Equivalent definition of the variation of the function on $[a,b]$ Let $f : \mathbb{R} \rightarrow \mathbb{R}$. The variation of $f$ over $[a,b] \subset \mathbb{R}$ is defined as
$$
V_{a}^{b}(f) := \sup_{ \substack{ a = t_0 < t_1 < \ldots < t_n = b \\ n \in \mathbb{N}} } \sum_{i=1}^{n} |f(t_i) - f(t_{i-1})|
$$
(i.e. the supremum is taken taken over all possible partitions $a = t_0 < t_1 < \ldots < t_n = b$, $n \in \mathbb{N}$ )?
Is it true that $$V_{a}^{b}(f) = \lim_{\delta_n \rightarrow 0} \sum_{i=1}^{n}|f(t_i) -f(t_{i-1})|$$
where $\delta_n = \max_{1 \leq i \leq n}(t_i - t_{i-1}).$
Of course, it can easily be shown by the triangle that the sum increases, as new points are added to the partition. But still, how can one actually show that the supremum is also the limit as the distance of the points goes to zero?
 A: As shown in the comment, the claim may not be true if $f$ is not continuous. The everywhere discontinuous Dirichlet function presents a counterexample.
However, if $f$ is continuous on $[a,b]$ then the claim is true. For any $\epsilon > 0$ there is a partition $P' = (x_0',x_1', \ldots, x_n')$ such that
$$V_a^b(f) - \frac{\epsilon}{2} < V(f;P') \leqslant V_a^b(f),$$
where 
$$V(f;P') = \sum_{k=1}^n |f(x_k') - f(x_{k-1}')|$$
Since $f$ is continuous on $[a,b]$ and, therefore, uniformly continuous there exists $\delta > 0$ such that if $|x-y| < \delta$, then $|f(x) - f(y)| < \epsilon/4n$.
Take any partition  $P= (x_0,x_1, \ldots, x_m)$ with $\|P\| < \delta$.  Let $Q = (y_0,y_1, \dots, y_p)$ be the common refinement $Q = P \cup P'$.  As you observed, it is easy to show using the triangle inequality  that $V(f;P) \leqslant V(f;Q)$ and $V(f;P') \leqslant V(f;Q).$
On the other hand, refining $P$ using $P'$ adds at most $n$ new points, and an interval of $Q$ is either identical to an interval of $P$ or is a subset.
Thus,
$$V(f;Q) = \sum_{[y_{k-1},y_k] = [x_{j-1},x_j]}|f(y_k) - f(y_{k-1})| + \sum_{[y_{k-1},y_k] \subset [x_{j-1},x_j]}|f(y_k) - f(y_{k-1})| $$
There first sum on the RHS is bounded above by $V(f;P)$ and there are at most $2n$ terms in the second sum, whence
$$V(f;Q) \leqslant  V(f;P) + 2n\frac{\epsilon}{4n} = V(f;P) + \frac{\epsilon}{2}.  $$
Thus,
$$V_a^b(f) - \epsilon < V(f;P') - \frac{\epsilon}{2}  \leqslant   V(f;Q) - \frac{\epsilon}{2} \leqslant V(f;P) \leqslant V_a^b(f) \\ \implies |V(f;P) - V_a^b(f)| < \epsilon$$
In this way, we have shown $V(f;P) \to V_a^b(f)$ as $\|P\| \to 0$, which implies that for a sequence of partitions $P_n$ where $\|P_n\| \to 0$, we have $V(f; P_n) \to V_a^b(f).$ 
