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Thread: Integral with another measure

  1. #1
    Super Member
    Mar 2006

    Integral with another measure

    Suppose that f is a non-negative integrable function. Let $\displaystyle \lambda (E) = \int _E f dm $ for all $\displaystyle E \in \mathbb M $, the set of all $\displaystyle \sigma $-algebra.

    Prove that $\displaystyle \int g d \lambda = \int fg dm \ \ \ \forall g $, g non-negative integrable.

    Proof so far.

    Now, from an earlier thread, I understand that $\displaystyle \lambda $ is a measure.

    Case 1) Suppose that g is a simple function, then $\displaystyle g= \sum ^n_{i=1} a_i 1_{E_i} $, Ei disjoint.

    Then $\displaystyle \int g d \lambda = \int \sum a_i1_{E_i}d \lambda = \sum ^n_{i=1} a_i \lambda (E_i)$

    Which gives $\displaystyle = \sum _{i=1}^n a_i \int _{E_i}f dm $

    Now, is this integrand equals to $\displaystyle \int gf dm $? I think it is but I just can't seem to work it out right.

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  2. #2
    MHF Contributor

    Aug 2008
    Paris, France
    You should simply write $\displaystyle \int_{E_i} f dm= \int 1_{E_i} f dm$, and use linearity of integral like you did before.
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