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

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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.

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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]

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