Let $ f [a,b] \rightarrow \mathbb{R}, [a,b] \subset \mathbb{R} $ and a partition $ P = \{ a=t_0 < t_1 < \cdots<t_n = b \} $ of $ [a,b]$. If:
$$ V(f,P)=\sum_{i=1}^n |f(t_i) - f(t_{i-1})| $$
is the variation of $f$ over $P$, then the total variation of $f$ over $[a,b]$ is:
$$ V_a^b f=\sup_P V(f,P) $$
If $f$ is continuous on $[a,b]$ then (Problems in Mathematical Analysis III, W.J.Kaczor M.T.Nowak, Exer. 1.2.15):
$$ \lim_{\mu (P) \rightarrow 0} V(f,P) = V_a^b f \tag 1 $$
where $ \mu(P) $ is the "mesh" of P. I am wondering if there are conditions other than the continuity of $f$ (e.g. if $f$ is of bounded variation?), under which $(1)$ is also valid.
Here's a counterexample to the suggestion that this works when $f$ is just a bounded variation function. Let $f:[0,1]\rightarrow\mathbb{R}$ with $f(1/i)=2^{-i},$ for all $i\geq 2,$ and $f(x)=0$ for all other $x\in[0,1].$ Then since each of these "spikes" adds $2^{1-i}$ to the total variation, $V_{0}^{1}f=1,$ which proves that $f$ is of bounded variation.
Now let $P'\subset(0,1)\cap(\mathbb{R}\setminus\mathbb{Q})$, $P=\{0,1\}\cup P'$, so that $P$ is a partition of $[0,1]$ with all of its points in the interior of $(0,1)$ irrational. Then $V(f,P)=0,$ since $f(t_{i})=0$ for all $t_{i}\in P.$ Clearly we may choose a sequence of partitions with this property such that $\mu(P)\rightarrow 0,$ but $V(f,P)\equiv 0$ cannot possibly converge to $V_{0}^{1}f=1.$