TITLE: Simple content measure example (explanation of the sum)
QUESTION [1 upvotes]: Let $\Omega$ be countably infinite. Construct:
$$\mathcal{A} := \{A \subset \Omega: \#A < \infty \text{ or } \#A^c < \infty\}$$
then $\mathcal{A}$ is an algebra. Now define the set function $\mu$ on $\mathcal{A}$ by:
$$
    \mu(A)=\left\{
                \begin{array}{ll}
                  0, A \text{ is finite}\\
                  \infty, A^c \text{ is finite}
                \end{array}
              \right.
$$
For your convenience the definition of a content measure is:
$\mu$ is content on set $\mathcal{A}$ (has to be a semiring) if $\mu$ is additive: $\mu \left(\cup_{i=1}^n A_i \right) = \sum_{i=1}^n \mu \left( A_i \right)$, for any mutually disjoint selection of finitely many $A_1,A_2,...,A_n \in \mathcal{A}$
Now they say this:
$\mu \left(\cup_{\omega \in \Omega} \{ \omega \} \right) = \mu(\Omega) = \infty$, which makes sense to me because $\Omega$ is countable infinite, so the complement: $\emptyset$ is finite, and according to our definition our function would output infinity. But the following does not make sense to me: $\sum_{\omega \in \Omega} \mu \left(\{ \omega \} \right) = 0$, and  that is because according to the definition of the set $\mathcal{A}$ there will be some elements $A \in \mathcal{A}$ that are countable infinite. And so there will be at least one $\omega \in \Omega$ that will give us $\mu(\omega)=\infty$ and yet, according to their statement: $\sum_{\omega \in \Omega} \mu \left(\{ \omega \} \right) = 0$, which implies that there is no such element. Where am I wrong?

REPLY [2 votes]: By definition each $\mu(\{\omega\})=0$ and so $\sum_{\omega\in\Omega}\mu(\{\omega\})=0$. This means that $\mu$ is not countably additive.
I think what you get wrong is that you can not give any value to $\mu(\omega)$ since $\omega$ is not a set. You have to measure sets not elements.