If $\mu_t\to\mu$ weakly, then $\limsup_t|\mu_t|(A)\le|\mu|(A)$ for all closed $A$

Question

Let $E$ be a metric space, $\mathcal M(E)$ denote the space of finite signed measures on $\mathcal B(E)$ equipped with the total variation norm $\left\|\;\cdot\;\right\|$, $(\mu_t)_{t\in I}$ be a net in $\mathcal M(E)$ and $\mu\in\mathcal M(E)$ with $(\mu_t)_{t\in I}\to\mu$ with respect to the topology of weak convergence of measures.

I would like to show $$\limsup_{t\in I}|\mu_t|(A)\le|\mu|(A)\;\;\;\text{for all closed }A\subseteq E\tag1.$$

Note that we are able to show a result of this kind for open subsets: $$\liminf_{t\in I}|\mu_t|(C)\ge|\mu|(C)\;\;\;\text{for all open }C\subseteq E.\tag2$$

$(2)$ is an easy consequence of the following result:

Lemma: If $\nu\in\mathcal M(E)$, then $$|\nu|(C)=\sup_{\substack{f\in C_b(E)\\|f|\le1\\\left.f\right|_{E\setminus C}=0}}\nu f\;\;\;\text{for all open }C\subseteq E\tag3.$$

A proof of $(3)$ can be found in this question.

From this Lemma we obtain $(2)$: Let $\varepsilon>0$. By $(3)$, $$\mu f>|\mu|(C)-\frac\varepsilon2\tag4$$ for some $f\in C_b(E)$ with $|f|\le1$ and $\left.f\right|_{E\setminus C}=0$. Since $(\mu_t)_{t\in I}\to\mu$, there is a $t_0\in I$ with $$|(\mu_t-\mu)f|<\frac\varepsilon2\tag5.$$ Thus, $$|\mu_t|(C)\ge\mu_t f>|\mu|(C)-\varepsilon\;\;\;\text{for all }t\ge t_0\tag6.$$ Hence, we are done.

Note that an easy consequence of $(2)$ is that $$(|\mu_t|)_{t\in I}\to|\mu|\Leftrightarrow\lim_{t\in I}\left\|\mu_t\right\|=\left\|\mu\right\|\tag7.$$

Answer 1

$\newcommand\R{\mathbb R}$What you want to show is false in general. We reason here similarly to this previous answer.

Indeed, let $E:=\R$. For all even natural $n$, let $\mu_n(dx):=\mu(dx)(1+c\cos nx)$, where $\mu$ is the uniform distribution on the interval $[0,1]$ and $c$ is a real number. Then, say by the Riemann–Lebesgue lemma, $\mu_n\to\mu$ (as $n\to\infty$).

On the other hand, $$|\mu_n|([0,1])=\int_0^1 dx\,|1+c\cos nx| \ge\int_0^1 dx\,(c|\cos nx|-1) \to\frac2\pi\,c-1>1=|\mu|([0,1])$$ if $c>\pi$. So, $$\limsup_n |\mu_n|([0,1])\not\le|\mu|([0,1]).$$