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

Latest revision as of 20:31, 3 April 2016

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:

Ring of sets, $\mathcal{R}$ - DONE - Alec (talk) 20:31, 3 April 2016 (UTC)
$\sigma$ -ring, $\mathcal{R}$
Measure, $\mu$ , countably additive extended real valued set function on a $\sigma$ -ring, $\mathcal{R}$ , $\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0}$
Pre-measure, ˉμ, countably additive extended real valued set function defined on a ring of sets, R, ˉμ:R→ˉR≥0
- Goal 1: "extend" a pre-measure, $\bar{\mu}$ to a measure, $\mu$ such that (for a ring of sets $\mathcal{R}$ ): $\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)]$ ^{[Question 1]}
Hereditary set system - a system of sets, say H, such that ∀A∈H∀B∈P(A)[B∈H]
- Hereditary $\sigma$ -ring, $\mathcal{H}$ ^{[Question 2]}
Outer-measure, μ∗ - extended real valued countably subadditive monotonic set function with μ∗(∅)=0
- Theorem: for a pre-measure, ˉμ on a ring 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
$\mu^*$ -measurable sets
Theorem: - the set of all $\mu^*$ -measurable sets is a $\sigma$ -ring and $\mu^*$ is a measure on this sigma-ring
Theorem: - every set in $\sigma_R(\mathcal{R})$ is $\mu^*$ -measurable and $\mu^*$ is a measure on this sigma-ring
Theorem: - the measure induced on the sigma-ring of μ∗-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the σ-ring generated by R^{[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!

Questions

Jump up ↑ Why specifically a measure? An outer-measure extends a measure to be able to measure every subset of every set in $\mathcal{R}$ - 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 $\implies$ subadditivity
Jump up ↑
Suppose H(S) is the hereditary system generated by a collection of subsets, S, and σR(S) the σ-ring generated by S, is it true that:
- $\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))$ ?
The book makes it clear that it intends to use:
- $\mathcal{H}(\sigma_R(S))$
Jump up ↑ I could phrase this better

Old notes

[Expand]

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

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
- complete measure - μ 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, 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:
  - $\forall A\in\mathbf{H}(\mathcal{R})[\mu^*(A)=\text{inf}\{\sum^\infty_{n=1}\mu(A_n)\ \vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R} \wedge A\subseteq \bigcup^\infty_{n=1}A_n\}]$ 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 $\mu$
μ∗-measurable - given an outer measure μ∗ on a hereditary σ-ring H a set A∈H is $\mu^*$ -measurable if:
- ∀B∈H[μ∗(B)=μ∗(A∩B)+μ∗(B∩A′)]
  - PROBLEM: How can we do complementation in a ring?^{[Solution 1]}
  - Solution: Note that S∩T′=S−T, so Halmos is really saying:
    - A∈H is μ∗-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 σR(R) is μ∗-measurable (the σ-ring generated by R)
- Alternatively: $\sigma_R(\mathcal{R})\subseteq\mathcal{S}$
Theorem - ∀A∈H(R)[μ∗(A)⏟Outer=inf{ˉμ(B) | A⊆B∈S}⏟Induced by ˉμ={ˉμ(B) | A⊆B∈σR(R)}⏟Induced by ˉμ]
- 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 μ is a σ-finite measure on a σ-ring R and if μ∗ 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

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

[1] Jump up ↑ Why specifically a measure? An outer-measure extends a measure to be able to measure every subset of every set in $\mathcal{R}$ - 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 $\implies$ subadditivity

[2] Jump up ↑ Suppose $\mathcal{H}(S)$ is the hereditary system generated by a collection of subsets, $S$ , and $\sigma_R(S)$ the $\sigma$ -ring generated by $S$ , is it true that:
$\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))$ ?
The book makes it clear that it intends to use:
$\mathcal{H}(\sigma_R(S))$

[3] $\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))$ ?

[4] $\mathcal{H}(\sigma_R(S))$

[3] Jump up ↑ I could phrase this better

[4] 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:
$\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)]$

[7] $A\in\mathcal{H}$ is $\mu^*$ -measurable if:
$\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)]$

[8] $\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)]$

[Question 1]

[Question 2]

[Question 3]

[Solution 1]

@@ Line 1: / Line 1: @@
 ==Skeleton==
-* [[Ring of sets]]
+If we were to cut out all the "extra" (and useful) theorems into just the core that let us get from [[pre-measure|pre-measures]] to [[measure|measures]] we would be left with what I call the "skeleton".
-* [[Sigma-ring]]
+The "core" of Halmos' measure theory book is the following:
-* [[additive set function]]
+# [[Ring of sets]], {{M|\mathcal{R} }} - '''DONE''' - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 20:31, 3 April 2016 (UTC)
-* ''measure'', {{M|\mu}} - [[extended real valued]], non negative, [[countably additive set function]] defined on a [[ring of sets]]
+# [[Sigma-ring|{{sigma|ring}}]], {{M|\mathcal{R} }}
-* ''hereditary'' system - a system of sets, {{M|\mathcal{E} }} such that if {{M|E\in\mathcal{E} }} then {{M|\forall F\in\mathcal{P}(E)[F\in\mathcal{E}]}}
+# [[Measure]], {{M|\mu}}, countably additive extended real valued set function on a {{sigma|ring}}, {{M|\mathcal{R} }}, {{M|\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }}
-** ''hereditary ring generated by''
+# [[Pre-measure]], {{M|\bar{\mu} }}, countably additive extended real valued set function defined on a ring of sets, {{M|\mathcal{R} }}, {{M|\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }}
-* ''subadditivity''
+#* '''Goal 1: ''' "extend" a pre-measure, {{M|\bar{\mu} }} to a measure, {{M|\mu}} such that (for a ring of sets {{M|\mathcal{R} }}): {{M|1=\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)]}}<!--
-* ''outer measure'', {{M|\mu^*}} (p42) - [[extended real valued]], non-negative, monotone and countably subadditive set function on an ''hereditary'' {{sigma|ring}} with {{M|1=\mu^*(\emptyset)=0}}
-** Theorem: If {{M|\mu}} is a measure on a ring {{M|\mathcal{R} }} and if:
+QUESTION 1 (regarding goal 1)
-*** {{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}}''
+--><ref group="Question">Why specifically a measure? An [[outer-measure]] extends a measure to be able to measure every subset of every set in {{M|\mathcal{R} }} - at the cost of it no longer being [[additive set function|additive]] but instead [[subadditive set function|subadditive]] - why do we want additivity so much? Why is subadditivity not good enough? Obviously it's a weaker property as additivity {{M|\implies}} subadditivity</ref><!--
-* {{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')]}}
+END OF QUESTION 1, back after pre-measure
-*** '''PROBLEM: How can we do [[complementation]] in a ring?'''
+-->
+# [[Hereditary set system]] - a system of sets, say {{M|H}}, such that {{M|1=\forall A\in H\forall B\in\mathcal{P}(A)[B\in H]}}
+#* [[Hereditary sigma-ring|Hereditary {{sigma|ring}}]], {{M|\mathcal{H} }}<ref group="Question">Suppose {{M|\mathcal{H}(S)}} is the hereditary system generated by a collection of subsets, {{M|S}}, and {{M|\sigma_R(S)}} the {{sigma|ring}} generated by {{M|S}}, is it true that:
+* {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}}?
+The book makes it clear that it intends to use:
+* {{M|\mathcal{H}(\sigma_R(S))}}
+</ref>
+# [[Outer-measure]], {{M|\mu^*}} - extended real valued countably subadditive monotonic set function with {{M|1=\mu^*(\emptyset)=0}}
+#* '''Theorem: ''' for a [[pre-measure]], {{M|\bar{\mu} }} on a ring {{M|\mathcal{R} }} the function:
+#** {{MM|1=\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0} }} given by {{MM|1=\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''
+# [[outer-measurable sets|{{M|\mu^*}}-measurable sets]]
+# '''Theorem: ''' - the set of all {{M|\mu^*}}-measurable sets is a {{sigma|ring}} and {{M|\mu^*}} is a [[measure]] on this sigma-ring
+# '''Theorem: ''' - every set in {{M|\sigma_R(\mathcal{R})}} is {{M|\mu^*}}-measurable and {{M|\mu^*}} is a [[measure]] on this sigma-ring
+# '''Theorem: ''' - the measure induced on the sigma-ring of {{M|\mu^*}}-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the {{sigma|ring}} generated by {{M|\mathcal{R} }}<ref group="Question">I could phrase this better</ref>
+#* 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!
+==Questions==
+<references group="Question"/>
+==Old notes==
+{{Begin Inline Theorem}}
+These notes were "too long" I need to compress it into steps.
+{{Begin Inline Proof}}
+{{:Notes:Halmos measure theory skeleton/Old page}}
+{{End Proof}}{{End Theorem}}

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

Latest revision as of 20:31, 3 April 2016

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