Notes:The set of all mu*-measurable sets is a ring

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Overview

In Measure Theory Halmos does something weird for section 11, theorem B. I have yet to "crack" what he's doing, and this is the point of this page.

Halmos' theorem:

Section 11 - Theorem B - page 46:

  • If μ:HˉR0 is an outer-measure on a hereditary sigma-ring, H and if S is the set of all μ-measurable sets then S is a sigma-ring.
  • Furthermore, if AH and if (En)n=1S is a sequence of pair-wise disjoint sets, and E:=n=1En then:
    • μ(AE)=n=1μ(AEn)

Context:

His proof

Step Comment
Notice we have:

μ(A(E1E2))=μ(AE1)+μ(AE2)

As A(E1E2)AE1 and A(E2E2)AE2 we see:
  • μ(A(E1E2))= μ((A(E1E2))E1=AE1) +μ((A(E1E2))E1=AE2) (as E1 is μ-measurable)

(This requires that they are disjoint for the subtraction side to be true)

We could just as well have used E2 to split.

It follows by induction that:
  • μ(An=1En)=n=1μ(AEn)

For every nN0

Agreed
Define:
  • Fn:=ni=1Ei
"Then it follows from theorem A" that:
  • μ(A)=μ(AFn)+μ(AFn)ni=1μ(AEi)+μ(AE)
First note that:
  • μ(AFn)=ni=1μ(AEi), this is equality. So:
    • μ(A)=ni=1μ(AEi)+μ(AFn)

Next note that AFnAFm for mn, so:

  • AEAFn, by the subadditive property of μ (of all outer-measures we see:
    • μ(AE)μ(AFn)

Thus:

  • μ(A)ni=1μ(AEi)+μ(AE)
"Since it is true for every n we obtain":
  • μ(A)n=1μ(AEi)+μ(AE)μ(AE)+μ(AE)
Witchcraft

Details