and let $$\Pi_\xi:=\{y\in{\mathbf{R}}^d:\langle y,\xi\rangle=0\}.$$ If $y\in\Pi_\xi$ and $E\in{\mathbf{R}}^d$, we define the one dimensional slice $$E_{\xi,y}:=\{t\in{\mathbf{R}}: y+t\xi\in E\}.$$ For $u:{\Omega}{\rightarrow}{\mathbf{R}}$, we define $u_{\xi,y}:{\Omega}_{\xi,y}{\rightarrow}{\mathbf{R}}$ as $$u_{\xi,y}(t):=u(y+t\xi), \quad t\in {\Omega}_{\xi,y}.$$ Functions in $GSBV({\Omega})$ can be characterized by one-dimensional slices (see [@Bra1 Thm. 4.1])

\[teo:braides\] Let $u\in GSBV({\Omega})$. Then for all $\xi \in\S^{d-1}$ we have $$u_{\xi,y}\in GSBV({\Omega}_{\xi,y})\quad \text{for } \H^{d-1}\text{-a.e. }y\in\Pi_\xi.$$ Moreover for such $y$, we have $$\begin{gathered}
 u'_{\xi,y}(t)=\langle{\nabla}u(y+t\xi),\xi\rangle\quad \text{for a.e. } t\in{\Omega}_{\xi,y},\\
 J_{u_{\xi,y}}=\{t\in {\mathbf{R}}:y+t\xi\in J_u\},\end{gathered}$$ and $$u_{\xi,y}(t^\pm)=u^\pm(y+t\xi)\quad \text{or}\quad u_{\xi,y}(t^\pm)=u^\mp(y+t\xi)$$ according to whether $\langle\nu_u,\xi\rangle>0$ or $\langle\nu_u,\xi\rangle<0$. Finally, for every Borel function $g:{\Omega}{\rightarrow}{\mathbf{R}}$, $$\label{eq:braidesricostruzione}
 \int_{\Pi_\xi}\sum_{t\in J_{u_{\xi,y}}} g_{\xi,y}(t)\dH^{d-1}(y)=\int_{J_u}g|\langle\nu_u,\xi\rangle|\dH^{d-1}.$$ Conversely if $u\in L^1({\Omega})$ and if for all $\xi \in\{e_1,\dots,e_d\}$ and almost every $y\in\Pi_\xi$ we have $u_{\xi,y}\in SBV({\Omega}_{\xi,y})$ and $$\int_{\Pi_\xi}|Du_{\xi,y}|({\Omega}_{\xi,y})\dH^{d-1}(y)<+{\infty}$$ then $u\in SBV({\Omega})$.

Rectifiable vector Measures
---------------------------

Let us introduce the linear operator $\perp$ that associates to each vector $v=(v_1,v_2)\in {\mathbf{R}}^2$ the vector $v^\perp=(-v_2,v_1)$ obtained via a $90^\circ$ counterclockwise rotation of $v$. Notice that the $\perp$ operator maps divergence free ${\mathbf{R}}^2$-valued measures onto curl free ${\mathbf{R}}^2$-valued measures. Let $O\subset{\mathbf{R}}^2$ be a simply connected and bounded open set. By Stokes Theorem, for any divergence free measure $\s\in\M(O,{\mathbf{R}}^2)$ there exists a function $u\in BV({\Omega})$ with zero mean value such that $\s= Du^\perp$. On the other hand for $u\in \P({\Omega})$ $\sigma:=Du^\perp$ 