The set of all [ilmath]\mu^*[/ilmath]-measurable sets is a ring

Stub grade: A*

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.

Statement

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

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:

• [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

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

Recall what we must show in order for [ilmath]\mathcal{S} [/ilmath] to be a ring of sets:

It is sufficient to show only:

• [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]

1. [ilmath]\mathcal{S} [/ilmath] is [ilmath]\cup[/ilmath]-closed, that is: [ilmath]\forall E,F\in\mathcal{S}[E\cup F\in\mathcal{S}][/ilmath], ie, that: [ilmath]\forall E,F\in\mathcal{S}\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))][/ilmath]
   • Let [ilmath]E,F\in\mathcal{S} [/ilmath] be given. We now know (by hypothesis):
     1. [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)][/ilmath] [ilmath]\underline{\mathbf{\color{black}{(\text{Eq: } 1.1 )} } }[/ilmath]
     2. and [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)][/ilmath]
     • Let [ilmath]A\in H[/ilmath] be given.

---

Notes

Note that [ilmath]\mu^*(A\cap E)=\mu^*(A\cap E\cap F)+\mu^*((A\cap E)-F)[/ilmath] [ilmath]\underline{\mathbf{\color{black}{(\text{Eq: } 1.2 )} } }[/ilmath] This is okay because by hypothesis:

• [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)][/ilmath] and
• given an [ilmath]A\in H[/ilmath], [ilmath]A\cap E\subseteq A[/ilmath], as [ilmath]H[/ilmath] is hereditary, [ilmath]A\cap E\in H[/ilmath] too.

We can simply substitute [ilmath]\color{black}{( 2 )}[/ilmath] into [ilmath]\color{black}{( 1 )}[/ilmath] to get:

• Blah #Proof
• [ilmath]\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]}[/ilmath]

Pre-notes

We wish to show: [ilmath]\forall E,F\in\mathcal{S}\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))][/ilmath]

• We wish to show: [ilmath]\forall E,F\in\mathcal{S}\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))][/ilmath] from only:
  1. [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)][/ilmath]
  2. and [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)][/ilmath]
• Let [ilmath]E,F\in \mathcal{S} [/ilmath] and [ilmath]A\in H[/ilmath] be given.
  • We need to combine the terms of the hypothesis (which involve just [ilmath]A[/ilmath] and [ilmath]E[/ilmath], or just [ilmath]A[/ilmath] and [ilmath]F[/ilmath]) into ones that involve [ilmath]A[/ilmath], [ilmath]E[/ilmath] and [ilmath]F[/ilmath] first of all. We could start with [ilmath]\mu^*(A\cap(E\cup F))[/ilmath] and use the subadditive property of outer-measures however this'll involve a [ilmath]\le[/ilmath] symbol, and to go "the other way" (obtain [ilmath]\ge[/ilmath] terms) to show equality looks very hard.
  • This suggests we need to start looking at the set-theory part, the operations of [ilmath]\cup[/ilmath], [ilmath]\cap[/ilmath] and [ilmath]-[/ilmath].
• First note [ilmath]A\cap(E\cup F)=(A\cap E)\cup(A\cap F)[/ilmath]
  • Again, we don't actually want to split [ilmath]\mu^*((A\cap E)\cup(A\cap F))[/ilmath] into [ilmath]\mu^*(A\cap E)+\mu^*(A\cap F)[/ilmath] using properties of [ilmath]\mu^*[/ilmath], for the reasons above. So we must look elsewhere.
• Note the intersection of sets is a subset of each set, so [ilmath]A\cap E\subseteq E[/ilmath] but more importantly, [ilmath]A\cap E\subseteq A[/ilmath].
  • Since [ilmath]A\in H[/ilmath] and [ilmath]H[/ilmath] is a hereditary system of sets, [ilmath]A\cap E\in H[/ilmath]
• We know by hypothesis (point 2) that: [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)][/ilmath]
  • Thus we have: [ilmath]\mu^*(A\cap E)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)[/ilmath]
• We can substitute this into part 1 of the hypothesis, and arrive at:
  • [ilmath]\forall A\in H[\mu^*(A)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)+\mu^*(A-E)][/ilmath]
• If we continue trying to use [ilmath](A\cap E)\cup (A\cap F)[/ilmath] we'll just arrive at the same term involving [ilmath]F[/ilmath] instead of [ilmath]E[/ilmath]. So we turn our attention to the other part:
  • [ilmath]A-(E\cup F)[/ilmath], clearly [ilmath]A-(E\cup F)=(A-E)-F[/ilmath] but by commutivity of the union, [ilmath]E\cup F=F\cup E[/ilmath] so we also have:
    • [ilmath]A-(E\cup F)=(A-F)-E[/ilmath], we can pick either one.
• In our substitution above, we have a [ilmath]\mu^*(A-E)[/ilmath] term at the end, [ilmath]A-E\subseteq A[/ilmath], so [ilmath]A-E\in H[/ilmath]. If we use the hypothesis on [ilmath]F[/ilmath]:

  • [ilmath]\forall A\in H[\mu^*(A)=\mu^*(A\cap F)+\mu^*(A-F)][/ilmath]

  • We can get something like [ilmath]\mu^*(A-E)=\mu^*((A-E)\cap F)+\mu^*((A-E)-F)[/ilmath]
• Let's do this new substitution to obtain: [ilmath]\forall A\in H[\mu^*(A-E)=\mu^*((A-E)\cap F)+\mu^*((A-E)-F)][/ilmath]
• Recall we already know: [ilmath]\forall A\in H[\mu^*(A)=\mu^*((A\cap E)\cap F)+\mu^*((A\cap E)-F)+\mu^*(A-E)][/ilmath] - we can combine these two to arrive at:
  • [ilmath]\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)][/ilmath]
    • Remember from above [ilmath](A-E)-F=A-(E\cup F)[/ilmath] so:
      • [ilmath]\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))][/ilmath]

Actual notes

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:

		A
		[ilmath]\alpha[/ilmath]
				[ilmath]\beta[/ilmath]
		[ilmath]\delta[/ilmath]		[ilmath]\gamma[/ilmath]
						E
	F
Ven diagram showing the regions (some cells still have borders, only the coloured ones matter)

• [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