Suppose $A_i, i = 1, 2, 3 \ldots$ is an increasing sequence of sets from a sigma algebra $\Sigma$. Let $\mu$ be a measure on $\Sigma$.
Is $\sup{\{\mu(A_i)\}} = \mu\bigg(\bigcup\limits_{i=1}^{\infty} A_i\bigg)$? The RHS is certainly an upper bound, but I can't figure out if it's the least upper bound. I have tried using the epsilon definition to prove that $\mu\bigg(\bigcup\limits_{i=1}^{\infty} A_i\bigg)$ is a supremum, but to no avail.
On
If $\mu$ is assumed to be countably additive.
We have $A_n\subset A_{n+1}$ for each $n.$ Then
(I). Suppose $\sup_n\mu(A_n)=M<\infty.$ Let $B_1=A_1$ and $B_{n+1}=A_{n+1}\setminus A_n$ and let $x_n=\mu(B_n).$ We have $$\sup_n\sum_{j=1}^n x_j=\sup_n\mu(A_n)=M <\infty$$ and each $x_j\geq 0.$ Therefore $$M=\sum_{j=1}^{\infty}x_j =\sum_{j=1}^{\infty}\mu(B_j).$$ Now since $B_j$ and $B_k$ are disjoint when $j\ne k$, we have $$M=\sum_{j=1}^{\infty}\mu(B_j)=$$ $$=\mu(\cup_{j=1}^{\infty}B_j)=$$ $$=\mu(\cup_{j=1}^{\infty}A_j).$$
(II). It should be obvious that $\mu(\cup_nA_n)\geq \sup_n\mu(A_n),$ because $\cup_n A_n\supset A_j$ for every $j.$ So if $\sup_n\mu(A_n)=\infty$ then $\mu(\cup_nA_n)=\infty.$
If $\mu$ is not countably additive this may fail. For example let $B$ be the $\sigma$-algebra of all subsets of $\Bbb N$ and let $\mu(A)=0$ if $A$ is finite and $\mu(A)=\infty $ if $A$ is infinite. And let $A_n=\{j\in \Bbb N:j\leq n\}.$
If we set $$B_i = A_i \backslash A_{i-1}$$ then the sets $B_i$ are pairwise disjoint and $$A_n = \bigcup_{i=1}^n B_i. \tag{1}$$ Using the $\sigma$-additivity of the measure $\mu$ we find $$\mu \left( \bigcup_{n \in \mathbb{N}} A_n \right) = \mu \left( \bigcup_{i \in \mathbb{N}} B_i \right) = \sum_{i=1}^{\infty} \mu(B_i) = \lim_{n \to \infty} \underbrace{\sum_{i=1}^n \mu(B_i)}_{\stackrel{(1)}{=} \mu(A_n)}.$$ Since $\mu(A_n)$ is increasing in $n$, this shows $$\mu \left( \bigcup_{n \in \mathbb{N}} A_n \right) = \sup_{n \in \mathbb{N}} \mu(A_n).$$ This property of the measure $\mu$ is called continuity from below.