Let μ , v n be measures on ( X , B ) with μ being σ-finite and s u p { v n ( X ) } &lt; ∞. Note that v = ∑ n ≥ 1 2 − n v n &lt;&lt; μ if and only if v n &lt;&lt; μ. Let the Lebesgue decomposition of v n with respect to μ be v n = v n , a + v n , s , where d v n , a d μ = f n .Note that v is a finite (hence σ-finite) measure, so it has a Lebesgue decomposition. I want to find the Lebesgue decomposition of v with respect to μ and calculate d v a d μ . I believe the Lebesgue decomposition will just be v = v a + v s = ∑ n ≥ 1 2 − n v n , a + ∑ n ≥ 1 2 − n v n , s . Then my question is this:Are Radon Nikodym derivatives countably additive? Can I say that d v a d μ = ∑ n ≥ 1 2 − n f n ? Thanks.

Question

Let   μ  ,      v    n   be measures on   (  X  ,      B    ) with   μ being   σ-finite and   s  u  p  {      v    n    (  X  )  }  &amp;lt;  ∞. Note that   v  =      ∑          n      ≥      1            2          −      n            v    n    &amp;lt;&amp;lt;  μ if and only if       v    n    &amp;lt;&amp;lt;  μ. Let the Lebesgue decomposition of       v    n   with respect to   μ be       v    n    =      v          n      ,      a        +      v          n      ,      s      , where             d              v                  n          ,          a                            d      μ        =      f    n  .Note that   v is a finite (hence   σ-finite) measure, so it has a Lebesgue decomposition. I want to find the Lebesgue decomposition of   v with respect to   μ and calculate             d              v        a                    d      μ      . I believe the Lebesgue decomposition will just be   v  =      v    a    +      v    s    =      ∑          n      ≥      1            2          −      n            v          n      ,      a        +      ∑          n      ≥      1            2          −      n            v          n      ,      s      . Then my question is this:Are Radon Nikodym derivatives countably additive? Can I say that             d              v        a                    d      μ        =      ∑          n      ≥      1            2          −      n            f    n  ? Thanks.

ol5ick189 · Accepted Answer

Suppose each       v    n   has decomposition   d      v    n    =      f    n      d  μ  +  d      λ    n   with       λ    n    ⊥  μ. You are interested in the decomposition for   v  =      ∑          n      =      1              ∞            v    n  . We have      v    n    (  S  )  =      ∑          n      =      1              ∞        (      ∫    S        f    n    d  μ  +      λ    n    (  S  )  )  =      ∫    S        ∑          n      =      1              ∞            f    n      d  μ  +      ∑          n      =      1              ∞            λ    n    (  S  )  .It is easy to show that   λ  =      ∑          n      =      1              ∞            λ    n    ⊥  μ. Thus  d  v  =      ∑          n      =      1              ∞            f    n      d  μ  +  d  λis the decomposition for   v.

Cara Duke · Accepted Answer

Whenever       f    n   are measurable functions on a measure space   (  X  ,  F  ,  μ  ) with       f    n    :  X  →  [  0  ,  ∞  ], then       ∫    X        ∑          n      =      1              ∞            f    n      d  μ  =      ∑          n      =      1              ∞            ∫          X            f    n      d  μ. This is a simple consequence of monotone convergence theorem and the additivity of the integral on nonnegative functions. I guess i&#039;m right

Let μ , v n </msub> be measures on ( X , <mr

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