Additivity of outer measure if one of the sets is open.Suppose A and G are disjoint subsets of R and G is open. Then | A ∪ G | = | A | + | G | .*Note that: | A| and l( I) denote the outer measure of A and the length of interval I respectively.I wonder that, for the part of proof to the case G = ( a , b ) and the deduced inequality: ∑ n = 1 ∞ l ( I n ) = ∑ n = 1 ∞ ( l ( J n ) + l ( L n ) ) + ∑ n = 1 ∞ l ( K n ) ≥ | A | + | G | How can the above inequality show that | A ∪ G | ≥ | A | + | G | ?I can only deduce from ( A ∪ G ) ⊂ ⋃ n = 1 ∞ I n , that: | A ∪ G | ≤ | ⋃ n = 1 ∞ I n | ≤ ∑ n = 1 ∞ l ( I n ) However, along with: ∑ n = 1 ∞ l ( I n ) ≥ | A | + | G | then both the inequalities are on the same side, so I cannot link them to prove what I desire. How did author reach his conclusion?

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Additivity of outer measure if one of the sets is open.Suppose   A and   G are disjoint subsets of       R   and   G is open. Then       |    A  ∪  G      |    =      |    A      |    +      |    G      |  .*Note that: |  A| and l(  I) denote the outer measure of   A and the length of interval   I respectively.I wonder that, for the part of proof to the case   G  =  (  a  ,  b  ) and the deduced inequality:                              ∑                      n            =            1                                ∞                          l        (                  I                      n                          )                            =                  ∑                      n            =            1                                ∞                          (        l        (                  J                      n                          )        +        l        (                  L                      n                          )        )        +                  ∑                      n            =            1                                ∞                          l        (                  K                      n                          )                                          ≥                  |                A                  |                +                  |                G                  |                    How can the above inequality show that       |    A  ∪  G      |    ≥      |    A      |    +      |    G      |  ?I can only deduce from   (  A  ∪  G  )  ⊂      ⋃          n      =      1              ∞            I          n      , that:      |    A  ∪  G      |    ≤      |          ⋃              n        =        1                    ∞                    I              n              |    ≤      ∑          n      =      1              ∞        l  (      I          n        )  However, along with:      ∑          n      =      1              ∞        l  (      I          n        )  ≥      |    A      |    +      |    G      |  then both the inequalities are on the same side, so I cannot link them to prove what I desire. How did author reach his conclusion?

marktje28 · Accepted Answer

Recall the definition of outer measure:      |    A  ∪  G      |    =  inf  {  ∑  l  (      I    n    )  :  A  ∪  G  ⊂  ∪      I    n    }  .Then you already know for all   {      I    n    } covering   A  ∪  G,  ∑  l  (      I    n    )  ≥      |    A      |    +      |    G      |    ,after taking infimum over these intervals you get the result.

Additivity of outer measure if one of the sets is open. Suppose A and G are disjoint subset

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