Notes:Halmos measure theory skeleton

Skeleton

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, $\mathcal{R}$ - DONE - Alec (talk) 20:31, 3 April 2016 (UTC)
2. $\sigma$-ring, $\mathcal{R}$
3. Measure, $\mu$, countably additive extended real valued set function on a $\sigma$-ring, $\mathcal{R}$, $\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0}$
4. Pre-measure, $\bar{\mu}$, countably additive extended real valued set function defined on a ring of sets, $\mathcal{R}$, $\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0}$
   • Goal 1: "extend" a pre-measure, $\bar{\mu}$ to a measure, $\mu$ such that (for a ring of sets $\mathcal{R}$): $\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)]$
5. Hereditary set system - a system of sets, say $H$, such that $\forall A\in H\forall B\in\mathcal{P}(A)[B\in H]$
   • Hereditary $\sigma$-ring, $\mathcal{H}$
6. Outer-measure, $\mu^*$ - extended real valued countably subadditive monotonic set function with $\mu^*(\emptyset)=0$
   • Theorem: for a pre-measure, $\bar{\mu}$ on a ring $\mathcal{R}$ the function:
     • $$\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0}$$ given by $$\mu^*:A\mapsto\text{inf}\left.\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \right\vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\}$$

     is an outer measure

7. $\mu^*$-measurable sets
8. Theorem: - the set of all $\mu^*$-measurable sets is a $\sigma$-ring and $\mu^*$ is a measure on this sigma-ring
9. Theorem: - every set in $\sigma_R(\mathcal{R})$ is $\mu^*$-measurable and $\mu^*$ is a measure on this sigma-ring
10. Theorem: - the measure induced on the sigma-ring of $\mu^*$-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the $\sigma$-ring generated by $\mathcal{R}$
    • 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!

Old notes

[Expand]

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

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Skeleton

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

Solutions