Let $(X, \mathcal{M})$ be a measurable space and $\mu : \mathcal{M} → [0, \infty]$. Show that $\mu$ is a measure. Let $(X, \mathcal{M})$ be a measurable space and $\mu : \mathcal{M} → [0, \infty]$ such that
(a) $µ (A) <\infty$ for some $A \in \mathcal{M}$
(b) $µ(A \cup B) = \mu(A) + \mu(B)$ if $A, B \in \mathcal{M}$ and $A \cap B = \emptyset$, and
(c) $\mu (\bigcup_{n=1}^{\infty} A_{n}) \leq \sum_{n=1}^{\infty}\mu(A_n) $ if $A_n \in \mathcal{M}$ for $n=1,2,3,...$
Show that $\mu$ is a measure.
In my context, we define a measure as follows:
Definition:
A function µ is called a positive measure, defined in a $\sigma$-algebra $\mathcal{M}$ with values in $[0, \infty]$ and which is countably additive. This means that if $\{E_i\}$ is a disjoint countable collection of elements of $\mathcal{M}$, then $\mu (\bigcup_{i=1}^{\infty} E_{i}) = \sum_{i=1}^{\infty}\mu(E_i)$. 
To avoid trivial cases, we will further assume that $\mu (A) < \infty $ for at least one $A \in \mathcal{M}$
My attempt:
From the definition, it only remains for me to prove that $\mu$ is countably additive. To do this, let $\{A_n\}$ be a disjoint countable collection of elements of $\mathcal{M}$. Then, $$ \bigcup_{n=1}^{\infty}A_n  \supset \bigcup_{n=1}^{N}A_n$$ And so
$$\mu \left(\bigcup_{n=1}^{\infty}A_n \right)  \geq \mu \left(\bigcup_{n=1}^{N}A_n \right) = \sum_{n=1}^{N}\mu(A_n)$$
Since the left side of this inequality is independent of $N$, we have $$\mu \left(\bigcup_{n=1}^{\infty}A_n \right)  \geq \sum_{n=1}^{\infty}\mu(A_n)$$
For part (c) we have the equality.
First of all I want to know if there are any errors in my reasoning and, on the other hand, I don't see where I used the hypothesis of part (b) in my proof; This really does have me worried. I appreciate any help you can give me.
 A: What you have done seems correct. In fact, what you have done shows that given any $\sigma$-algebra $\mathcal{M}$ and any map $\mu : \mathcal{M} \to [0, \infty]$ with the property (b) alone, it follows that given a disjoint collection $\{A_i\}_i$, it is the case that $$\sum_{i = 1}^\infty \mu(A_i) \le \mu\left(\bigcup_{i = 1}^\infty A_i\right).$$
(As pointed in the comments, you used (b) to conclude that $$\mu \left(\bigcup_{n=1}^{N}A_n \right) \color{red}{=} \sum_{n=1}^{N}\mu(A_n),$$
which really follows by induction and (b).)
Property (c) was then just the minimum required for you to conclude equality.
As an exercise, you could try to come up with an example to show that (c) is really needed and doesn't just follow from (b).

I have ignored (a) in the above. Clearly, it is only there to avoid the case that $\mu$ is identically $\infty$, which is what your definition also demands. However, note that it is equivalent to demanding that $\mu(\varnothing) = 0$, assuming either (b) or countable additivity. (How?)
