Let $\Vert f \Vert_{BV} := |f(a)| + \mathrm{Var}(f)$ for all $f \in BV([a,b])$. We are given that this is a norm. Show that $BV([a,b])$ is a complete normed space, hence a Banach space.
My approach:
Let be $f\in BV([a,b])$ and chose some arbitrary $t\in]a,b[$. We consider the partition $t_0=a<t_1=t<t_2=b$. Then, we know that $$|f(t)|-|f(a)|\leq |f(t)-f(a)|+|f(b)-f(t)|\leq V(f)\\\implies |f(t)|\leq |f(a)|+V(f)\implies \Vert f\Vert_{\infty}\leq \Vert f\Vert_{BV}.$$ (Note that this also holds when $t=a$ or $t=b$)
We know that the space of bounded functions $B([a,b])$ equipped with the supremum norm $\Vert\cdot\Vert_{\infty}$ is complete. So if I take an arbitrary Cauchy sequence $(f_n)\in BV([a,b])$ then it converges in $B([a,b])$ with respect to $\Vert \cdot\Vert_{\infty}$. Let be $f$ its limit (in the space $B([a,b])$).
Let be $a=t_0<t_1<_\cdots<t_N=b$ an arbitrary partitition of $[a,b]$ and $(f_n)$ a Cauchy sequence such that for all $m,n>n_0$ we have $V(f_n-f_m)\leq\Vert f_n-f_m\Vert_{BV}<\frac{\epsilon}{3}$.
1.) Then,
$$ \sum\limits_{i=1}^N|f(t_i)-f(t_{i-1})|\\\leq \sum\limits_{i=1}^N|f(t_i)-f_n(t_i)|+\sum\limits_{i=1}^N|f_n(t_i)-f_m(t_{i-1})|+\sum\limits_{i=1}^N|f_m(t_{i-1})-f(t_{i-1})|\\ \leq \sum\limits_{i=1}^N|f(t_i)-f_n(t_i)|+V(f_n-f_m)+\sum\limits_{i=1}^N|f_m(t_{i-1})-f(t_{i-1})|\\<\frac{\epsilon}{3}+\sum\limits_{i=1}^N|f(t_i)-f_n(t_i)|+\sum\limits_{i=1}^N|f_m(t_{i-1})-f(t_{i-1})|. $$ We know that for all $t\in[a,b]$ we have $\lim\limits_{m\to\infty}f_m(t)=f(t)$, so that taking the limits $m,n\to\infty$ finally yields: $$ \sum\limits_{i=1}^N|f(t_i)-f(t_{i-1})|\leq \frac{\epsilon}{3}+ 0+0<\infty\implies f\in BV([a,b]) $$ because the partitition was arbitrarily chosen.
2.) If necessary we increase $n_0$ such that $|f_n(a)-f(a)|<\frac{\epsilon}{3}$ for all $n>n_0$. Then, $$ |f_n(a)-f(a)|+\sum\limits_{i=1}^N |f_n(t_i)-f(t_i)-\left(f_n(t_{i-1})-f(t_{i-1})\right)|\\ \leq \frac{\epsilon}{3}+\sum\limits_{i=1}^N |f_n(t_i)-f_m(t_i)|+\sum\limits_{i=1}^N |f(t_i)-f_m(t_i)|\\+\sum\limits_{i=1}^N |f(t_{i-1})-f_m(t_{i-1})|+\sum\limits_{i=1}^N |\left(f_n(t_{i-1})-f_m(t_{i-1})\right)|\cdots $$ Taking the limit $m\to\infty$ yields: $$ \cdots\leq \frac{\epsilon}{3}+\frac{\epsilon}{3}+0+0+\frac{\epsilon}{3}\implies \Vert f_n-f\Vert_{BV}<\epsilon $$ because the partitition was arbitrarily chosen.
So $BV([a,b])$ is a Banach space.
I know that there is already a nice solution to this question (https://math.stackexchange.com/a/211071/579544) but I would like to know if my approach also works? (My tutor told me that I can't apply the limits like I did. However, he didn't explain it why?!)
EDIT:
1.) For all $m,n>n_0$ we have$$ V(f_m)-V(f_n)\leq V(f_m-f_n)\implies V(f_m)\leq V(f_m-f_n)+V(f_n)\leq \frac{\epsilon}{3}+V(f_n)\\ \implies \sum\limits_{i=1}^N|f_m(t_i)-f_m(t_{i-1})|\leq \frac{\epsilon}{3}+V(f_n). $$ We know that for all $t\in[a,b]$ we have $\lim\limits_{m\to\infty}f_m(t)=f(t)$, so that taking the limit $m\to\infty$ yields: $$ \sum\limits_{i=1}^N|f(t_i)-f(t_{i-1})|\leq \frac{\epsilon}{3}+V(f_n)<\infty\implies f\in BV([a,b]) $$ because the partitition was arbitrarily chosen.
2.) If necessary we increase $n_0$ such that $|f_n(a)-f(a)|<\frac{\epsilon}{3}$ for all $n>n_0$. Then, $$ |f_n(a)-f(a)|+\sum\limits_{i=1}^N |f_n(t_i)-f(t_i)-\left(f_n(t_{i-1})-f(t_{i-1})\right)|\\ \leq \frac{\epsilon}{3}+\sum\limits_{i=1}^N |f_n(t_i)-f_m(t_i)-\left(f_n(t_{i-1})-f_m(t_{i-1})\right)|\\+\sum\limits_{i=1}^N |f(t_i)-f_m(t_i)|+\sum\limits_{i=1}^N |f(t_{i-1})-f_m(t_{i-1})| $$ Taking the limit $m\to\infty$ and recalling that $\Vert f_n-f_m\Vert_{BV}<\frac{\epsilon}{3}$ yields: $$ \cdots\leq \frac{\epsilon}{3}+\frac{\epsilon}{3}+0+0\implies \Vert f_n-f\Vert_{BV}<\epsilon $$ because the partitition was arbitrarily chosen.
