I am wondering: given a function $u \in BV(\Theta)$ where $\Theta$ is an open subset in $\mathbb{R}^n$ and a Borel subset $B \in \mathcal{B(\Theta)},$ is the function $w \colon= u \chi_B$ still in $BV(\Theta)$? I guess so, since clearly $w \in L^1(\Theta)$ and the total variation $V(u\chi_B,\Theta) \leq V(u,\Theta) < + \infty,$ where \begin{equation} V (u, \Theta) := \sup \left\{ \int_\Theta u\, {\rm div}\, \psi : \psi\in C^\infty_c (\Theta, \mathbb R^n), \, \|\psi\|_{C^0}\leq 1\right\}\, . \end{equation} Is this right?
Can we also conclude that for the distributional derivative it holds that \begin{equation} Dw(\cdot)=D\chi_Bu(\cdot)=Du(\,\cdot \cap B) ? \end{equation}
I started with noting that similarly $\tilde w \colon = u \chi_{B^c} \in BV(\Theta)$ and so $\forall \phi \in C_c^{1}(\Theta)$ and $i=1, \dots n:$ \begin{align*} -\int_{\Theta} \phi \, dD_iu &= \int_{\Theta} u \frac{\partial \phi}{\partial x_i} \, dx = \int_{\Theta} w \frac{\partial \phi}{\partial x_i} \, dx + \int_{\Theta} \tilde w \frac{\partial \phi}{\partial x_i} \, dx \\ &= -\int_{ \Theta} \phi \, d D_iw -\int_{ \Theta} \phi \, d D_i\tilde w = - \int_{ \Theta} \phi \, d ( \underbrace{D_iw + D_i \tilde w}_{=D_i\left( w + \tilde w \right)} ) \\ &= -\int_{\Theta} \phi \, d\left(D_i(w+\tilde w) \right) \end{align*} and therefore $D_i w + D_i \tilde w = D_i(w+ \tilde w) = D_i u = D_i u(\cdot \cap B )+ D_i u(\cdot \cap B^c ),$ from which we get $D_i w = D_i u(\cdot \cap B )$ and analogously $D_i \tilde w = D_i u (\cdot \cap B )$ for any $i,$ i.e. \begin{equation} Dw = Du (\cdot \cap B ), \qquad D \tilde w = D u (\cdot \cap B^c ). \end{equation}
Or should I write $Dw(\cdot \cap B) = Du (\cdot \cap B )$ and analogously for tilde?
Thanks for any input.