A counter-example for a problem related to weak-star convergent

We will answer the following question:

[Q.] Let ${\mu_{n}}$ be a sequence of regular Borel measures on a set ${X}$ , which converges to a measure ${\mu}$ on ${X}$ in weak-star. Is it possible to show that ${|\mu_{n}|}$ converges to ${|\mu|}$ in weak-star?

Recall the definition of weak-star convergence first in this context. Since the dual space of ${C_{0}(X)}$ is the Borel regular measures on ${X}$ (see the Appendix C of book [Conway 1990, A course in functional analysis]), ${\mu_{n}}$ is convergent to ${\mu}$ in weak-star, if

$\displaystyle \int_{X} f(x) \mu_{n}(dx) \rightarrow \int_{X} f(x) \mu(dx), \ \forall f \in C_{0}(X).$

[A.] The answer is NO. Let ${X=[-1,1]}$ , and ${\mu_{n} = \frac 1 2 (\delta_{1/n} - \delta_{-1/n})}$ , where ${\delta}$ is the dirac measure. Note that ${\mu_{n}}$ converge to zero measure ${\mu = 0}$ in weak-star, but ${|\mu_{n}| = \frac 1 2 (\delta_{1/n} + \delta_{-1/n})}$ converge to ${\delta_{0}}$ . In particular, the limit of total variation of the sequence is not equal to total variation of the limit measure, i.e. ${|\mu|(X)= 0 \neq 1 = |\delta_{0}|(X)}$ .