Difference between revisions of "Notes:Halmos measure theory skeleton"

Revision as of 20:05, 22 March 2016

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 1]}
    • 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

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