I've been told the properties of a $\sigma$-algebra $\mathcal{F}$ over some set $\Omega$ are:
- $\Omega \in \mathcal{F}$
- $\mathcal{A} \in \mathcal{F} \implies \mathcal{A}^c \in \mathcal{F}$
- $\mathcal{A}_1, \mathcal{A}_2, \mathcal{A}_3, \ldots \in \mathcal{F} \implies (\bigcup_{i=1}^{\infty} \mathcal{A}_i) \in \mathcal{F}$, i.e., countable unions of members of $\mathcal{F}$ are also in $\mathcal{F}$.
My questions are:
- Why do we have to specify "countable" unions? Obviously this means that finite unions of members are also in the $\sigma$-algebra, as well as countably infinite unions; but it seems meant to contrast uncountably infinite unions. If countably infinite unions doesn't get you outside the set, how would uncountably infinite unions do it? What does it really mean to take uncountably infinite unions? In case it doesn't really make sense to do so, why do we specify countable? In the case that it does, can you give an example of taking uncountable unions of elements of a $\sigma$-algebra where the result is outside the $\sigma$-algebra?
- Is the only difference between an algebra and a $\sigma$-algebra over $\Omega$ the "countable" part? That is, an algebra is closed over finite unions?
Best Answer
For (1), remember that our goal is to define a measure, which will be a function $\mu : \mathcal{F} \to [0,\infty]$ satisfying
These axioms are supposed to encode the idea the idea of "area" ("measure") of a set $A$. So the area of the empty set is $0$ and the area of a bunch of separate pieces should be the sum of the areas of those pieces.
Now the countability requirement for $\sigma$-algebras comes directly from the countable sums in the definition of a measure $\mu$! So we should ask: why do we restrict ourselves to countable unions in the definition of a measure?
Well, let's look at the usual notion of area in the plane. This is formalized by the notion of lebesgue measure but an informal understanding will be enough for this discussion. It should be believable that the area of any single point should be $0$. That is, $\mu(\{x\}) = 0$ for any point $x$. But we also know that the unit square $[0,1] \times [0,1]$ should have area $1 \times 1 = 1$! So far so good.
Now for the potential contradiction. If we allowed arbitrary sums in definition of $\mu$ (and thus if we required arbitrary unions in our definition of $\sigma$-algebra), we would be able to write the unit square as a disjoint union of all its points: $[0,1] \times [0,1] = \bigcup_{x \in [0,1] \times [0,1]} \{ x \}$! Then our axiom would say that
$$1 = \mu([0,1] \times [0,1]) = \mu \left ( \bigcup_{x \in [0,1] \times [0,1]} \{ x \} \right ) = \sum_{x \in [0,1] \times [0,1]} \mu(\{x \}) = \sum_{x \in [0,1] \times [0,1]} 0 = 0$$
So our measure can't possibly play nicely with arbitrary unions! It turns out that, since we usually work with (countable) sequences of numbers, functions, regions, etc. that countable unions are enough (and don't cause this issue!), and so we work with only countable unions from the start.
As a brief aside, you might ask what happens if there's an infinity bigger than countable, but smaller than $\Big | [0,1] \times [0,1] \Big |$ (that is, if the continuum hypothesis is false). It turns out that in this case unions of "small uncountable" families is a fairly subtle set theoretic question, which is another reason to not introduce it in a class that doesn't assume some knowledge of axiomatic set theory.
As for (2): yes. Notice that any finite union $A_1 \cup A_2 \cup \cdots \cup A_n$ is also a countable union $A_1 \cup A_2 \cup \cdots \cup A_n \cup \emptyset \cup \emptyset \cup \cdots$, so we can always recover our finite operations from our countable ones. In particular, every $\sigma$-algebra is an algebra, and the only ~bonus power~ that a $\sigma$-algebra has is the countable unions.
I hope this helps ^_^