If $f\colon[a,b]\rightarrow \mathbb{R}$ and $P=(a=x_0, x_1, ...,x_n=b)$ is any partition of $[a,b]$ let $v_f(P)$ be defined as $$v_f(P) = \sum_{k=1}^n |f(x_k) - f(x_{k-1})| $$ and the total variation on $[a,b]$ be defined as :
$$V_f[a,b] = \sup \{ v_f(P): \text{P is a partition of} [a,b]\}$$
Question: If $f \in BV[a,b] $ show that there exist a sequence $(P_n)$ of partitions of $[a,b]$ such that $V_f[a,b] = \lim (v_f(P_n))$
My attempt:
From the definition of $V_f[a,b]$ and the definition of supremum,
For every $\epsilon > 0 \ \ \exists$ a partition P of [ a,b] such that
$V_f[a,b] - \epsilon \leq v_f(P) \leq V_f[a,b]$
In particular, if $\epsilon = \frac{1}{n}$ , there exists a sequence of partitions $(P_n)$ of $[a,b]$ such that $ v_f[a,b] - \frac{1}{n} \leq v_f(P_n) \leq V_f[a,b]$ taking the limit as $n \rightarrow \infty$ we get: $v_f(P_n) = V_f[a,b]$ by the squeeze theorem.
I believe that I'm missing something.Like defining $(P_n)$ in terms of P ? I'm not sure. Any help?
