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](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures).



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> I would like to show $$\limsup\_{t\in I}|\mu\_t|(A)\le|\mu|(A)\;\;\;\text{for all closed }A\subseteq E\tag1.$$
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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:



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> **Lemma**: If $\nu\in\mathcal M(E)$, then $$|\nu|(C)=\sup\_{\substack{f\in C\_b(E)\\|f|\le1\\\left.f\right|\_{E\setminus C}=0}}\nu f\;\;\;\text{for all open }C\subseteq E\tag3.$$
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A proof of $(3)$ can be found in [this question](https://math.stackexchange.com/q/3967941/47771).


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 $$(|\mu\_t|)\_{t\in I}\to|\mu|\Leftrightarrow\lim\_{t\in I}\left\|\mu\_t\right\|=\left\|\mu\right\|\tag7.$$



If $\mu_t\to\mu$ weakly, then $\limsup_t|\mu_t|(A)\le|\mu|(A)$ for all closed $A$
$\newcommand\R{\mathbb R}$What you want to show is false in general. We reason here similarly to [this previous answer](https://mathoverflow.net/a/380107/36721).


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,
$$\limsup\_n |\mu\_n|([0,1])\not\le|\mu|([0,1]).$$


