Measure and additivity I am learning Measure Theory on my own, so please forgive me if my question is naive.
Finitely additive and $\sigma$-additive measures are defined in a natural way on finite algebras and $\sigma$-algebras, respectively. 
I know that every $\sigma$-additive function is additive, and that every $\sigma$-algebra is an algebra. Hence, we can safely define a $\sigma$-additive measure on a finite algebra. More, the finite measure and the $\sigma$-additive measure coincide. 
I can see two ways of doing it:
1) Add infinitely many empty sets to satisfy the requirement of denumerability.
For example, $$\mu(E\cup F)=\mu(E\cup F\cup\emptyset\cup\emptyset\cup\ ...)=\mu E+\mu F+\mu 0+\mu 0+\ ...=\mu E+\mu F$$
2) Simply restrict the the $\sigma$-additive measure $\mu$ to the finite algebra $\mathcal{A}$. Say, $\nu=\mu{\left|\mathcal{A}\right.}$.
My question is the following: Can we define a finitely additive measure on a $\sigma$-algebra? 
Maybe, partitioning our $\sigma$-algebra into finitely many equivalence classes could be a good start. But, still, at least one cell will be infinite... So the question is still pending, and I was wondering if somebody could help me to understand that point. Thanks.
 A: Fact. A σ-additive measure(which also is a finitely additive measure) always exists on an arbitrary $\sigma$-algebra  $\cal{F}$ of subsets of $\Omega(\Omega \neq \emptyset)$.
Proof. Let fix an element  $x \in \Omega$. For each $Y \in \cal{F}$  we set $\mu(Y)=1$ if $x \in Y$ and 
$\mu(Y)=0$ otherwise. Then $\mu$ is  σ-additive measure provided that if $(A_n)$ is sequence of pairwise disjoint measurable subsets then $\mu(\cup_{n \in N}A_n)=\sum_{n \in N}\mu(A_n)$.
Indeed, here are possible only two cases:
Case 1.$x \in  \cup_{n \in N}A_n$
Then there will be $n_0$ such that $x \in A_{n_0}$ which by the definition of $\mu$, means that $\mu(\cup_{n \in N}A_n)=1$, $\mu(A_{n_0})=1$ and  $\mu(A_{n})=0$ for each $n \neq n_0$. Hence we have
$$
 \mu(\cup_{n \in N}A_n)=1=1+0=\mu(A_{n_0})+\sum_{n \in N \setminus \{n_0\}}\mu(A_n)=\sum_{n \in N}\mu(A_n).
$$
Case 2.$x \notin  \cup_{n \in N}A_n$
Then by the definition of $\mu$, we have that $\mu(\cup_{n \in N}A_n)=0$  and  $\mu(A_{n})=0$ for each $n \in N$. Hence the validity of the equality 
$$
 \mu(\cup_{n \in N}A_n)=0=\sum_{n \in N}\mu(A_n)
$$
is obvious.
