I am trying to prove that if $\{A_i\}$ is an increasing sequence of sets from some sigma algebra with $\mu(A_i) < \infty$ $\forall i$, where $\mu$ is some arbitrary measure, then $$\lim_{i\to\infty} \mu(A_i) = \mu\bigg(\bigcup\limits_{i=1}^{\infty} A_i\bigg)$$
I would like to prove this using monotone convergence for sequences of real numbers, i.e. the theorem that the limit of an increasing sequence which is bounded above is its supremum. However, I am not sure if $\mu\bigg(\bigcup\limits_{i=1}^{\infty} A_i\bigg)$ is actually the supremum of $\{\mu(A_i)\}$. It is surely an upper bound, but I can't prove that it's the least upper bound.
If it is the supremum, how can I prove it?
Define $f_n:=\chi_{A_n}$ and $f:=\chi_{\cup_{i=1}^\infty A_i}$. Then since $(A_n)_{n\in\mathbb{N}}$ is an increasing sequence of sets, we have that $$0\le f_n\uparrow f.$$ Then, by monotone convergence theorem $$\mu(A_n)=\int f_n\operatorname{d}\mu\uparrow \int f\operatorname{d}\mu =\mu(\cup_{i=1}^\infty A_i).$$