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Let $X$ be a set of finite (positive) measure. Let $C$ be the collection of all finite subsets of $X$ and their complements in $X$. Is $C$ an algebra of sets? Is $C$ a $\sigma$-algebra? Explain.

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Ancient Proverb: OP gives orders, OP needs no answers. –  Did Mar 24 '12 at 15:31
    
What measure is defined on $X$? –  Davide Giraudo Mar 24 '12 at 15:31
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For algebra of sets, look at the definition. It will be easy for you to verify that the conditions of the definition are satisfied. For $\sigma$-algebra, the answer in general is no. Let $X$ be the positive integers. The measure is irrelevant, but if you want one, put mass $1/2$ at $1$, mass $1/4$ at $2$, and so on. Now is $C$ closed under countable union? –  André Nicolas Mar 24 '12 at 15:40

2 Answers 2

To answer a question of this kind, we need to know what the terms used in the question mean. Often, that's all we need to know!

What is an algebra of subsets of $X$? The details of the definition differ from book to book. I will use one of the standard definitions. You may have to adapt the answer to your local definition.

An algebra of subsets of $X$ is a non-empty collection $\mathcal{A}$ of subsets of of $X$ which is closed under the usual Boolean operations. Specifically, (i) if $S\in \mathcal{A}$, then the complement of $S$ is in $\mathcal{A}$; (ii) If $S$ and $T$ are in $\mathcal{A}$, then $S\cup T$ and $S\cap T$ are in $\mathcal{A}$.

So first we check whether if $S\in C$, then the complement of $S$ is in $C$. This is built in. If $S$ is finite, then the complement of $S$ is in $C$. If $S$ is a set whose complement is finite, then again the complement of $S$, a finite set, is in $C$ by the definition of $C$.

Now we check that if $S$ and $T$ are in $C$, then the union and intersection of $S$ and $T$ are in $C$. There are several cases. Maybe $S$ and $T$ are finite. Then their union is finite, so is in $C$. Maybe one of the sets (say $S$) is finite and the complement of $T$ is finite. Then since $T\subseteq S\cup T$, the complement of $S\cup T$ is finite, so $S\cup T \in C$. Or maybe the complement of each of $S$ and $T$ are each finite. Then again the complement of $S\cup T$ is finite.

Now we should deal with $S\cap T$. The work is more or less mechanical, like the work for $S\cup T$.


Next we deal with the $\sigma$-algebra question. Here the answer will be "not necessarily." (Indeed the answer is "no iff $X$ is infinite.") But to answer the question it is enough to come up with one example where $C$ is not a $\sigma$-algebra.

Let $X$ be the set of positive integers. The measure is irrelevant, but if you want one, put mass $\frac{1}{2^n}$ at the point $n$. Now let $E$ be the set of even integers. Then $E$ is not finite, and its complement is not finite. So $E$ is not in $C$. But $E$ is a countable union of finite (one-point) sets. So $C$ is not closed under the operation of countable union, and therefore is not a $\sigma$-algebra.

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What alternative definitions of algebra of sets are there? At least in the context of measure theory, all definitions I know are equivalent. –  Michael Greinecker Apr 15 '12 at 18:13
    
To a beginning student, equivalent is not enough, since the student has to answer the question using the local definition. –  André Nicolas Apr 15 '12 at 18:21
    
Thank you, I was just curious whether there is something should learn. –  Michael Greinecker Apr 15 '12 at 18:31

A $\sigma$-algebra has the properties that it contains the empty set, is closed with respect to countable union and is closed with respect to taking complements.

(i) You have $\varnothing \in C$ since $\mu (\varnothing) = 0$ is finite.

(ii) $C$ is closed with respect to taking complements by definition.

So two out of three properties of a sigma algebra are satisfied.

(iii) You want $C$ to be closed with respect to countable (infinite) union. Here's a counter example: Let $X = \mathbb Z$. Pick $A_i = \{ 2i \}$. Then $\bigcup_i A_i$ is infinite and so is its complement hence $C$ is not closed with respect to countable union.

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By induction C is closed with respect to countable union is probably a joke. –  Did Mar 24 '12 at 15:45
    
@DidierPiau Yes you're right. Thanks for the downvote. –  Rudy the Reindeer Mar 24 '12 at 15:49
    
Re-reading this answer, I fail to see why any measure should be mentioned at all: as explained by Davide and André in the comments, the question is not about measures but only about the structure of a collection of subsets of X. In particular, the whole paragraph (iii) of this answer is a mystery to me. –  Did Apr 15 '12 at 18:32
    
@Didier Yes, right again. –  Rudy the Reindeer Apr 15 '12 at 19:07

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