Difference between revisions of "Notes:Halmos measure theory skeleton"
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*** {{M|1=\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 {{M|\mu^*}} is an ''extension'' of {{M|\mu}} to an outer measure on {{M|\mathbf{H}(\mathcal{R})}} | *** {{M|1=\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 {{M|\mu^*}} is an ''extension'' of {{M|\mu}} to an outer measure on {{M|\mathbf{H}(\mathcal{R})}} | ||
** {{M|\mu^*}} is the ''outer measure induced by the measure {{M|\mu}}'' | ** {{M|\mu^*}} is the ''outer measure induced by the measure {{M|\mu}}'' | ||
+ | * {{M|\mu^*}}-measurable - given an ''outer measure'' {{M|\mu^*}} on a hereditary {{sigma|ring}} {{M|\mathcal{H} }} a set {{M|A\in\mathcal{H} }} is ''{{M|\mu^*}}-measurable'' if: | ||
+ | ** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')]}} | ||
+ | *** '''PROBLEM: How can we do [[complementation]] in a ring?''' |
Revision as of 19:26, 22 March 2016
Skeleton
- Ring of sets
- Sigma-ring
- additive set function
- measure, μ - extended real valued, non negative, countably additive set function defined on a ring of sets
- hereditary system - a system of sets, E such that if E∈E then ∀F∈P(E)[F∈E]
- hereditary ring generated by
- subadditivity
- outer measure, μ∗ (p42) - extended real valued, non-negative, monotone and countably subadditive set function on an hereditary σ-ring with μ∗(∅)=0
- Theorem: If μ is a measure on a ring R and if:
- ∀A∈H(R)[μ∗(A)=inf{∑∞n=1μ(An) | (An)∞n=1⊆R∧A⊆⋃∞n=1An}] then μ∗ is an extension of μ to an outer measure on H(R)
- μ∗ is the outer measure induced by the measure μ
- Theorem: If μ is a measure on a ring R and if:
- μ∗-measurable - given an outer measure μ∗ on a hereditary σ-ring H a set A∈H is μ∗-measurable if:
- ∀B∈H[μ∗(B)=μ∗(A∩B)+μ∗(B∩A′)]
- PROBLEM: How can we do complementation in a ring?
- ∀B∈H[μ∗(B)=μ∗(A∩B)+μ∗(B∩A′)]