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Lebesgue measure on sigma algebra, help ........... Which of the following are sigma algebras? reply with justification please.

  1. All subsets in rational numbers
  2. { {0},{1},{0,1} }in space {0,1}
  3. all intervals [x,y) x,y elements of [0,1] and all their unions in the space [0,1)
  4. all subsets of [0,1]
  5. all open subsets in real line(with usual metric)
  6. all finite subsets and all subsets with finite complement in rationals.

please help thank you.

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    @Berci What does your *no* refer to?2012-09-30

2 Answers 2

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  1. Not a sigma algebra because if we take any subset from the set of subsets of Q denoted by A, we find that the complement of the set we have taken is irrational which is not present in the set A. Hence not a sigma algebra.

  2. The null set is missing in the sets given. Note that for set theory, the null set is a subset of any set by convention whereas in case of sigma algebras, the null set is a requirement. Therefore for a set to be a sigma algebra, the null set must be there. It is not a convention but a requirement for a set to be a sigma algebra.

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  1. Revision: Null set is missing, so this is not a $\sigma$-algebra. Thanks, Arthur.

  2. The collection you describe generates the Borel Sets. Is there a Borel set that is not in this collection.

  3. Yes.

  4. Intersect $(-1/n, 1/n)$ and you have your answer.

  5. this is not a $\sigma$-algebra. Enumerate the rationals and let $A$ be all elements with even index and $B$ be all elements with odd index. This lies in the $\sigma$-algebra generated by the finite and cofinite subsets of the rationals, but it is neither finite nor cofinite.

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    Why? from the definition of sigma-algebra.2012-09-30