Sorry, guys, this kind of question perhaps is too easy for you, but i really can't see where my mistake is.
Suppose we have a addtive set function $\mu$ defined on class of sets $\Sigma$ with range on the extended non-negative real values.
Immediately, we can conclude that, for every $E_i \in \Sigma$, $i=1,...,n$, $\mu(\cup_{i=1}^n E_i) \leq \mu(E_1) + ... + \mu(E_n)$
Now, suppose $\Sigma$ being a $ \sigma $-field and take a sequence $\{E_i\}_{i \geq 1}$. Why should i not conclude, by taking the limit over n in the above innequality, that $\mu(\cup_{i=1}^{\infty} E_i) \leq \sum_{i=1}^{\infty} \mu(E_i)$ ?? (i'm thinking in left term as a monotone sequence $(a_n)$ and right term as a monotone sequence $(b_n)$ also. this should implies subadditivity or not?!? )
This question worries me, because in attempt to prove $\sigma$-additivity, the reverse innequality is get just by monotonicity of the additive non-negative set function. Also, i know we usually suppose regularity by borel sets to prove subadditivity, so, i know there exists a mistake here, but, as i said, i really cannot see where it is.
Thank you a lot
Let $\Sigma$ be the $\sigma$-algebra $\mathscr{P} ( \mathbb{R} )$, and let $\mu \colon \Sigma \to \bar{\mathbb{R}}_+$ be defined by: \begin{align} \mu (A) = \begin{cases} 0 &\text{if } A \text{ finite } \\ +\infty &\text{otherwise} \end{cases} \end{align}
Given disjoint subsets $E_1, \ldots, E_n$ of $\mathbb{R}$:
either all of them are finite, in which case their union is also finite
or one of them is infinite, and then the union is infinite
In both cases, we have $$ \mu ( E_1 \cup \cdots \cup E_n ) = \mu ( E_1 ) + \cdots + \mu ( E_n ) $$
However, this cannot be extended to a countable number of subsets. Write $\mathbb{Q} = \cup_{r \in \mathbb{Q}}\ \{r\}$. Then $\mu (\mathbb{Q}) = +\infty$ since $\mathbb{Q}$ is infinite, but
$$ \sum_{r \in \mathbb{Q}} \mu (\{r\}) = \sum_{r \in \mathbb{Q}} 0 = 0 $$