Does $\mu_n \overset{*}{\rightharpoonup} \mu$ (or $\mu_n \rightharpoonup \mu$) imply $\{[\mu_n] \mid n \in \mathbb N\}$ is bounded?

Question

Let

• $X$ be a metric space,
• $\mathcal M(X)$ the space of all finite signed Borel measures on $X$,
• $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions on $X$, and
• $\mathcal C_0(X)$ be the space of real-valued continuous functions on $X$ that vanish at infinity.

Then $\mathcal C_b(X)$ and $\mathcal C_0(X)$ are real Banach space with supremum norm $\|\cdot\|_\infty$. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$. Then $(\mathcal M(X), [\cdot])$ is a Banach space. Let $\mu_n,\mu \in \mathcal M(X)$. We define weak convergence by $$ \mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X), $$ and weak$^*$ convergence by $$ \mu_n \overset{*}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_0 (X). $$

Now let $\mu_n,\mu \in \mathcal M(X)$. If $\mu_n \rightharpoonup \mu$, then $\mu_n (X) \to \mu (X)$ and thus $\{\mu_n(X) \mid n \in \mathbb N\}$ is bounded. On the other hand, $\mu_n \overset{*}{\rightharpoonup} \mu$ does not necessarily imply $\mu_n (X) \to \mu (X)$.

My questions:

1. Does $\mu_n \rightharpoonup \mu$ imply $\{[\mu_n] \mid n \in \mathbb N\}$ is bounded?
2. Does $\mu_n \overset{*}{\rightharpoonup} \mu$ imply $\{[\mu_n] \mid n \in \mathbb N\}$ is bounded?

Thank you so much!