A measure μ : B R → [ 0 , ∞ ] is locally finite if μ ( K ) &lt; ∞ for all K compact. Show that any locally finite measure is σ-finite. Give an example of a σ-finite measure on R , which is not locally finite.Attempt: Since R = ⋃ n ∈ Z [ n , n + 1 ] and each [ n , n + 1 ] is compact, we have μ ( [ n , n + 1 ] ) &lt; ∞ for each n and μ is σ-finite. Is this part of the problem correct?For the second part, I have no idea. I wanted to take counting measure on a collection of subsets of R , but I know this measure is locally finite on the integers and not so with the usual topology. But I know this measure is not sigma finite on R . Any suggestions?

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A measure       μ    :      B                  R              →  [  0  ,      ∞    ] is locally finite if       μ    (  K  )  &amp;lt;      ∞   for all K compact. Show that any locally finite measure is   σ-finite. Give an example of a   σ-finite measure on       R  , which is not locally finite.Attempt: Since       R    =      ⋃          n      ∈              Z              [  n  ,  n  +  1  ] and each   [  n  ,  n  +  1  ] is compact, we have   μ  (  [  n  ,  n  +  1  ]  )  &amp;lt;  ∞ for each   n and   μ is   σ-finite. Is this part of the problem correct?For the second part, I have no idea. I wanted to take counting measure on a collection of subsets of       R  , but I know this measure is locally finite on the integers and not so with the usual topology. But I know this measure is not sigma finite on       R  . Any suggestions?

Layla Love · Accepted Answer

Your solution to the first part is correct. For the second part, e.g.   μ  (  X  )  =  m  (  tan  ⁡  (  X  ∩  (  −  π      /    2  ,  π      /    2  )  )  ) where   m is the Lebesgue measure.The intuition is clear: We ignore any part of   X outside   (  −  π      /    2  ,  π      /    2  ), and since tan is a bijection from   (  −  π      /    2  ,  π      /    2  ) to       R  , the Lebesgue measure on       R   induces a natural measure on   (  −  π      /    2  ,  π      /    2  ). To formally prove this won&#039;t be hard. In particular,   μ  (  [  π      /    2  ,  π      /    2  ]  )  =  ∞, so   μ is not locally finite.If this is not intuitive enough. We can do the following modified version of the counting measure, let   A  =  {  1  ,  1      /    2  ,  1      /    3  ,  1      /    4  ,  ⋯  }, and define   μ  (  X  )  =      |    A  ∩  X      |  , then   μ is   σ-finite but not locally finite as   μ  (  [  0  ,  1  ]  )  =  ∞.

Davon Irwin · Accepted Answer

No, your answer to 1. is not completely correct. For   σ-additivity, you are in need of disjoint sets, i. e. consider   (  n  ,  n  +  1  ] (with parenthesis). These sets are not compact themselves, but are subsets of compact sets   [  n  ,  n  +  1  ]. Now, since   μ      (  (  n  ,  n  +  1  ]  )  ≤  μ      (  [  n  ,  n  +  1  ]  ), you get the result by an inequality. EDIT: I am wrong.

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