Why countable unions, intersections etc.? I was just wondering why one always insists on countability when it comes to the definition of a $\sigma$-algebra in measure theory. I mean, measure theory works as it does, but is there a deeper reason why e.g. it is reasonable to demand that the countable union of measurable sets is measurable again? Why not an uncountable union? Thanks in advance!
 A: Another reason is that in a summable family $(x_i)_{i\in I}$, it can be shown the set $\;\{i\in I\mid x_i\neq 0\}$  is at most countable. Thus the theory of summable families comes down to the theory of series.
A: Countable has this very peculiar property of infinity.
It can be indexed, such that every point is only a finite distance away. Namely, it can be indexed using $\Bbb N$. And $\Bbb N$ has that great property that given any $k\in\Bbb N$, you only need to travel finitely many steps to meet with $k$.
Why is that important? Because (unfortunately) mathematics and mathematicians take their intuition from the real world. This means that you develop probability and measure from finite experiments. Often finite experiments have "unimportant residue", which is to say that the more times you run something, the closer that residue gets to $0$. So when you take a limit, you get this squeaky clean result.
Well... sometimes, anyway.
But this is where countable unions come from. First we cared about running an experiment $n$ times, then we also noticed that when $n\to\infty$ things look nicer. But what does it mean $n\to\infty$ when running an experiment? It means that we have a countable sequence of events, and we want to know whether the limit of those events is an events. We sure want it to be. So why not make it? And so $\sigma$-algebras are born.
(All that being said, there are uses - interesting uses - to algebras which have closure of more than just countable unions. But that usually gets under set theory, rather than measure theory or probability.)
A: If you insist on uncountable unions, then the measure of the unit interval is the sum of uncountably many measures of the individual points of the interval. These measures are either 0 (in which case $\mu(I) = 0$, which is bad) or nonzero, in which case $\mu(I)$ would be $\infty$,. 
A possibly more interesting question is whether finitely-additive measures would be interesting. Bumby and Ellentuck wrote an interesting (to me) paper on the topic, although perhaps part of the interest was that it was the first "real math research paper" I ever read. :)
A: I hope this works (check this): Counterexample why it is a bad idea to consider uncountable unions:
Take $\lambda \in [0,1] \setminus \mathbb{Q}$ and define $A_\lambda := \{\lambda + r \,|\,r\in\mathbb{Q}\} \cap [0,1]$. Then the sets $A_\lambda$ are
pairwise disjoint.
Now we have
$$\mu(\bigcup_{\lambda \in [0,1] \setminus \mathbb{Q}} A_\lambda) = \mu([0,1]) \neq \sum_{\lambda \in [0,1] \setminus \mathbb{Q}} \mu(A_\lambda) = 0.$$
A: It's quite simple: Allowing only finite union yields too weak results [cf. Jordan measure] and allowing uncountable unions makes the system of measurable sets too big. If you for example assume that all singletons are measurable, then uncountable unions would imply that any set is measurable. This already doesn't work well with the Lebesgue measure, as the example of Vitali sets show.
