Finitely Additive Set Function Inequality Let $ \mu $ be a non negative, finitely additive set function on the field $ \mathcal F $ . If ${\{A_i\}}_{i \in \Bbb N} $ are disjoint sets in $ \mathcal F $ and $\bigcup_{i=1}^\infty A_i \in \mathcal F $, show that $ \mu (\bigcup_{i=1}^\infty A_i ) \geq  \sum_{i=1}^\infty \mu (A_i) $
Here is what I have done so far,
$ \bigcup_{i=1}^\infty A_i = \{ (\bigcup_{i=1}^\infty A_i) \setminus \bigcup_{i=1}^m A_i \} \bigcup \;(\bigcup_{i=1}^m A_i)   $ 
This is a finite disjoint union and, 
$ \{(\bigcup_{i=1}^\infty A_i) \setminus \bigcup_{i=1}^m A_i\} \in \mathcal F  $ and $ \bigcup_{i=1}^m A_i \; \in \mathcal F $ since $ \mathcal F $ is a field. 
Then the finitely additive property of $ \mu $ gives us that,
$ \mu (\bigcup_{i=1}^\infty A_i) = \mu \left((\bigcup_{i=1}^\infty A_i) \setminus \bigcup_{i=1}^m A_i \right) + \mu (\bigcup_{i=1}^m A_i) = \mu \left((\bigcup_{i=1}^\infty A_i) \setminus \bigcup_{i=1}^m A_i \right) + \sum_{i=1}^m \left( \mu( A_i) \right) \geq \sum_{i=1}^m \left( \mu( A_i) \right) $ 
Therefore,
$ \mu (\bigcup_{i=1}^\infty A_i) \geq  \sum_{i=1}^m \left( \mu( A_i) \right)  $
Since this is true for every $m \in \Bbb N $ am I allowed to say $ \mu (\bigcup_{i=1}^\infty A_i) \geq  \sum_{i=1}^\infty \left( \mu( A_i) \right)  $ ??
I have seen that done before like with the proof for Bessel's Inequality but my teacher made a point in class that just because something holds for every natural number doesn't mean its true for infinity.
 A: Your idea is essentially this: we know that for every $m$ we have $\bigcup_{i=1}^{m} A_i \subseteq \bigcup_{i=1}^\infty A_i$, so also (by monotonicity of $\mu$) $\mu(\bigcup_{i=1}^m A_i) \le \mu(\bigcup_{i=1}^\infty A_i)$. 
(You reprove the monotonicity ($A \subseteq B$ in the field, then $\mu(A) \le \mu(B)$) for this specific case, but you need not do that, if it was proved before: if $A \subseteq B$, both in $\mathcal{F}$, then $\mu(B) = \mu((B \setminus A) \cup A) = \mu(B \setminus A) + \mu(A) \ge \mu(A)$ as $\mu$ is non-negative and the union is disjoint.)
As the (finite) union is disjoint, for all $m$ we know, as you state correctly, that $a_m := \sum_{i=1}^m \mu(A_i) \le \mu(\bigcup_{i=1}^\infty A)$.
Now the $a_m$ are an increasing sequence of reals, bounded above by the fixed real number $\mu(\bigcup_{i=1}^\infty A_i)$, so their limit exist, and is by definition of convergent series equal to $\sum_{i=1}^\infty \mu(A_i)$ and this limit is still bounded above by $\mu(\bigcup_{i=1}^\infty A_i)$, essentially because all sets of the form $(-\infty, c]$ are closed in the reals.
I wrote it down a bit pedantically, perhaps, but yes, this argument is perfectly valid. The essential idea is that the finite sums etc. are all in the reals so you use all the standard facts for the reals there. You have proved an inequality of real numbers, using monotonicity of $\mu$, which you need not re-prove, from axioms for $\mu$ and then proceed from there. 
Added after comment: if $\mu$ can have two co-domains, without extra information, namely all positive reals $[0,+\infty)$ (no infinite values) or $[0,+\infty]$, in the extended reals, so $+\infty$ is also a value that is allowed (and sums are adapted in the logical way).
If the former is the case, $\mu(\bigcup_{i=1}^\infty A_i) < +\infty$ by definition, as we take the $\mu$ of a set in the field (!). If the latter, then the above argument also holds, and even if $\mu(\bigcup_{i=1}^\infty A_i) = +\infty$ we have nothing to prove, as all numbers are $\le +\infty$ and the statement becomes trivially true (no need to go through the proof): any series of $\mu(A_i)$ values is either some finite positive number of also $+\infty$, and both are $\le +\infty$..
