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Suppose $f:[a,b] \rightarrow \mathbb{R}$

Prove $f$ is of bounded variation on $[a,b] \iff$ there is a $M \in \mathbb{R}$ such that, if $\{(a_k,b_k)\}_{1}^{n}$ is any finite set of disjoint open subintervals of $[a,b]$, then \begin{equation*} \sum_{k=1}^{n} |f(b_k)-f(a_k)| \leq M \end{equation*} Also show that the supremum of all such sums is equivalent to $V_f(a,b)$.

$(\Rightarrow) f \in BV[a,b]$ means that $\sum_{k=1}^{n} |f(x_k)-f(x_{k-1})| \leq M$ for all partitions on $[a,b]$.

I think if we consider the partition which contains all the $a_k$'s and $b_k$'s then this has to be true (seeing the sets are disjoint), by definition of being of bounded variation. However I'm not confident in that. Furthermore I don't know how to prove the converse or the additional part.

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    Corresponding to the set of intervals $(a_{n}, b_{n}) $ we can always create a partition $P$ of $[a, b] $ by putting together all the points $a_{n}, b_{n} $ in a set together with end points $a, b$. Clearly if $f$ is of BV then the sum $V_{f} (P) $ is bounded by some $M$ and the sum in your question does not exceed $V_{f} (P) $ so it is also bounded by $M$. The converse can also be proved in similar manner by noting that $V_{f} (P) $ is also a type of some as mentioned in your question and thus is bounded. – 2017-02-16
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    You can now easily prove that supremum of sums of the type in your question is equal to $V_{f} (a, b) $. – 2017-02-16

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