Notes:The set of all mu*-measurable sets is a ring
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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∗:H→ˉR≥0 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 A∈H and if (En)∞n=1⊆S is a sequence of pair-wise disjoint sets, and E:=⋃∞n=1En then:
- μ∗(A∩E)=∞∑n=1μ∗(A∩En)
Context:
- The previous theorem proved was that S is a ring of sets.
- I have proved that μ∗|S - the restriction (function) of μ∗ to S - is a pre-measure.
His proof
Step | Comment |
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Notice we have:
μ∗(A∩(E1∪E2))=μ∗(A∩E1)+μ∗(A∩E2) |
As A∩(E1∪E2)⊆A∩E1 and A∩(E2∪E2)⊆A∩E2 we see:
(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:
For every n∈N≥0 |
Agreed |
Define:
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"Then it follows from theorem A" that:
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First note that:
Next note that A−Fn⊇A−Fm for m≥n, so:
Thus:
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"Since it is true for every n we obtain":
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Witchcraft |