For what it's worth: I find Rudin's book very good; the exposition is beautiful and I often look up stuff in it. So I use it mainly as a reference. I don't know how good it serves as a "beginner's book". I haven't done any of the exercises, but scanning over them I see there are some darn hard ones.
Anyway, I always like to use several sources at the same time when learning a subject. As such I would recommend Rudin to be one of them, but probably not if it's your only source.
The following exercise is fun, but admittedly it took me quite some time; more than a couple of minutes!
lavinia said:
First some easy ones.
- Show that any infinite sigma algebra is uncountable.
I once gave this one to a Ph.D. in probability theory and he was unaware of it - my point about this book - although he solved it in a couple of minutes.
More to come.
Let \Sigma be a sigma-algebra on a set X. Define the following relation on X:
x\sim y:\Leftrightarrow (\forall A\in\Sigma: x\in A\Leftrightarrow y\in A).
This is obviously an equivalence relation, basically because "iff" is. Hence we get a partition P=\{[x_i]\ |\ i\in I\} of X. We have a nice decription of the equivalence classes, namely
[x]=\bigcap \{A\in\Sigma\ |\ x\in A\},
i.e. it is the smallest set contained in all things in our sigma-algebra that contain x. Indeed, because a sigma-algebra is closed under complements, we have:
(y is in that intersection) iff (for all A\in\Sigma with x\in A we have y\in A) iff (for all B\in\Sigma with x\notin B we have y\notin B).
So indeed
(y is in that intersection) iff (y ~ x).
Now suppose that \Sigma is countable; we will show it must be finite. This means each [x] is in fact a
countable intersection, hence [x]\in\Sigma. So we get the map I\to \Sigma given by i\mapsto [x_i]; it is injective by definition of I (it indexes the partition). Hence I is also countable. But then we get a map
2^I\to \Sigma
J\mapsto \bigcup_{i\in J}[x_i]
(because such a union is countable by our previous remark, hence in the sigma-algebra). Again by definition of I it is injective. It is also surjective; in fact the inverse is given by
\{i\in I\ |\ b\in[x_i]\text{ for some }b\in B\}\mapsfrom B.
Hence 2^{|I|}=|\Sigma|. If I were infinite then \Sigma would be uncountable. Hence I is finite and consequently \Sigma is finite.