Let $U$ be a bounded smooth domain of $\mathbb{R}^d$. We write $m$ for the Lebesgue measure on $U$. A function $f \in L^1(U,m)$ has bounded variation in $U$ if \begin{align*} V(f,U):=\sup \left\{\int_{U}f\,\text{div}\varphi\,dm : \varphi \in C^1_{c}(U;\mathbb{R}^d),\, \sup_{x \in U}|\varphi(x)| \le 1. \right\}<\infty. \end{align*} Here, $|\varphi(x)|=\{\varphi_1(x)^2+\cdots+\varphi_d(x)^2\}^{1/2}$. We denote by $BV(U)$ the set of functions of bounded variation in $U$. For $f \in BV(U)$, we set \begin{align*} \|f\|_{BV(U)}=\|f\|_{L^1(U,m)}+V(f,U), \end{align*} where $\|f\|_{L^1(U,m)}$ is the $L^1$-norm of $f$.
The following theorem is well-known (now $U$ is a bounded smooth domain):
Let $\{f_n\}_{n=1}^\infty$ be a bounded sequence in $BV(U)$. Then, there exists a subsequence $\{f_{n_k}\}_{k=1}^\infty$ and $f \in L^1(U,m)$ such that $\lim_{k \to \infty}f_{n_k}=f$ in $L^1(U,m)$.
My question
Let $\{f_n\}_{n=1}^\infty$ be a (not necessarily bounded) sequence in $BV(U)$. We assume that $\{f_n\}_{n=1}^\infty$ is a bounded sequeene in $L^1(U,m)$, and for any $\varphi \in C_{c}^1(U;\mathbb{R}^d)$, \begin{align*} \lim_{n \to \infty}\int_{U}f_n\,\text{div}\varphi\,dm=\int_{U}f\,\text{div}\varphi\,dm\quad {\rm (A)} \end{align*} for some $f \in BV(U)$. Of course, this does not imply that $\{f_n\}_{n=1}^\infty$ is a bounded sequence in $BV(U)$ (the condition (A) is not sufficient to use the uniform boundedeness principle). However, $C_{c}^\infty(U;\mathbb{R}^d)$ (with the topology induced by the seminorms) is a complete metric space. Then, the condition (A) and the uniform boundedness principle should show that $\lim_{n \to \infty}f_n =f$ in a strong sense.
What does this convergence mean? If the convergence is metrizable, please tell me the metric.
