If $\mu_n \overset{\ast}{\rightharpoonup}\mu$, then $\mu^+_n \overset{\ast}{\rightharpoonup} \mu^+$ and $\mu^-_n \overset{\ast}{\rightharpoonup}\mu^-$ Let $X$ be a Polish space and $\mu, \mu_n$ finite signed Borel measures on $X$. Assume that $\mu_n \overset{\ast}{\rightharpoonup} \mu$, i.e.,
$$
\int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu
$$
for all bounded continuous functions $f:X \to \mathbb R$. Let $\mu = \mu^+ - \mu^-$ and $\mu_n = \mu_n^+ - \mu_n^-$ be their Jordan decompositions.

Is it true that $\mu^+_n \overset{\ast}{\rightharpoonup} \mu^+$ and $\mu^-_n \overset{\ast}{\rightharpoonup} \mu^-$?


No, this is not true. Unless $X$ is trivial, you can find a (nonatomic) measure $\lambda$ on $X$ and a sequence $(f_n) \subset L^2(\lambda)$
with $f_n \rightharpoonup 0$ and (say) $f^+_n \rightharpoonup 1$, both in $L^2(\lambda)$. Now, consider $\mu_n = f_n \lambda$.
Finally, if $g_n \rightharpoonup g$ in $L^2(\lambda)$, then
$$
\int_X (g_n - g) h \, \mathrm{d}\lambda \to 0
$$
for all $h \in L^2(\lambda)$.
Thus, if $\lambda$ is finite, this holds for all bounded, continuous functions $h$. Hence, the measures $g_n \lambda$ converge weak-$*$ towards $g \lambda$.
