Relation between function of B.V and A.C. function? 
Define the $B.V.$ norm $$\|f\|_{B.V}=V(f)+|f(a)|$$ on the closed interval $[a,b]$.
  How to prove the set of absolutely continuous functions in $[a,b]$ is a closed subspace of $BV[a,b]$?

I have shown every absolutely continuous function is of bounded variation. I am stuck in proving it actually forms a closed subspace. 
Could someone kindly help? Thanks very much!
 A: Assume that $f_n\to f$ in the BV norm, and $f_n$ are absolutely continuous functions, while $f$ is just a BV function. We wish to show that $f$ is absolutely continuous, i.e. $\forall \varepsilon>0$, $\exists \delta>0$ such that for all $a\leq a_1\leq b_1 \leq a_2\leq b_2\leq \dots\leq a_k\leq b_k\leq b$, $k\in \mathbb  N$ with $\sum (b_i-a_i)<\delta$ we have $\sum |f(b_i)-f(a_i)|<\varepsilon$.
By assumption $V(f_n-f)\to 0$, and $V(f_n-f_m)\to 0$ as $m,n\to \infty$, since $f_n$ is a Cauchy sequence. In particular, we can choose large $N$ such that $V(f_n-f_m)<\varepsilon$ and $V(f_n-f)<\varepsilon$ for all $n,m\geq N$. Since $f_N$ is absolutely continuous, we can choose $\delta$ small enough such that $\sum |f_N(b_i)-f_N(a_i)|<\varepsilon$. Then for $n\geq N$ we have
$$\sum |f_n(b_i)-f_n(a_i)| \leq \sum |f_N(b_i)-f_N(a_i)|+ \sum |f_n(b_i)-f_n(a_i) -f_N(b_i)+f_N(a_i)|$$
$$\leq \varepsilon +V(f_n-f_N)\leq 2\varepsilon$$
Therefore, similarly we obtain
$$\sum |f(b_i)-f(a_i)| \leq \sum |f_n(b_i)-f_n(a_i)| + V(f_n-f)$$
$$\leq 2\varepsilon+ \varepsilon= 3\varepsilon $$
and this completes the proof that $f$ is absolutely continuous.
