be open. Suppose $\{ f_n \}$ is a sequence of least gradient functions in $\Omega$ convergent in $L^1_{loc}(\Omega)$ to $f$. Then $f$ is of least gradient in $\Omega$.

An identical result for solutions of least gradient problem is impossible, as the trace operator is not continuous in $L^1$ topology. We need an additional assumption regarding traces. A correct formulation would be:

\[stabilnosc\] Suppose $f, f_n \in L^1(\partial\Omega)$. Let $u_n$ be a solution of LGP for $f_n$, i.e. $Tu_n = f_n$. Let $f_n \rightarrow f$ in $L^1(\partial\Omega)$ and $u_n \rightarrow u$ in $L^1(\Omega)$. Assume that also $Tu = f$. Then $u$ is a solution of LGP for $f$.

To deal with regularity of solutions of LGP, it is convenient to consider superlevel sets of $u$, i.e. sets of the form $\partial \{ u > t \}$ for $t \in \mathbb{R}$. It follows the the two subsequent results:

\[lem:jednoznacznoscnadpoziomic\] Suppose $u_1, u_2 \in L^1(\Omega)$. Then $u_1 = u_2$ a.e. iff for every $t \in \mathbb{R}$ the superlevel sets of $u_1$ and $u_2$ are equal, i.e. $\{ u_1 > t \} = \{ u_2 > t \}$ up to a set of measure zero.

\[twierdzeniezbgg\] $($[@BGG Theorem 1]$)$\
Suppose $\Omega \subset \mathbb{R}^N$ is 