# Show that countable additivity holds(complex measures).

I have an exercise where I have showed that for a particular secuence of complex measures $\{\nu_n\}_n$, we have that $\lim_{n \rightarrow \infty}\nu_n(A)$ converges for all measureable sets A in our sigma-algebra.

I need to show that $\nu(A)=\lim_{n \rightarrow \infty}\nu_{n}(A)$ is a complex measure.

That is: I need to show that the value is a complex number, and that countable additivity holds. I have allready shown that we have convergence, so all I need to do is shown countable additivity.

Let $\{A_i\}_i$ be a seguence of disjoint subsets, we have that:

$\nu(\cup A_i)=\lim_{n \rightarrow \infty}\nu_n(\cup A_i)=\lim_{n \rightarrow \infty}[\Sigma_{i=1}^\infty\nu_n({A_i)}]$.

The problem is moving the limit inside the sum? My only idea for this is looking at the sum as an integral over the natural numbers:

$f_n(i)=\nu_n(A_i)$, hence

$=\lim_{n \rightarrow \infty}\int_{\mathbb{N}}f_n(i)d\lambda$, where $\lambda$ is the counting measure on $\mathbb{N}$. But, in order to move the limit inside I need to use the dominated convergence theorem. But I do not know if $|f_n(i)|=|\nu_n(A_i)|$ is dominated by an integrable function?