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Question on Definition of Cover of a Set

  1. Feb 11, 2014 #1
    So when we have an open cover of a set X means we have a collection of sets [itex]\{ E_\alpha\}_{\alpha \in I}[/itex]

    such that [itex] X \subset \bigcup_{\alpha \in I} E_\alpha [/itex].

    My question comes from measure theory, on the question of finite [itex]\sigma[/itex] -measures,

    The definition I'm readying says [itex]\mu[/itex] is [itex]\sigma [/itex] - finite if there exists sets [itex] E_i \in \mathcal{A}[/itex] for [itex]i = 1,2, ...[/itex] such that [itex]\mu(E_i) < \infty[/itex] for each [itex]i[/itex] and [itex]X = \bigcup_{i=1}^\infty E_i[/itex].

    Can I say that μ is σ - finite if there exists an open cover for X such that μ(Ei)< ∞ for each i? Is my understanding correct?
     
    Last edited: Feb 11, 2014
  2. jcsd
  3. Feb 12, 2014 #2
    Make it a countable open cover, and you're good to go.
     
  4. Feb 12, 2014 #3

    jgens

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    Gold Member

    It merits note that the converse claim "If μ is σ-finite, then there exists a countable open cover {Ek} with μ(Ek) < ∞" (where X is a topological space and μ is a measure over the Borel algebra) is generally false.
     
  5. Feb 12, 2014 #4
    As an example, consider the "rational counting" measure [itex]\mu:\mathcal B_{\mathbb R}\to [0,\infty][/itex] on the real line. So [itex]\mu(A) = |A\cap\mathbb Q|[/itex]. The measure [itex]\mu[/itex] is [itex]\sigma[/itex]-finite, as witnessed by the countable Borel cover [itex]\{\{r\}:\enspace r\in\mathbb Q\}\cup\{\mathbb R\setminus\mathbb Q\}[/itex]. However, every nonempty open set has infinite [itex]\mu[/itex]-measure.
     
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