Notes:Halmos measure theory skeleton

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If we were to cut out all the "extra" (and useful) theorems into just the core that let us get from pre-measures to measures we would be left with what I call the "skeleton". The "core" of Halmos' measure theory book is the following:

  1. Ring of sets, [ilmath]\mathcal{R} [/ilmath] - DONE - Alec (talk) 20:31, 3 April 2016 (UTC)
  2. [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath]
  3. Measure, [ilmath]\mu[/ilmath], countably additive extended real valued set function on a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath], [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]
  4. Pre-measure, [ilmath]\bar{\mu} [/ilmath], countably additive extended real valued set function defined on a ring of sets, [ilmath]\mathcal{R} [/ilmath], [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]
    • Goal 1: "extend" a pre-measure, [ilmath]\bar{\mu} [/ilmath] to a measure, [ilmath]\mu[/ilmath] such that (for a ring of sets [ilmath]\mathcal{R} [/ilmath]): [ilmath]\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)][/ilmath][Question 1]
  5. Hereditary set system - a system of sets, say [ilmath]H[/ilmath], such that [ilmath]\forall A\in H\forall B\in\mathcal{P}(A)[B\in H][/ilmath]
  6. Outer-measure, [ilmath]\mu^*[/ilmath] - extended real valued countably subadditive monotonic set function with [ilmath]\mu^*(\emptyset)=0[/ilmath]
    • Theorem: for a pre-measure, [ilmath]\bar{\mu} [/ilmath] on a ring [ilmath]\mathcal{R} [/ilmath] the function:
      • [math]\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0}[/math] given by [math]\mu^*:A\mapsto\text{inf}\left.\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \right\vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\}[/math]
      is an outer measure
  7. [ilmath]\mu^*[/ilmath]-measurable sets
  8. Theorem: - the set of all [ilmath]\mu^*[/ilmath]-measurable sets is a [ilmath]\sigma[/ilmath]-ring and [ilmath]\mu^*[/ilmath] is a measure on this sigma-ring
  9. Theorem: - every set in [ilmath]\sigma_R(\mathcal{R})[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable and [ilmath]\mu^*[/ilmath] is a measure on this sigma-ring
  10. Theorem: - the measure induced on the sigma-ring of [ilmath]\mu^*[/ilmath]-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath][Question 3]
    • We haven't shown anything to do with equality of these two sigma-rings! We only show that the outer measure each induce is the same!


  1. Why specifically a measure? An outer-measure extends a measure to be able to measure every subset of every set in [ilmath]\mathcal{R} [/ilmath] - at the cost of it no longer being additive but instead subadditive - why do we want additivity so much? Why is subadditivity not good enough? Obviously it's a weaker property as additivity [ilmath]\implies[/ilmath] subadditivity
  2. Suppose [ilmath]\mathcal{H}(S)[/ilmath] is the hereditary system generated by a collection of subsets, [ilmath]S[/ilmath], and [ilmath]\sigma_R(S)[/ilmath] the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]S[/ilmath], is it true that:
    • [ilmath]\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))[/ilmath]?
    The book makes it clear that it intends to use:
    • [ilmath]\mathcal{H}(\sigma_R(S))[/ilmath]
  3. I could phrase this better

Old notes

These notes were "too long" I need to compress it into steps.


  • Ring of sets
  • Sigma-ring
  • additive set function
  • measure, [ilmath]\mu[/ilmath] - extended real valued, non negative, countably additive set function defined on a ring of sets
    • complete measure - [ilmath]\mu[/ilmath] is complete if:
      • [ilmath]\forall A\in\mathcal{R}\forall B[B\subseteq A\wedge\mu(A)=0\implies B\in\mathcal{R} ][/ilmath]
  • hereditary system - a system of sets, [ilmath]\mathcal{E} [/ilmath] such that if [ilmath]E\in\mathcal{E} [/ilmath] then [ilmath]\forall F\in\mathcal{P}(E)[F\in\mathcal{E}][/ilmath]
    • hereditary ring generated by
  • subadditivity
  • outer measure, [ilmath]\mu^*[/ilmath] (p42) - extended real valued, non-negative, monotone and countably subadditive set function on an hereditary [ilmath]\sigma[/ilmath]-ring with [ilmath]\mu^*(\emptyset)=0[/ilmath]
    • Theorem: If [ilmath]\mu[/ilmath] is a measure on a ring [ilmath]\mathcal{R} [/ilmath] and if:
      • [ilmath]\forall A\in\mathbf{H}(\mathcal{R})[\mu^*(A)=\text{inf}\{\sum^\infty_{n=1}\mu(A_n)\ \vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R} \wedge A\subseteq \bigcup^\infty_{n=1}A_n\}][/ilmath] then [ilmath]\mu^*[/ilmath] is an extension of [ilmath]\mu[/ilmath] to an outer measure on [ilmath]\mathbf{H}(\mathcal{R})[/ilmath]
    • [ilmath]\mu^*[/ilmath] is the outer measure induced by the measure [ilmath]\mu[/ilmath]
  • [ilmath]\mu^*[/ilmath]-measurable - given an outer measure [ilmath]\mu^*[/ilmath] on a hereditary [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{H} [/ilmath] a set [ilmath]A\in\mathcal{H} [/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
    • [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')][/ilmath]
      • PROBLEM: How can we do complementation in a ring?[Solution 1]
      • Solution: Note that [ilmath]S\cap T'=S-T[/ilmath], so Halmos is really saying:
        • [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
          • [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B - A)][/ilmath]
  • Theorem: if [ilmath]\mu^*[/ilmath] is an outer measure on a hereditary [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{H} [/ilmath] and if [ilmath]\mathcal{S} [/ilmath] is the class of all [ilmath]\mu^*[/ilmath]-measurable sets, then [ilmath]\mathcal{S} [/ilmath] is a ring of sets
  • Theorem: (p46) - [ilmath]\mathcal{S} [/ilmath] is a sigma-ring
  • Theorem: - Every set of [ilmath]\mu^*=0[/ilmath] belongs to [ilmath]\mathcal{S} [/ilmath] and the set function [ilmath]\bar{\mu}:\mathcal{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] is a complete measure (AKA: [ilmath]\bar{\mu} [/ilmath] is the measure induced by [ilmath]\mu^*[/ilmath]).
  • Theorem - Every set in [ilmath]\sigma_R(\mathcal{R})[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable (the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath])
    • Alternatively: [ilmath]\sigma_R(\mathcal{R})\subseteq\mathcal{S} [/ilmath]
  • Theorem - [ilmath]\forall A\in\mathbf{H}(\mathcal{R})[\underbrace{\mu^*(A)}_{\text{Outer} }=\underbrace{\text{inf}\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\mathcal{S}\} }_{\text{Induced by } \bar{\mu} }=\underbrace{\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\sigma_R(\mathcal{R})\} }_{\text{Induced by } \bar{\mu} }][/ilmath]
    • That is to say that:
      • Outer measure induced by [ilmath]\bar{\mu} [/ilmath] on [ilmath]\sigma_R(\mathcal{R})[/ilmath] AND
      • the outer measure induced by [ilmath]\bar{\mu} [/ilmath] on [ilmath]\mathcal{S} [/ilmath]
      • agree with [ilmath]\mu^*[/ilmath]
  • completion of a measure - p55
  • Theorem: if [ilmath]\mu[/ilmath] is a [ilmath]\sigma[/ilmath]-finite measure on a [ilmath]\sigma[/ilmath]-ring [ilmath]\mathbb{R} [/ilmath] and if [ilmath]\mu^*[/ilmath] is the outer measure it induces, then:
    • The completion of the extension of [ilmath]\mu[/ilmath] to [ilmath]\sigma_R(\mathcal{R})[/ilmath] is identical with [ilmath]\mu^*[/ilmath] on the class of all [ilmath]\mu^*[/ilmath]-measurable sets


  1. We only have to deal with [ilmath]A\cap B'[/ilmath] which could just be Halmos abusing notation. If we take it literally ("[ilmath]B':=\{x(\in?)\vert x\notin B\}[/ilmath]") we may be able to work through this. Note that the intersection of sets is a subset of each set so [ilmath]A\cap B'\subseteq A[/ilmath] and is in fact [ilmath]=A-B[/ilmath], so what Halmos is really saying is:
    • [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
      • [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)][/ilmath]