A Sigma Algebra that Contains Intervals (a,infinity) Contains Every IntervalDate: 09/21/2010 at 11:03:55 From: saba Subject: segma- algebra (real analysis) Prove that if a sigma-algebra of the subset of real numbers contains all intervals of the form (a,infinity), then it contains all intervals. I know that the sigma-algebra of a set is a Lebesgue measurable set, which contains all intervals and all open sets. Now, if a sigma-algebra contains an interval of the form (a,infinity), then the complement of this interval also, by definition, is contained by the sigma-algebra. I know that every interval is measurable. And I know that the collection of measurable sets is a sigma-algebra. But if I start with the interval (a,infinity), how do I prove that other intervals are contained by the sigma-algebra? Date: 09/22/2010 at 21:11:23 From: Doctor Jordan Subject: Re: segma- algebra (real analysis) Hi Saba, As I understand it, you want a proof that if F is a sigma-algebra (the elements of which are subsets of R) that contains all intervals of the form (a,infinity), then F contains all intervals. So we want to show for any real numbers a < b that the interval (a,b) is an element of F. We are given that (a,infinity) and (b,infinity) are elements of F. The complement of (b,infinity) is (-infinity,b], and this is an element of F because F is a sigma-algebra. The intersection of (a,infinity) and (-infinity,b] is (a,b], and this is an element of F because F is a sigma-algebra. However, we want to show that (a,b) is an element of F, not that (a,b] is an element of F. At this point, I got stuck for a while. If we knew that [b,infinity) were an element of F, then its complement (-infinity,b) would be an element of F. The intersection of (a,infinity) and (-infinity,b) is (a,b), which is what we want to be an element of F. Therefore, to show that (a,b) is an element of F, it suffices to show that [b,infinity) is an element of F. Because F is a sigma-algebra, if A_1,A_2,... are elements of F, then their intersection is an element of F; namely, a sigma-algebra is closed under countable intersections. Let A_n = (b - 1/n,infinity) for n = 1,2,.... Each set A_n is an element of F, because it is an interval from some real number to infinity. Therefore the intersection of the A_n's is an element of F. The intersection of the A_n's is [b,infinity). Check this by showing that every element of the intersection of the A_n's is in [b,infinity), and that every element of [b,infinity) is in the intersection of the A_n's. Therefore [b,infinity) is an element of F, and hence (a,b) is an element of F. Thus if every interval of the form (a,infinity) is an element of F, then every interval is an element of F. - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ Date: 09/22/2010 at 22:17:21 From: saba Subject: Thank you (segma- algebra (real analysis)) Thank you very very very much. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/