TITLE: Real analysis: collection of sets - sigma-algebra or not?
QUESTION [0 upvotes]: 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.

REPLY [1 votes]: 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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