# Lebesgue measure sigma algebra

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.

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In each case, you must specify what the entire space is. You did this in #2; this is a sigma algebra. This leaves an ambiguity in #1. –  ncmathsadist Sep 30 '12 at 17:24
Welcome to M.SE! You will receive better answers if you provide context about where you have encountered a problem, what your background is and what you have tried so far. As to your question, it has little to do with Lebesgue measure. A $\sigma$-algebra has to contain $\emptyset$ and the whole space- it has to contain the complement of every set it contains. And it has to contain the union of every sequence of sets it contains. Try to check this in each case. –  Michael Greinecker Sep 30 '12 at 17:24
No, a $\sigma$-algebra contains $\emptyset$. –  Berci Sep 30 '12 at 17:25
@Berci What does your no refer to? –  Michael Greinecker Sep 30 '12 at 17:27

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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The mathjax program automatically numbers and it renumbered my replies incorrectly. I have fixed that. –  ncmathsadist Sep 30 '12 at 17:41
Are you certain about 2? –  Arthur Fischer Sep 30 '12 at 17:44
Yes. This is the power set of $\{0,1\}$ in the space $\{0,1\}$. The power set on any set is a $\sigma$-algebra on that set. –  ncmathsadist Sep 30 '12 at 18:26
Note that the empty set is not in the collection under consideration. –  Arthur Fischer Sep 30 '12 at 18:28
Yes Arthur, normally it has to contain null set in the sigma algebra set. –  pinky Sep 30 '12 at 18:44