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A measure related with Reimann integral

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]( \mathbb{R}^k , \mathcal{A} , m_{k} )[/itex] be a Lebesgue measurable space, i.e., [itex]m_{k}=m[/itex] is a Lebesgue measure. Let [itex]f: \mathbb{R^k} \to \mathbb{R}[/itex] be a [itex]m[/itex]-integrable function. Define a function [itex]\mu : \mathcal{A} \to [0,\infty][/itex] by $$ \mu(A) := \int_{A} f(x) dx $$ with [itex]A \in \mathcal{A}[/itex]. Then is [itex]\mu[/itex] a measure?

    Qeustion. Is there any error in the below working? And does it need any additional hypothesis to show $mu$ is a measure? Or is it acutally not a measure?

    2. Relevant equations



    3. The attempt at a solution
    When [itex]A[/itex] is an empty set, the integral is obviously 0. Now suppose [itex]A_{0} , \dots , A_{n} \in \mathcal{A}[/itex] as they are disjoint. Then $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) := \int_{\mathbb{R}^{k}} \chi_{\bigcup_{k=0}^{n} A_{k}}(x)f(x)dx $$ and we get $$ \chi_{\bigcup_{k=0}^{n} A_{k}} = \sum_{k=0}^{n} \chi_{A_k}$$ as [itex]A_{k}[/itex] are disjoint. Therefore, we have $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) = \sum_{k=0}^{n} \mu(A_{k}). $$ Thus taking [itex]n \to \infty[/itex] we get the required result.
     
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  3. Oct 8, 2012 #2

    micromass

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    I'm missing somewhere that [itex]f\geq 0[/itex] a.e.

    You have to justify why

    [tex]\lim_{n\rightarrow +\infty} \mu ( \bigcup_{k=0}^{n} A_{k} )=\mu ( \bigcup_{k=0}^{+\infty} A_{k} )[/tex]
     
  4. Oct 8, 2012 #3
    Thanks for your answer, and as you have told me I should assume that [itex]f(x) \geq 0 [/itex] a.e.. And for the last part you mentioned, I think Dominated Convergence Theorem would suffice, woudn't it? I used it in my solution..
     
  5. Oct 8, 2012 #4

    micromass

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    Yes, the dominated convergence would suffice here.
     
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