I have come across a lemma which states:
Let $\mu$ be a finite content on an algebra $\mathcal{A}$.
$\mu$ is a pre-measure if and only if $\mu$ is continuous on the empty set, i.e., for every sequence $(A_i)_{i \geq 1}$ from $\mathcal{A}$ with $A_i \downarrow \emptyset$, we have $\mu(A_i) \rightarrow 0$ as $i \rightarrow \infty$.
Could someone provide an example of a finite content $\mu$ that is not continuous on the empty set?
Let $\mathcal{U}$ be a free ultrafilter on $\Bbb{N}$. There is an associated finitely-additive measure $\mu_\mathcal{U}$ on subsets of $\Bbb{N}$ such that $\mu_\mathcal{U}(S)=1$ for $S\in\mathcal{U}$ and $\mu_\mathcal{U}(S)=0$ for $S\not\in\mathcal{U}$.
Let $A_n=\Bbb{N}\setminus\{0,\ldots,n-1\}$. Then $\mu_\mathcal{U}(A_n)=1$ but $\bigcap_{n\in\Bbb{N}} A_n = \emptyset$.
If you're not working in $\mathsf{ZFC}$, it's possible for there to be no free ultrafilters on $\Bbb{N}$. For a finitely-additive finite measure to be discontinuous at $\emptyset$, it must be defined on an infinite set. The space of finitely-additive finite measures defined for all subsets of $\Bbb{N}$ is the (Banach space) dual of $\ell^\infty(\Bbb{N})$. In $\mathsf{ZF}$ it's possible the dual of $\ell^\infty(\Bbb{N})$ reduces to $\ell^1(\Bbb{N})$. The members of $\ell^1(\Bbb{N})$, interpreted as measures on $\Bbb{N}$ are all $\sigma$-additive so are continuous at $\emptyset$.
Since $\mu$ is not required to be defined for all subsets of $\Bbb{N}$, you can take the algebra $\mathcal{A}$ to be the algebra consisting of all finite or co-finite subsets of $\Bbb{N}$. Define $\mu(S)=1$ for $S$ co-finite and $\mu(S)=0$ for $S$ finite and you get the same example, but no need for free ultrafilters.