Suppose $\left\{\phi_{n}\right\}$ is a sequence of $\sigma$-additive set functions defined on a $\sigma$-ring $\mathscr{S}$ and that $\lim \phi_{n}(E)=\phi(E)$ exists for all $E$ in $\mathscr{S}$. Show that $\phi$ is finitely additive on $\mathscr{S}$. If either
(i) $\phi_{n}(E) \rightarrow \phi(E)$ uniformly on $\mathscr{S}$ with $\phi(E)>-\infty$ for all $E \in \mathscr{S}$; or
(ii) $\phi_{1}(E)>-\infty, \phi_{n}(E)$ monotone increasing for all $E \in \mathscr{S}$; show that $\phi$ is $\sigma$-additive on $\mathscr{S}$.
My solution attempts:
I tryied doing something lik thiss: $ \phi\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty \phi(A_i) $
Since $\phi_n(E)$ is a monotone increasing sequence for all $E$ in $\mathscr{S}$, we can apply the Monotone Convergence Theorem for set functions. This theorem states that if $\{\phi_n\}$ is a sequence of set functions, and $\phi_n(E)$ is monotone increasing and converges pointwise to $\phi(E)$ for all $E$ in $\mathscr{S}$, then:
$ \phi\left(\lim_{n\to\infty} E_n\right) = \lim_{n\to\infty} \phi(E_n) $
Now, let $B_m = \bigcup_{i=1}^m A_i$. We have:
$ B_1 \subset B_2 \subset B_3 \subset \ldots $
These sets form an increasing sequence, and we can use the Monotone Convergence Theorem:
$ \phi\left(\lim_{m\to\infty} B_m\right) = \lim_{m\to\infty} \phi(B_m) $
Butt $\lim_{m\to\infty} B_m = \bigcup_{i=1}^\infty A_i$ by definition. So, we have:
$ \phi\left(\bigcup_{i=1}^\infty A_i\right) = \lim_{m\to\infty} \phi(B_m) $
Gosh, let's consider the right-hand side:
$ \lim_{m\to\infty} \phi(B_m) = \lim_{m\to\infty} \phi\left(\bigcup_{i=1}^m A_i\right) $
Using finite additivity of $\phi$, we can write:
$ \lim_{m\to\infty} \phi\left(\bigcup_{i=1}^m A_i\right) = \sum_{i=1}^\infty \phi(A_i) $
Therefore, we've got that $ \phi\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty \phi(A_i) $
Is all of that alright?