I'm given a measurable space $(\Omega, \mathcal{F})$ and a finitely additive measure $m$ that satisfies $m(\Omega) = 1$. I'm asked to give counterexamples for continuity from above and below.
My attempt (discontinuity from above): take $\Omega = \mathbb{N}$ and $\mathcal{F} = \mathcal{P}(\mathbb{N})$ the set of subsets of $\mathbb{N}$. Define: $$ m(A) = \begin{cases} 0, & A \text{ finite} \\ 1, & \text{otherwise} \end{cases} $$ Now, take $A_n = \{ j \in \mathbb{N} \ |\ j \geq n \} $ which satisfies $$ A_1 \supset A_2 \supset \dots, \quad \bigcap_{i=1}^{\infty} A_i = \emptyset \quad (A_n \downarrow \emptyset) $$ Then, it obviously holds \begin{align*} m\left(\bigcap_i^{\infty} A_i\right) &= m(\emptyset) = 0 \\ \lim_{n \to \infty} m(A_n) &= \lim_{n \to \infty} m\left( \{ j \geq n, j \in \mathbb{N} \} \right) = \lim_{n \to \infty} 1 = 1 \neq m\left( \bigcap_i^{\infty} A_i \right) \end{align*} since, $\forall n \geq 1$, $A_n$ is a co-finite set. Is my approach correct?
No. Actually, your finite additive measure $m$ is not well defined.
For example: by additivity $m(3n+1)+m(3n+2) = m(3n+1\cup3n+2)$ which imply $1+1 = 1$ which is false.