Why sigma algebra and not other systems of sets? I got a short question. Why we are considering sigma-algebras and not other systems of sets in measure-theory? Why not dynkin-system for example?
 A: The Dynkin-system is poor for the purpose of measure theory. In measure theory (probability theory) calculating the measure of a sets over which a sequence of functions converges is important.
Let, for instance, $\{f_n\}$ is a sequence of measurable functions. Then te set
$$\{x: \lim_{n \rightarrow \infty}f_n(x)=f(x)\}$$ cannot be expressed in general in a Dynkin system.
Why not? Because
$$\{x: \lim_{n \rightarrow \infty}f_n(x)=f(x)\}=\bigcap_{n=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m=N}^{\infty}\{x:|f_m(x)-f(x)|\le\frac{1}{n}\}.$$ 
Such a set cannot be expressed in a Dynkin system unless it contains at least the finite intersections of its sets.
Saying that a Dynkin system is too special deserves a down vote!
A: For instance, you want "measurable functions" to be a ring, hence you want $1_A\cdot1_B=1_{A\cap B}$ to be "measurable" if $A,B$ are "measurable". But this means that you want "measurable sets" to be closed under finite intersection.
But a Dynkin system closed under finite intersection is a $\sigma$-algebra.
Added
The fact is that you first want to define what "being measurable set" means, and then you want to say that a function $\Omega\rightarrow\mathbb{R}$ is "measurable" if the counterimage of open sets is "measurable". Which implies that a characteristic function $1_A$ is "measurable" if and only if $A$ is "measurable". Now the point is that the more you want (both algebraically and analytically) on the set of "measurable function", the more you need on the "measurable sets".
There exists, anyways, a theory of non-$\sigma$-algebras, corresponding to finitely additive measures, which (I've been told) have some use in number theory.
