
#1
Feb2713, 07:30 PM

P: 391

Hi, All:
I think the following deals with continuity of measure, but I'm not 100%: Let I:=[0,1] , and let A_{n} be a sequence of pairwisedisjoint measurable sets whose union is I ( is me? :) ) . Let {B_{j}} be a sequence of measurable subsets of I , so that, for μ the standard Lebesgue measure: Lim_{j→∞} μ( A_{n}\cap B_{j} )=0 , for all n . I want to show that above implies that : Lim_{j→∞} μ(B_{j})=0 .(**) This is what I have: We know that Ʃμ(A_{n})=1 . So we must have some A_{no} in the collection with μ(A_{no})=a>0. ( I am assuming that the A_n's must all be of the form [a,b) , with A_{1}=[0,a) A_{2}=[a,b) , etc. , plus a {1} thrownin ) 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=ce+e< μ(B_{j})< c+e+e In particular, μ(B_{j})>c>0 . Now, I can find an open set O_{j}, for each j , with μ(C_{j})=μ(O_{j}) . I know the quantification here is tricky; I am then using that: O_{j}= \/(c_{j}_{i} ,d_{j}_{i}) And, since m(B_{j})>c for all j>N , there is an index for the j's  use j=1 without loss of generality  such that m(c_{1},d_{1})>0 Now, this interval (c_{1},d_{1}) must intersect some interval A_{n} , 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 yx>0 , contradicting the assumption condition (**) that Lim_{j→∞} μ(B_{j})=0. I think I'm on the right track, but not 100%. Please critique. Thanks. 



#2
Feb2713, 10:22 PM

P: 3,173

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.




#3
Mar913, 09:13 AM

Sci Advisor
P: 1,168

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. 


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