Prove the normed space of bounded variation functions is complete 
Let $\Vert f \Vert = |f(0)| + \mathrm{Var}f$ for all $f \in BV([0,1])$; we are given that it is a norm. Show that $BV([0,1])$ is a complete normed space with this norm.

I have shown that any Cauchy sequence in $BV([0,1])$ must converge to some function pointwise, but I am stuck at proving that the function must have bounded variation.
Could someone help me?
 A: Let $\{f_n\}_{n=1}^{ +\infty}    $ be a Cauchy sequence for $\lVert\cdot\rVert$. In particular, the sequence of real numbers $\{f_n(0)\}$ is Cauchy, hence converges to a real number we call $f(0)$. Now, considering the partition $t_0=0<1=t_1$, we have 
$$\lVert f_k-f_j\rVert\geqslant\operatorname{Var}(f_k-f_j)\geqslant \left|f_k(1)-f_j(1)\right|-\left|f_k(0)-f_j(0)\right|,$$
proving that $\left\{f_k(1)\right\}$ is Cauchy, hence converges to a real number called $f(1)$. Now for $t\in(0,1)$, we consider the partition $t_0:=0<t=:t_1<t_2:=1$ to get that $\{f_k(t)\}$ is Cauchy, hence it converges to a number called $f(t)$. Now, to conclude, we need two things:


*

*$f$ is of bounded variation. Indeed, let $t_0=0<t_1<\dots<t_n=1$ be a partition of $[0,1]$. Then 
$$\sum_{j=0}^{n-1}\left|f(t_{j+1})-f(t_j)\right|\leqslant \sum_{j=0}^{n-1}\left|f(t_{j+1})-f_N(t_{j+1})\right|+\sum_{j=0}^{n-1}\left|f(t_j)-f_N(t_j)\right|+\operatorname{Var}(f_N).$$
Let $n_0$ be an integer such that $\operatorname{Var}(f_j-f_k)\leqslant 1$ if $j,k\geqslant n_0$. Then for each $N$, $\operatorname{Var}\left(f_N\right)\leqslant \max\left\{1+\operatorname{Var}\left(f_{n_0}\right),\operatorname{Var}(f_1),\dots,\operatorname{Var}\left(f_{n_0-1}\right)\right\}=:M$. Consequently, 
$$\sum_{j=0}^{n-1}\left|f(t_{j+1})-f(t_j)\right|\leqslant \sum_{j=0}^{n-1}\left|f(t_{j+1})-f_N(t_{j+1})\right|+\sum_{j=0}^{n-1}\left|f(t_j)-f_N(t_j)\right|+M.$$
Taking in last displayed equation the $\limsup_{N\to +\infty}$, we get that $f$ is of bounded variation. Indeed, for each $j\in\{0,\dots,n\}$,  $f_N(t_j)\to f(t_j)$. Therefore, 
$$\sum_{j=0}^{n-1}\left|f(t_{j+1})-f(t_j)\right|\leqslant M,$$
and $M$ is independent on the choice of the partition.

*$\lVert f-f_N\rVert\to 0$. We have by definition $f_n(0)\to f(0)$ so we have to show that $\operatorname{Var}(f_n-f)\to 0$. Let $\varepsilon>0$. We can find $N=N(\varepsilon)$ such that if $m,n\geqslant N$ and $0=t_0<t_1<\dots<t_l=1$ is a partition of $[0,1]$ then 
$$\sum_{j=0}^{l-1}|(f_m-f_n)(x_{j+1})-(f_m-f_n)(x_j)|\leqslant\varepsilon.$$
Let us take $\limsup_{m\to +\infty}$ in the previous inequality. Since $f_m(x_j)\to f(x_j)$ for each $j\in \{0,\dots,l\}$, we get for each $n\geqslant N(\varepsilon)$:
$$\sum_{j=0}^{l-1}|(f-f_n)(x_{j+1})-(f-f_n)(x_j)|\leqslant\varepsilon.$$
Since this inequality is true for any partition, we get $\operatorname{Var}(f-f_n)\leqslant \varepsilon$ for $n \geqslant N(\varepsilon)$. 
