Continuity of Measure ( I Think)

  • Context: Graduate 
  • Thread starter Thread starter WWGD
  • Start date Start date
  • Tags Tags
    Continuity Measure
Click For Summary
SUMMARY

This discussion focuses on the continuity of measure, specifically analyzing the implications of a sequence of pairwise-disjoint measurable sets An whose union is the interval I = [0,1]. The user aims to demonstrate that if Limj→∞ μ(An ∩ Bj) = 0 for all n, then Limj→∞ μ(Bj) must also equal 0. The argument involves contradiction, leveraging the properties of Lebesgue measure and the structure of measurable sets. The conclusion is that the assumption of a non-zero limit leads to a contradiction, affirming the original claim.

PREREQUISITES
  • Understanding of Lebesgue measure and its properties
  • Familiarity with measurable sets and their intersections
  • Knowledge of limits and convergence in mathematical analysis
  • Ability to work with sequences of sets and their measures
NEXT STEPS
  • Study the properties of Lebesgue measure in detail
  • Learn about the concept of measure zero and its implications
  • Explore the proof techniques used in measure theory, particularly contradiction
  • Investigate the relationship between measurable sets and their limits
USEFUL FOR

Mathematicians, students of analysis, and anyone studying measure theory who seeks to deepen their understanding of continuity of measure and its applications in real analysis.

WWGD
Science Advisor
Homework Helper
Messages
7,806
Reaction score
13,120
Hi, All:

I think the following deals with continuity of measure, but I'm not 100%:

Let I:=[0,1] , and let An be a sequence of pairwise-disjoint measurable sets
whose union is I ( is me? :) ) . Let {Bj} be a sequence of measurable subsets
of I , so that, for μ the standard Lebesgue measure:

Limj→∞ μ( An\cap Bj )=0 , for all n .

I want to show that above implies that : Limj→∞ μ(Bj)=0 .(**)

This is what I have:

We know that Ʃμ(An)=1 . So we must have some Ano in the
collection with μ(Ano)=a>0.

( I am assuming that the A_n's must all be of the form [a,b) , with A1=[0,a)

A2=[a,b) , etc. , plus a {1} thrown-in )

Now, I am trying to argue by contradiction , assuming that the limit above in (**) equals
some c+e ; e->0 , though I am not sure of how to show that the limit actually exists,
tho I am assuming for now that it does:

So, assuming limit in (**) exists and equals c+e ( e->0) , we have that there is an
integer N such that for all j>N :

c=c-e+e< μ(Bj)< c+e+e

In particular, μ(Bj)>c>0 .

Now, I can find an open set Oj, for each j , with

μ(Cj)=μ(Oj) .

I know the quantification here is tricky; I am then using that: Oj= \/(cji ,dji)

And, since m(Bj)>c for all j>N , there is an index for the j's --

use j=1 without loss of generality -- such that m(c1,d1)>0

Now, this interval (c1,d1) must intersect some interval

An , and the intersection must be of one of the forms:

[x,y) , (x,y] , or (x,y) . In either case, the measure of the intersection is

y-x>0 , contradicting the assumption condition (**) that

Limj→∞ μ(Bj)=0.

I think I'm on the right track, but not 100%. Please critique.

Thanks.
 
Physics news on Phys.org
Can you translate your question to Latex or PDF, cause it's hard to read through ascii. I am not 20 anymore that I have the patience to read that way.
 
Use the fact that the tail of the sum m(Ai) goes to zero, since the total sum is 1

Then use the fact that, by the limit condition, there is a K>0 with

Lim_n->0 (Bk /\An)=0 , for all k>K . Then , from the fact that m(Ai)->0 ,

Use Bk=(Bk /\ U Ai) , to conclude that m(Bk)->0.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K