Show that $V[a,b]$ is a Banach space Let $V[a,b]$ be the space of all complex-valued functions of bounded variation on $[a,b]$, which we treat naturally as a $\Bbb C$-vector space. In addition, we equip it with the norm:
$$\|x\|:=|x(a)|+Vx,$$
in which $Vx$ denotes the total variation of $x(t)$ on $[a,b]$. I have to show that $(V[a,b],\|\cdot\|)$ is a Banach space.
I think I'm almost there but stuck in one critical step, here are my thoughts:
Suppose $\{x_n\}$ is a Cauchy sequence in $V[a,b]$, then it is clear that both $|x_m(a)-x_n(a)|$ and $V(x_m-x_n)$ turn small when $m,n$ turn large. In particular, $\lim_{n\to\infty}x_n(a)$ exists; and given any $\epsilon>0$, we can find:
1). $N_1$ s.t. $\forall m,n>N_1$, $|x_m(a)-x_n(a)|<\epsilon/2$ and $V(x_m-x_n)<\epsilon/2$.
2). $N_2>N_1+1$ s.t. $\forall m,n>N_2$, $|x_m(a)-x_n(a)|<\epsilon/2^2$ and $V(x_m-x_n)<\epsilon/2^2$.
2). $N_3>N_1+1$ s.t. $\forall m,n>N_3$, $|x_m(a)-x_n(a)|<\epsilon/2^3$ and $V(x_m-x_n)<\epsilon/2^3$.
$$\cdots$$
This progress may continue indefinitely and we can pick $n_k\in (N_k,N_{k+1})$ so that 
$$|x_{n_{k+1}}(a)-x_{n_k}(a)|<\epsilon/2^k,\, V(x_{n_{k+1}}-x_{n_k})<\epsilon/2^k.$$
Define $g_N=x_{n_1}+\sum_{k=1}^N (x_{n_{k+1}}-x_{n_k})$. Note that for all $t\in [a,b]$ we have
\begin{align}
 V(x_{n_{k+1}}-x_{n_k})&\ge |(x_{n_{k+1}}-x_{n_k})(b)-(x_{n_{k+1}}-x_{n_k})(t)|+|(x_{n_{k+1}}-x_{n_k})(t)-(x_{n_{k+1}}-x_{n_k})(a)|\\
&\ge |(x_{n_{k+1}}-x_{n_k})(t)-(x_{n_{k+1}}-x_{n_k})(a)| \\
&\ge |(x_{n_{k+1}}-x_{n_k})(t)|-|(x_{n_{k+1}}-x_{n_k})(b)|.
\end{align}
Hence we have
$$|(x_{n_{k+1}}-x_{n_k})(t)|\le |(x_{n_{k+1}}-x_{n_k})(a)|+V(x_{n_{k+1}}-x_{n_k})<\epsilon/2^{k-1}.$$
Hence by Weierstrass M-test $g_N$ converges absolutely uniformly to some function $g$ on $[a,b]$. The hardest part is: how to show $g_N\to g$ by $\|\cdot\|$?
Naively I guess the following should be right:
$$V(g-g_N)=V(\lim_{m\to\infty}g_m-g_N)=\lim_{m\to\infty}V(g_m-g_N).$$
However, even the uniform convergence isn't enough to justify the interchange of limits! So I really don't know any method else to get around it. Any help? Thanks in advance!
 A: Fix a BV-norm Cauchy sequence $(g_n)$, $\varepsilon>0$, and $j,k\in\Bbb N$ such that $||g_j-g_k||<\varepsilon$. Let $R>0$ such that $||g_n||\leq R$ for all $n$.
You have shown that $(g_n)$ converges uniformly to a bounded function $g$ on $[a,b]$. (Here is another proof of this fact which is a bit easier: Let $t\in [a,b]$. We have
$$
|g_j(t)-g_k(t)|\leq |g_j(t)-g_j(a)|+|g_j(a)-g_k(a)|+|g_k(a)-g_k(t)|\\
\leq V(g_j)+|g_j(a)-g_k(a)|+V(g_k)<3\varepsilon
$$
and so $(g_n)$ is uniformly Cauchy.)
Let $a=x_1<x_2<\dots<x_n=b$ be a partition of $[a,b]$. Let $N\in\Bbb N$ large enough so that $|g_k(x_i)-g(x_i)|<\varepsilon/n$ for all $i$ whenever $k\geq N$. (Pointwise convergence is enough for this.) Assuming $k\geq N$, we have
$$
\sum_{i=1}^{n-1}|g(x_{i+1})-g(x_i)|\leq \sum_{i=1}^{n-1}|g(x_{i+1})-g_k(x_{i+1})|+\sum_{i=1}^{n-1}|g_k(x_{i+1})-g_k(x_i)|+{\sum_{i=1}^{n-1}|g_k(x_{i})-g(x_i)|}\\
\leq 2\varepsilon + \sum_{i=1}^{n-1}|g_k(x_{i+1})-g_k(x_i)|.
$$
Now we have 
$$
\sum_{i=1}^{n-1}|g(x_{i+1})-g(x_i)|\leq R+2\varepsilon
$$
and we conclude that $g\in V[a,b]$.
Now, by similar calculations assuming both $j\geq N$, we obtain
$$
\sum_{i=1}^{n-1}|(g_k-g)(x_{i+1})-(g_k-g)(x_i)|\leq \sum_{i=1}^{n-1}|(g_k-g_j)(x_{i+1})-(g_k-g_j)(x_i)|+2\varepsilon\\\leq V(g_j-g_k)+2\varepsilon<3\varepsilon.
$$
This is independent of the choice of partition, so $V(g_k-g)\leq 3\varepsilon$.
This certainly needs cleaning up somewhat, but hopefully the key ideas are clear enough.
