The set of all $\mu^*$-measurable sets is a ring

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Statement

$\mathcal{S}$, the set of all $\mu^*$ measurable sets, is a ring of sets.

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

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

Proof

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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.

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I've thought about it, we know:

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

And want to show:

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

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

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

• $\alpha\ :=\ A-(E\cup F)$
• $\beta\ :=\ (A\cap E)-F$
• $\gamma\ :=\ A\cap E\cap F$
• $\delta\ :=\ (A\cap F)-E$

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

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

Proof #1

• Let $E,F\in S$ be given
  • Let $A\in H$ be given. We wish to show $\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))$ or $\mu^*(A)=\mu^*(\beta\gamma\delta)+\mu^*(\alpha)$
    • Notice $\mu^*(\beta\gamma\delta)=\mu^*(\beta\gamma\delta\cap E)+\mu^*(\beta\gamma\delta-E)$
      • By tidying up the sets, we see this is: $\mu^*(\beta\gamma\delta)=\mu^*(\beta\gamma)+\mu^*(\delta)$
    • So $\mu^*(\beta\gamma\delta)=\mu^*(\beta\gamma)+\mu^*(\delta)$
    • Notice $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)$, or $\mu^*(A)=\mu^*(\beta\gamma)+\mu^*(\alpha\delta)$
      • Re-arranging this, we see $\mu^*(\beta\gamma)=\mu^*(A)-\mu^*(\alpha\delta)$
    • Substituting this in: $\mu^*(\beta\gamma\delta)=\mu^*(A)-\mu^*(\alpha\delta)+\mu^*(\delta)$
      • Notice we can use $F$ to split the $\alpha\delta$ into $\alpha\delta\cap F=\delta$ and $\alpha\delta-F=\alpha$
      • So $\mu^*(\alpha\delta)=\mu^*(\delta)+\mu^*(\alpha)$
    • Substituting this back in we see: $\mu^*(\beta\gamma\delta)=\mu^*(A)-\mu^*(\alpha)-\mu^*(\delta)+\mu^*(\delta)$
    • Simplifying: $\mu^*(\beta\gamma\delta)=\mu^*(A)-\mu^*(\alpha)$
    • Rearranging: $\mu^*(A)=\mu^*(\beta\gamma\delta)+\mu^*(\alpha)$
    • However notice:
      1. $\beta\gamma\delta=A\cap(E\cup F)$ and
      2. $\alpha=A-(E\cup F)$
    • So we have:
      • $\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))$
  • Since $A\in H$ was arbitrary we have: $\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))]$
    • So $E\cup F\in S$
• Since $E,F\in S$ were arbitrary, we have $\forall E,F\in S[E\cup F\in S]$
  • As a formula: $\forall E,F\in S\big[\forall A\in H[\mu^*(A)=\mu^*(A\cap(E\cup F))+\mu^*(A-(E\cup F))]\big]$ Caution:$\forall$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 $[\ ]$ however nothing is given so far - hence the brackets

This completes the proof.

Proof #2

Gist is the same, I did this on paper:

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

The result follows

See also

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