Let ( X , A , ν ) be a measure space. Suppose that the measure of X is bounded. Show that for disjoint A 1 , A 2 , … lim n → ∞ ν ( A n ) = 0.Since X is bounded from countable additivity we have that ν ( ⋃ n = 1 ∞ A n ) = ∑ n = 1 ∞ ν ( A n ) ≤ ν ( X ) &lt; ∞ .Thus from Borel-Cantelli ∑ n = 1 ∞ ν ( A n ) &lt; ∞ ⟹ ν ( lim sup n → ∞ A n ) = 0.Now I also have that ν ( lim sup n → ∞ A n ) ≥ lim sup n → ∞ ν ( A n )which implies that lim sup n → ∞ ν ( A n ) = 0.How can I deduce from here that lim n → ∞ ν ( A n ) = 0? It seems I would need to use the fact that 0 = lim sup n → ∞ ν ( A n ) ≥ lim inf n → ∞ ν ( A n ) ≥ 0so lim inf n → ∞ ν ( A n ) = 0 also and this would imply that lim n → ∞ ν ( A n ) = 0 because of?

Question

Let   (  X  ,      A    ,  ν  ) be a measure space. Suppose that the measure of   X is bounded. Show that for disjoint       A    1    ,      A    2    ,  …       lim          n      →      ∞        ν  (      A    n    )  =  0.Since   X is bounded from countable additivity we have that  ν      (          ⋃              n        =        1            ∞              A      n        )    =      ∑          n      =      1        ∞    ν  (      A    n    )  ≤  ν  (  X  )  &amp;lt;  ∞  .Thus from Borel-Cantelli      ∑          n      =      1        ∞    ν  (      A    n    )  &amp;lt;  ∞    ⟹    ν  (      lim sup          n      →      ∞            A    n    )  =  0.Now I also have that  ν  (      lim sup          n      →      ∞            A    n    )  ≥      lim sup          n      →      ∞        ν  (      A    n    )which implies that      lim sup          n      →      ∞        ν  (      A    n    )  =  0.How can I deduce from here that       lim          n      →      ∞        ν  (      A    n    )  =  0? It seems I would need to use the fact that  0  =      lim sup          n      →      ∞        ν  (      A    n    )  ≥      lim inf          n      →      ∞        ν  (      A    n    )  ≥  0so       lim inf          n      →      ∞        ν  (      A    n    )  =  0 also and this would imply that       lim          n      →      ∞        ν  (      A    n    )  =  0 because of?

Ryan Newman · Accepted Answer

You are overcomplicating things. Just because you need to prove a statement about measure theory doesn&#039;t mean you have to only throw measure theory at it. Forget Borel-Cantelli and lim sup, just focus on what is in front of you.Since   X is bounded from countable additivity we have that  ν  (      ⋃          n      =      1        ∞        A    n    )  =      ∑          n      =      1        ∞    ν  (      A    n    )  ≤  ν  (  X  )  &amp;lt;  ∞  .Yes, and that is all you need from measure theory.Basically, what you now proved is that for       a    n    =  ν  (      A    n    ), you know that the series      ∑          n      =      1        ∞        a    n  converges.What the statement you are trying to prove says is that the sequence       a    n   converges. You don&#039;t need any measure theory to prove that. It&#039;s plain old first-year calculus.

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