Difference in Sigma additivity and finite additivity A set function $\mu(A)$ is called a measure if 1. ... 2 ... 3. $\mu$ is additive in the sense that if $A$ is a set in $\mathscr{S}_{\mu}$ such that $$A = \bigcup_{k=1}^{n} A_{k}, $$ where $A_{1}, \cdots , A_{n}$ are pairwise disjoni sets in $\mathscr{S}_{\mu}$, then $\mu(A) = \sum_{k=1}^{n} \mu(A_{k})$
A measure $\mu$ with domain of definition $\mathscr{S}_{\mu}$ is said to be $\textbf{sigma-additive}$ if $$\mu(A) = \sum_{k=1}^{\infty} \mu(A_{n})$$ for all sets $A , A_{1}, \cdots , A_{n}, \cdots \in \mathscr{S}_{\mu} $ satisfying $$A = \bigcup_{n=1}^{\infty} A_{n}, \; A_{i} \cap A_{j} = \emptyset \; (i \neq j)$$

What is the difference in these two definitions additivity? The $\sigma$-additive definition has the sum going to infinity, but that seems to be about it?
 A: Don't know if it helps but perhaps an example may provide some intuition for the importance of $\sigma$-additivity? Suppose for example that you have a finite positive $\sigma$ additive measure $\mu$ on ${\Bbb R}$ then a very nice property is the following continuity property:
$$ \lim_{x\rightarrow 0^+} \mu( \ (0,x)\ ) = 0$$
The non-finite measure $\mu(A) = \int_A \frac{1}{|x|} dx$ does not verify this so why does it work for a finite measure?  When you start working with measures such continuity is often needed and for this you really need countably additivity. Although the statement uses a limit of $x$ being a real number it suffices to look at $x=1/n$ and note that $\sigma$-additivity implies:
$$  \mu( \ (0,1] \ ) = \sum_{n\geq 1} \mu ( \ ( \frac{1}{n+1},\frac{1}{n}  ] \ ) < +\infty$$
The mere fact that the sum is convergent implies that the 'tail'-sum $\mu ( (0,\frac{1}{n}] )$ must go to zero as $n\rightarrow \infty$. Finite additivity does not suffice to prove this.
A: Yes, sigma additivity differs from finite additivity only by the fact that infinite series instead of finite sums are allowed. 
However, this makes a large difference in practice and finitely additive measures may be stranger than you may think at first glance. For example there exists a finitely additive, translation invariant measure on the integers such that the set of all integers has measure 1, i. e.  something like a finitely additive translation invariant probability measure. Note that all singletons will have zero measure which implies that all finite sets also have zero measure. Sets with measure in between zero and one have some "positive density", e. g.  the set of all even numbers would have measure 1/2...
