Denote by $M$ the topological dual of the Banach space $C[0,1]$. Then $M$ can be identified with the space of signed measures on $[0,1]$. Denote by $M^+ \subseteq M$ the positive cone of $M$ consisting of positive measures. Each $\mu \in M$ has the Jordan decomposition $\mu = \mu^+ - \mu^-$ with $\mu^\pm \in M^+$ and set $|\mu| := \mu^+ + \mu^-$. Equip $M$ with the weak* topology such that a sequence $\mu_n \in M$ converges to $\mu$ iff $\int_0^1 f d\mu_n \to \int_o^1 f d\mu$ for all $f \in C[0,1]$.
Assume that $\mu_n \to 0$. Then $|\mu_n|$ does not need to converge ($|.| : M \to M^+$ is not continuous) but $|\mu_n|$ has a convergent subsequence (from $\mu_n \to 0$ it follows that the total variation norms $|| \, |\mu_n| \, || = || \mu_n ||$ are uniformly bounded and we can apply the Helly selection theorem on $|\mu_n|$).
Question. Is there a sequence $\sigma_n \in M^+$ (of positive measures) such that $|\mu_n| + \sigma_n$ converges weakly*?
Probably hopeless idea: remove the convergent subsequence of $|\mu_n|$, providing a (finite or infinite) sequence of remaining elements of $|\mu_n|$ which in turn has a convergent subsequence and proceed iteratively. If this iteration terminates after finitely many steps, we can cook up such a sequence $\sigma_n$ explicitely. But the problem is that this iteration does not need to stop in finitely many steps. So I think that this idea is hopeless and gives probably a hint that such a sequence $\sigma_n$ of positive measures need not exist.