If SS is an infinite σ\sigma algebra on XX then SS is not countable

I am going over a tutorial in my real analysis course. There is
an proof in which I don’t understand some parts of it.

The proof relates to the following proposition:

(S – infinite σ-algebra on X) S is uncountable.

Proof:

Assume: S={Ai}+i=1. xX:Bx:=xAiAi. [Note: BxS (Bx – countable intersection].

Lemma: BxByBx=By.

Proof(of lemma):

zBxByBzBxBy.

1.xBzxBxBzBxBzSBxBzBx
(contradiction:  definition of Bx) Bz=Bx

2.yBzyByBz  ByBzS  ByBzBy(contradiction: definition of By) Bz=By Bx=By 

Consider: {Bx}xX. If: there are finite sets of the
form Bx then: S is a union of a finite number of disjoint sets
S is finite there is an infinite number of sets of
the form Bx. |iANBxi|0.(contradiction)

There are couple of things I don’t understand in this proof:

  1. Why the fact that we found a set (BxBz) in S
    containing x and is strictly contained in Bx a contradiction
    ?

  2. Why if there are only a finite number of different sets of the
    form Bx then S is a union of a finite number of disjoint
    sets and is finite ?

Answer

  1. Because Bx is supposed to be the intersection of all measurable sets containing x, but you’ve found a measurable set containing x strictly inside Bx.

  2. Because for any measurable set T, we have T=xTBx. Thus, if there are n distinct sets of the form Bx, then there are at most 2n elements of S.

Attribution
Source : Link , Question Author : Belgi , Answer Author : Zev Chonoles

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