We prove the continuity from below property of measures: if $A_1 \subseteq A_2 \subseteq \cdots$ are measurable sets, then $\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)$

Proof: Define $B_1 = A_1$ and for $n \geq 2$, let $B_n = A_n \setminus A_{n-1}$. Then the sets $B_n$ are pairwise disjoint, and $\bigcup_{n=1}^{\infty} B_n = \bigcup_{n=1}^{\infty} A_n$

By countable additivity of $\mu$: $\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \mu\left(\bigcup_{n=1}^{\infty} B_n\right) = \sum_{n=1}^{\infty} \mu(B_n)$

Now observe that for each $k$: $A_k = \bigcup_{n=1}^{k} B_n$

By finite additivity: $\mu(A_k) = \sum_{n=1}^{k} \mu(B_n)$

Therefore: $\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \sum_{n=1}^{\infty} \mu(B_n) = \lim_{k \to \infty} \sum_{n=1}^{k} \mu(B_n) = \lim_{k \to \infty} \mu(A_k)$

This completes the proof.