2    )  ≤  m  (  E  )Hence, the proof is complete.

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