Let $F$ be a $\sigma$-algebra and let $\mu: F \rightarrow [0,\infty]$ be an additive function.
I have defined
$U_n := \bigsqcup_{j=n}^{\infty}C_j$
where $(C_j)$ is a pairwise disjoint family of elements of $F$
Is it possible to show $\mu (U_n) \rightarrow 0$ (as $n \rightarrow \infty$) or is this not necessarily true?
Since the $C_n$-s being disjoint implies that $\bigcap_{n\in\Bbb N}U_n=\emptyset$, this is true if $\mu$ is $\sigma$-additive and one of the $U_n$-s has finite measure. This is done using the fact that $\sigma$-additivity is equivalent to finite additivity + the fact that, for all increasing sequences of sets $\{A_n\}_{n\in\Bbb N}$, $\mu\left(\bigcup_{n\in\Bbb N}A_n\right)=\lim_{n\to\infty}\mu(A_n)$: then you only need to consider $A_n=X\setminus U_n$.
For a counterexample with $\sigma$-additive $\mu$ and all the $U_n$-s of infinite measure, consider $U_n=\{x\in\Bbb R\,:\, x\ge n\}$ (and the $C_n$-s defined accordingly, namely $C_n=U_{n}\setminus U_{n-1}$).
For a counterexample with a finite finitely additive measure, consider this measure: essentially, call $\lambda$ the Lebesgue measure, let $\phi:\ell^\infty\to\Bbb R$ be a Banach limit and call $\mu_\phi(A)=\phi\left(\frac1{1+k}\lambda(A\cap [0,1+k])\,:\,k\in\Bbb N\right)$. $\mu_\phi$ can be defined for all Lebesgue measurable subsets of $\Bbb R$. Again, consider the sequenece of sets $U_n=\{x\in\Bbb R\,:\, x\ge n\}$: you have $\mu(U_n)=1$ for all $n$.