measure theory question?

Question

How to prove the following claim from Wikipedia?
http://en.wikipedia.org/wiki/Atom_(measure_theory)

"One can prove that if μ is a non-atomic measure and A is a measurable set with μ(A) > 0, then for any real number b satisfying μ(A) > b > 0
there exists a measurable subset B of A such that μ(B) = b"

Answer 1

@ ampian:

Corben's proof is correct-- adding on sets of arbitrary but tiny measure in such a cumulative manner works. The constructed union has measure <=b by construction. To see that it is b, note that you can find a finite union of the B_k that has measure arbitrarily smaller than b. Since the measure of the countable union is larger than that of any of the finite unions, it has measure b.

Answer 2

Follow the definition of atom:

Formally, given a measure space (X,Σ) and a finite measure μ on that space, a set A in Σ is called an atom if
μ (A) >0
and for any measurable subset B of A with
μ(A) > μ (B)
one has μ(B) = 0.

To prove:
" if μ is a non-atomic measure and A is a measurable set with μ(A) > 0, then for any real number b satisfying μ(A) > b >
there exists a measurable subset B of A such that μ(B) = b"

Proof:
since μ is a non-atomic measure, a set A in it is not an atom, so with μ(A) > 0, the opposite of being an atom happens, namely: for any b>0 ,here exists measurable subset B of A with
μ(A) > μ (B) =b > 0

Answer 3

I don't agree with the proof proposed above. The negation of being an atom is There exists B such as μ(B)<μ(A).

But since B (that I prefer to call A1) exists, you can construct An, and μ(An)->0.

Now there is k such as μ(Ak) is just below b.

Lets call Ak= B0.

Now we will repeat this construction but constructing a new sequence which terms will be choosen in A\B0.

And you choose the first term Bk' such as μ(Bk')
You call Bk' = B1 and you do it again choosing the elements in A\{B0 U B1}...

Finally the set U(n=0 to infinity) Bn =B.
This is a measurable set (by axioms on measurables sets.)

And you have μ(B)=b.

Answer 4

It seems the proof by the second author is still having some problems. Of course, you constructed a sequence of subsets with increasing measure, also the sequence is bounded above by b. However, it is only enough to show the limit exists but not enough to show the limit is exactly b.

Answer 5

degree theory undergirds effortless branch, that is an financial/pastime theoretic subject remember. effortless branch incorporates dividing a source (regularly time-venerated as a "cake") between countless events with distinctive fee structures. the linked fee structures are represented by way of countably additive, non-adverse danger measures, defined on the "cake" (that is, formally, a series). countless effortless branch methods have distinctive standards on the measures-- some require, as an occasion, that the measures be non-atomic, or that the measures be truthfully non-provide up with understand to a minimum of one yet another or with understand to Lebesgue degree.

Answer 6

i don't know but i want to mention that vlee proof is wrong.