Difference between revisions of "The set of all mu*-measurable sets is a ring"

Revision as of 08:34, 29 May 2016 (view source) Alec (Talk | contribs) m (→‎See also: Finished the proof, in a logical way.) ← Older edit		Latest revision as of 08:36, 29 May 2016 (view source) Alec (Talk | contribs) m (Removed weird Halmos stuff.)
Line 6:		Line 6:
	:* See the page [[Mu*-measurable set|{{M|\mu^*}}-measurable set]] for more information.		:* See the page [[Mu*-measurable set|{{M|\mu^*}}-measurable set]] for more information.
	==Proof==		==Proof==
−	{{Warning|Below is just some disjointed set of notes I did (and a test for a new template) and NOT complete, I've saved the page so I can save my work}}	+	{{Requires writeup}}
−	{{Begin Notebox}}Recall what we must show in order for {{M|\mathcal{S} }} to be a [[ring of sets]]:	+
−	{{Begin Notebox Content}}	+
−	It is sufficient to show only:	+
−	* {{M|\forall A,B\in \mathcal{S}[A\cup B\in \mathcal{S}]}}	+
−	* {{M|\forall A,B\in \mathcal{S}[A-B\in \mathcal{S}]}}	+
−	{{End Notebox Content}}{{End Notebox}}	+
−	# {{M|\mathcal{S} }} is ''{{M|\cup}}-closed'', that is: {{M|\forall E,F\in\mathcal{S}[E\cup F\in\mathcal{S}]}}, ie, that: {{M|1=\forall E,F\in\mathcal{S}\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))]}}	+
−	#* Let {{M|E,F\in\mathcal{S} }} be given. We now know (by hypothesis):	+
−	#*# {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)]}} {{TrialEq|def=1.1}}	+
−	#*# and {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)]}}	+
−	#** Let {{M|A\in H}} be given.	+
−	<hr/>	+
−	===Notes===	+
−	Note that {{M|1=\mu^*(A\cap E)=\mu^*(A\cap E\cap F)+\mu^*((A\cap E)-F)}} {{TrialEq|def=1.2}}	+
−	This is okay because by hypothesis:	+
−	* {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)]}} and	+
−	* given an {{M|A\in H}}, {{M|A\cap E\subseteq A}}, as {{M|H}} is [[hereditary (measure theory)|hereditary]], {{M|A\cap E\in H}} too.	+
−	We can simply substitute {{TrialEq|link=2}} into {{TrialEq|link=1}} to get:	+
−	* Blah [[#Proof]]	+
−	* {{M|1=\underline{\mathbf{\color{black}{(\text{Eq: } 1.2 ):} }\ \ \ \ \forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(x_n,\ell)<\epsilon]} }}	+
−	===Pre-notes===	+
−	{{Begin Blue Notebox|plain}}	+
−	We wish to show: {{M|1=\forall E,F\in\mathcal{S}\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))]}}	+
−	{{Begin Blue Notebox Content}}	+
−	* We wish to show: {{M|1=\forall E,F\in\mathcal{S}\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))]}} from only:	+
−	*# {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)]}}	+
−	*# and {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)]}}	+
−	* Let {{M|E,F\in \mathcal{S} }} and {{M|A\in H}} be given.	+
−	** We need to combine the terms of the hypothesis (which involve just {{M|A}} and {{M|E}}, or just {{M|A}} and {{M|F}}) into ones that involve {{M|A}}, {{M|E}} and {{M|F}} first of all. We could start with {{M|\mu^*(A\cap(E\cup F))}} and use the [[subadditive]] property of [[outer-measure|outer-measures]] however this'll involve a {{M|\le}} symbol, and to go "the other way" (obtain {{M|\ge}} terms) to show equality looks very hard.	+
−	** This suggests we need to start looking at the set-theory part, the operations of {{M|\cup}}, {{M|\cap}} and {{M|-}}.	+
−	* First note {{M|1=A\cap(E\cup F)=(A\cap E)\cup(A\cap F)}}	+
−	** Again, we don't actually want to split {{M|\mu^*((A\cap E)\cup(A\cap F))}} into {{M|\mu^*(A\cap E)+\mu^*(A\cap F)}} using properties of {{M|\mu^*}}, for the reasons above. So we must look elsewhere.	+
−	* Note [[the intersection of sets is a subset of each set]], so {{M|A\cap E\subseteq E}} but more importantly, {{M|A\cap E\subseteq A}}.	+
−	** Since {{M|A\in H}} and {{M|H}} is a [[hereditary system of sets]], {{M|A\cap E\in H}}	+
−	* We know by hypothesis (point 2) that: {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)]}}	+
−	** Thus we have: {{M|1=\mu^*(A\cap E)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)}}	+
−	* We can substitute this into part 1 of the hypothesis, and arrive at:	+
−	** {{M|1=\forall A\in H[\mu^*(A)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)+\mu^*(A-E)]}}	+
−	* If we continue trying to use {{M|(A\cap E)\cup (A\cap F)}} we'll just arrive at the same term involving {{M|F}} instead of {{M|E}}. So we turn our attention to the other part:	+
−	** {{M|A-(E\cup F)}}, clearly {{M|1=A-(E\cup F)=(A-E)-F}} but by [[commutivity of the union]], {{M|1=E\cup F=F\cup E}} so we also have:	+
−	*** {{M|1=A-(E\cup F)=(A-F)-E}}, we can pick either one.	+
−	* In our substitution above, we have a {{M|\mu^*(A-E)}} term at the end, {{M|A-E\subseteq A}}, so {{M|A-E\in H}}. If we use the hypothesis on {{M|F}}:	+
−	*:* {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)]}}	+
−	** We can get something like {{M|1=\mu^*(A-E)=\mu^*((A-E)\cap F)+\mu^*((A-E)-F)}}	+
−	* Let's do this new substitution to obtain: {{M|1=\forall A\in H[\mu^*(A-E)=\mu^*((A-E)\cap F)+\mu^*((A-E)-F)]}}	+
−	* Recall we already know: {{M|1=\forall A\in H[\mu^*(A)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)+\mu^*(A-E)]}} - we can combine these two to arrive at:	+
−	** {{M|1=\forall A\in H[\mu^*(A)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)+\mu^*((A-E)\cap F)+\mu^*((A-E)-F)]}}	+
−	*** Remember from above {{M|1=(A-E)-F=A-(E\cup F)}} so:	+
−	**** {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap E\cap F)+\mu^*((A\cap E)-F)+\mu^*((A-E)\cap F)+\mu^*(A-(E\cup F))]}}	+
−	{{End Notebox Content}}{{End Notebox}}	+
−		+
−	==Actual notes==	+
	I've thought about it, we know:		I've thought about it, we know:
	# {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)]}} and		# {{M|1=\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)]}} and

---

Latest revision as of 08:36, 29 May 2016

Statement

[ilmath]\mathcal{S} [/ilmath], the set of all [ilmath]\mu^*[/ilmath] measurable sets, is a ring of sets^[1].

Recall that given an outer-measure, [ilmath]\mu^*:H\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath], where [ilmath]H[/ilmath] is a hereditary [ilmath]\sigma[/ilmath]-ring that we call a set, [ilmath]A\in H[/ilmath] [ilmath]\mu^*[/ilmath]-measurable if^[1]:

• [ilmath]\forall B\in H[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)][/ilmath].
• See the page [ilmath]\mu^*[/ilmath]-measurable set for more information.

Proof

---

This requires one or more proofs to be written up neatly and is on a to-do list for having them written up. This does not mean the results cannot be trusted, it means the proof has been completed, just not written up here yet. It may be in a notebook, some notes about reproducing it may be left in its place, perhaps a picture of it, so forth.

---

I've thought about it, we know:

1. [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)][/ilmath] and
2. [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)][/ilmath]

And want to show:

• [ilmath]\forall A,B\in \mathcal{S}[A\cup B\in \mathcal{S}][/ilmath]
• [ilmath]\forall A,B\in \mathcal{S}[A-B\in \mathcal{S}][/ilmath]

Consider ANY [ilmath]3[/ilmath] sets, [ilmath]A,\ E[/ilmath] and [ilmath]F[/ilmath]. Rather than dealing with "complicated" and non-unique expressions (eg [ilmath](A\cap E)-F=(A-F)\cap E[/ilmath] which are the same despite looking different, and there's no canonical or natural form for these sets), let us instead define:

• [ilmath]\alpha\ :=\ A-(E\cup F)[/ilmath]
• [ilmath]\beta\ :=\ (A\cap E)-F[/ilmath]
• [ilmath]\gamma\ :=\ A\cap E\cap F[/ilmath]
• [ilmath]\delta\ :=\ (A\cap F)-E[/ilmath]

Note that [ilmath]\forall \Omega\in\{\alpha,\ \beta,\ \gamma,\ \delta\}[/ilmath][ilmath]\Big[\big(\mu^*(\Omega)=\mu^*(\Omega\cap E)+\mu^*(\Omega-E)\big)[/ilmath][ilmath]\wedge[/ilmath][ilmath]\big(\mu^*(\Omega)=\mu^*(\Omega\cap F)+\mu^*(\Omega-F)\big)\Big][/ilmath], as any such [ilmath]\Omega[/ilmath] is a subset of [ilmath]A[/ilmath]. And we have the above for all [ilmath]A\in H[/ilmath]. As [ilmath]H[/ilmath] is a hereditary system of sets, we have it for all subsets of a given [ilmath]A[/ilmath] too.

As a matter of notation, we write [ilmath]\Omega_1\Omega_2[/ilmath] for [ilmath]\Omega_1\cup\Omega_2[/ilmath], so for example [ilmath]\beta\gamma\delta=A\cap(E\cup F)[/ilmath]

Proof #1

• Let [ilmath]E,F\in S[/ilmath] be given
  • Let [ilmath]A\in H[/ilmath] be given. We wish to show [ilmath]\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))[/ilmath] or [ilmath]\mu^*(A)=\mu^*(\beta\gamma\delta)+\mu^*(\alpha)[/ilmath]
    • Notice [ilmath]\mu^*(\beta\gamma\delta)=\mu^*(\beta\gamma\delta\cap E)+\mu^*(\beta\gamma\delta-E)[/ilmath]
      • By tidying up the sets, we see this is: [ilmath]\mu^*(\beta\gamma\delta)=\mu^*(\beta\gamma)+\mu^*(\delta)[/ilmath]
    • So [ilmath]\mu^*(\beta\gamma\delta)=\mu^*(\beta\gamma)+\mu^*(\delta)[/ilmath]
    • Notice [ilmath]\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)[/ilmath], or [ilmath]\mu^*(A)=\mu^*(\beta\gamma)+\mu^*(\alpha\delta)[/ilmath]
      • Re-arranging this, we see [ilmath]\mu^*(\beta\gamma)=\mu^*(A)-\mu^*(\alpha\delta)[/ilmath]
    • Substituting this in: [ilmath]\mu^*(\beta\gamma\delta)=\mu^*(A)-\mu^*(\alpha\delta)+\mu^*(\delta)[/ilmath]
      • Notice we can use [ilmath]F[/ilmath] to split the [ilmath]\alpha\delta[/ilmath] into [ilmath]\alpha\delta\cap F=\delta[/ilmath] and [ilmath]\alpha\delta-F=\alpha[/ilmath]
      • So [ilmath]\mu^*(\alpha\delta)=\mu^*(\delta)+\mu^*(\alpha)[/ilmath]
    • Substituting this back in we see: [ilmath]\mu^*(\beta\gamma\delta)=\mu^*(A)-\mu^*(\alpha)-\mu^*(\delta)+\mu^*(\delta)[/ilmath]
    • Simplifying: [ilmath]\mu^*(\beta\gamma\delta)=\mu^*(A)-\mu^*(\alpha)[/ilmath]
    • Rearranging: [ilmath]\mu^*(A)=\mu^*(\beta\gamma\delta)+\mu^*(\alpha)[/ilmath]
    • However notice:
      1. [ilmath]\beta\gamma\delta=A\cap(E\cup F)[/ilmath] and
      2. [ilmath]\alpha=A-(E\cup F)[/ilmath]
    • So we have:
      • [ilmath]\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))[/ilmath]
  • Since [ilmath]A\in H[/ilmath] was arbitrary we have: [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))][/ilmath]
    • So [ilmath]E\cup F\in S[/ilmath]
• Since [ilmath]E,F\in S[/ilmath] were arbitrary, we have [ilmath]\forall E,F\in S[E\cup F\in S][/ilmath]
  • As a formula: [ilmath]\forall E,F\in S\big[\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))]\big][/ilmath] Caution:[ilmath]\forall[/ilmath]s commute, hence the brackets, I believe (I scratched a quick proof somewhere) that this is equivalent to the formula without the outer set of [ilmath][\ ][/ilmath] however nothing is given so far - hence the brackets

This completes the proof.

Proof #2

Gist is the same, I did this on paper:

• [ilmath]\mu^*(\alpha\gamma\delta)[/ilmath] (as [ilmath]\alpha\gamma\delta=A-(E-F)[/ilmath]), split using [ilmath]F[/ilmath] to get [ilmath]\mu^*(\alpha)+\mu^*(\gamma\delta)[/ilmath]
• We need a [ilmath]\gamma\delta[/ilmath] term, but [ilmath]\gamma\delta=A\cap F[/ilmath] so by def of [ilmath]F[/ilmath]:
  • [ilmath]\mu^*(\gamma\delta)=\mu^*(A)-\mu^*(\alpha\beta)[/ilmath]
• Now: [ilmath]\mu^*(\alpha\gamma\delta)=\mu^*(A)-\mu^*(\alpha\beta)+\mu^*(\alpha)[/ilmath]
• [ilmath]\alpha\beta[/ilmath] can be split by [ilmath]F[/ilmath], so [ilmath]\mu^*(\alpha\beta)=\mu^*(\alpha)+\mu^*(\beta)[/ilmath], thus:
• [ilmath]\mu^*(\alpha\gamma\delta)=\mu^*(A)-\mu^*(\alpha)-\mu^*(\beta)+\mu^*(\alpha)[/ilmath]

The result follows

See also

• The set of all [ilmath]\mu^*[/ilmath]-measurable sets is a [ilmath]\sigma[/ilmath]-ring
• The restriction of an outer-measure to the set of all [ilmath]\mu^*[/ilmath]-measurable sets is a measure

References

1. ↑ ^{1.0 1.1} Measure Theory - Paul R. Halmos

v • d • e   Measure Theory Overview of the basic and important objects studied in measure theory Set-structures ring of sets, algebra of sets, [ilmath]\sigma[/ilmath]-ring, [ilmath]\sigma[/ilmath]-algebra, hereditary [ilmath]\sigma[/ilmath]-ring measurable space Measures Pre-measure [ilmath]\leadsto[/ilmath] outer-measure [ilmath]\leadsto[/ilmath] measure (see Extending pre-measures to measures (via: Extending pre-measures to outer-measures [ilmath]\rightarrow[/ilmath] "Restricting outer-measures to measures" maybe?). Alternatively: inner-measure can be used. measure space Functions measurable map, simple function (standard representation) Spaces Integrals Integrating a simple function [ilmath]\leadsto[/ilmath] Integral of a positive function [ilmath]\leadsto[/ilmath] Integrating all measurable functions