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)}.


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