Proof the additivity of a measure. 
Theorem : If $\{A_{1},A_{2},A_{3},....\}$ is  a  countable  disjoint  collection of  measurable sets, then  $$\mu \left(\bigcup_{i=1}^{\infty} A_{i}\right) =  \sum_{i=1}^{\infty} \mu (A_{i}).$$

Now  the  proof  goes  like  this:
Given  that  $\mu(S)=\int_{I} \chi_{S}$, let $T_{n} = \cup_{i=1}^{n} A_{i}$, $\chi_{n}=\chi_{T_{n}}$, $T=\cup_{i=1}^{\infty} A_{i}$ . Since $\mu$ is  finitely  additive  we  have  $$\mu (T_{n}) = \sum_{i=1}^{n} \mu (A_{i}),$$
$\mu(T_{n}) \le \mu (T_{n+1})$, and $\{\mu(T_{n})\}$ is a increasing sequence   bounded  above  by $\mu(T)$, so $\mu(T_n)$ is convergent since it is a sequence of real numbers. Now each $\chi_{n}$ is non-negative everywhere on $I$ and the  series  $\sum_{i=1}^{\infty} \int_{I} \chi_{n}$  converges (because the sequence of partial sums $\{\mu(T_{n})\}$  converges). Then  the  series  $\sum_{i=1}^{\infty} \chi_{n}$  converges  to, say, $\chi_{0}$ and $$\int_{I} \chi_{0} =\int_{I} \sum_{n=1}^{\infty} \chi_{n} = \sum_{n=1}^{\infty} \int_{I} \chi_{n}$$  by Levi's  theorem. Now the last summand equals $$\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\int_{I} \chi_{i}=\lim_{n\rightarrow\infty}\mu(T_{n}).$$
Now  if  only  I  could  show  $\chi_{0}=\chi_{T}$   it  would  be  proved  that $\mu(T)=\lim_{n\rightarrow\infty}\mu(T_{n})$ and the  statement  would  be  proved. Or  have  I  made  other  silly  choices  in  the  proofs  that  should  be  altered  to  make  it  work?
 A: There are at least three issues in you argument.


*

*Since, for all $n$, $T_{n} \subseteq T_{n+1}$, and $T=\cup_{i=1}^{\infty} A_{i}=\cup_{i=1}^{\infty} T_{i}$, it is easy to show that $\sum_{i=1}^{\infty} \chi_{i}\neq \chi_{T}$. In fact, for any $x\in T$,  $\sum_{i=1}^{\infty} \chi_{i}(x)=+\infty$.

*Levi's Theorem (Monotone convergence) does not hold if $\mu$ is just finitely additive (and not $\sigma$-aditive). So you can not apply Levi's Theorem. 

*You can not prove that $\mu$ is $\sigma$-aditive just form tha fact that  $\mu$ is finitely additive (WITHOUT any additional condition). Why? Because there are examples where $\mu$ is finitely additive, but not $\sigma$-aditive.
Remark: To solve issue 1, you may work with $\chi_{A_{n}}$, $\sum_{i=1}^n \chi_{A_i} = \chi_{T_n}$  and $\sum_{i=1}^{\infty} \chi_{A_i} = \chi_T$. HOWEVER, solving issue 1 is not enough to prove the result. 
Remark 2: Assuming that the Lebesgue Integral is built already and the μ is the lebesgue integral of the function χ, one way to prove the theorem is: 
Let  $\ $ $T_{n} = \cup_{i=1}^{n} A_{i}$ $\ $, $T=\cup_{i=1}^{\infty} A_{i}$ . Since $\{A_{1},A_{2},A_{3},....\}$ is  a  disjoint  sets collection of  measurable sets and $\mu$ is  finitely  additive  we  have  $$\mu (T_{n}) = \sum_{i=1}^{n} \mu (A_{i})$$
We have that $\chi_{T_n}=\sum_{i=1}^n\chi_{A_i}$ and $\chi_T=\sum_{i=1}^\infty\chi_{A_i}$. It easy to see that  $\chi_{T_n}$ are positive functions converging monotonically to $\chi_T$. So, applying Levi's  theorem, we have
$$ \mu \left ( \sum_{i=1}^\infty A_i \right )=\mu(T)=\int \chi_T = \lim_{n \to \infty} \int \chi_{T_n}= \lim_{n \to \infty} \mu(T_n) = \lim_{n \to \infty} \sum_{i=1}^{n} \mu (A_{i})$$
