weak-*-convergence of measures ==> convergence of the total mass?

Let $X = [0,1]$. Let $\mu_n$ be a sequence of regular signed Borel measures on $X$, which converges to a measure $\mu$ on $X$ in weak-star, i.e. for any $f\in C_0(X)$, we have $\int_X f \mu_n(dx) \to \int_X f \mu(dx)$.

Does this imply that the total variation norm (=total mass) $|\mu|(X)$ converge?

That means: We know $\mu= \mu^+ - \mu^-$ (Jordan decomposition, $\mu^+$ and $\mu^-$ positive measures...). Then we defined $|\mu|(X)=\mu^+(X) - \mu^-(X)$.

Other formulation of question: Does then $|\mu_n|(X) \to |\mu|(X)$ ?

• Here's a sledehammer: The Josefson-Nissenzweig Theorem states that for every infinite dimensional Banach space $X$, there exists a weak* null sequence of norm-one elements in $X^*$. If $(f_n)$ is such a sequence in $X^*$, then the sequence $f_1,0,f_2,0,\ldots$ converges weak* to zero, but the sequence of their norms do not. – David Mitra Mar 13 '14 at 12:06

I think the answer is negative. An example would be the sequence $\mu _n = \delta _{\frac12 - \frac1n}-\delta _{\frac12 + \frac1n}$, which converges in weak* topology to the trivial measure $\mu$, $\mu ([0;1]) = 0$.