I know that $BV([0,1])$ is a Banach space when equipped with $||f||_1=|f(t_0)|+Var(f)$ with $t_0\in[0,1]$ or with $||f||_2=||f||_\infty+Var(f)$. Basically, if $(f_n)_n\subset BV([0,1])$ is such that $f_n\rightarrow f$ pointwise and $(f_n)_n$ is Cauchy in the sense of $Var(\cdot)$, then $f\in BV([0,1])$ and $Var(f-f_n)\rightarrow 0$.
As $BV([0,1])\subset L^1([0,1])$, I was wondering if $BV([0,1])$ is a Banach space when equipped with $||f||=\int_0^1|f(t)|dt+Var(f)$. I have shown that any Cauchy sequence with respect to this norm has a subsequence that converges a.e. to a function in $L^1([0,1])$. This conclusion does note allow me to use the idea of the first paragrah because total variation is not preserved for functions that are equals a.e., for instance, $\chi_{\mathbb{Q}\cap[0,1]}=0$ a.e., however $Var(\chi_{\mathbb{Q}\cap[0,1]})=\infty$ and $Var(0)=0$. So, I want to know whether $||\cdot||$ makes $BV([0,1])$ a Banach space or not.