Let $I=[a,b]$, and consider some $f\in BV(I)$ (the space of all functions with bounded total variation on $I$). Let $$\mathcal Q=\{[a_k,b_k]:k\in\mathbb N,~\operatorname{cl}\left(\bigcup_{k=1}^\infty[a_k,b_k]\right)=I,~(a_k,b_k)\cap(a_j,b_j)=\emptyset~\forall i\neq j \}.$$ In their book, Laws of Chaos: Invariant Measures and Dynamical Systems in One Dimension, Gora and Boyarsky claim in a problem that it is possible for $$\sum_{k=1}^\infty V_{[a_k,b_k]}(f)<V_I(f).$$ I struggle to see how this is true, and believe I have a proof that $$\sum_{k=1}^\infty V_{[a_k,b_k]}(f)=V_I(f).$$ Now this question is pretty much the same as this one, but I do not think the accepted answer there is actually correct, as the example given shows that we can have $$V_{[a,c)}(f)+V_{[c,d]}(f)<V_I(f),$$ but the variation of a function over half open intervals is very different from that over closed intervals, so this does not actually address the question. In fact we know that we always have $$V_{[a,c]}(f)+V_{[c,d]}(f)=V_I(f).$$ My supposed proof for the countable case is a standard approximation argument showing inequalities both ways, but it is quite long to type out, and Boyarski and Gora are much wiser than me so my proof is most probably wrong. Thus before I go through the effort of typing it out I'd appreciate some input from the learned members here, because I might be being silly and missing an easy counterexample.
EDIT: On inspecting the fantastic answer by mathcounterexamples.net I realised that I misinterpreted the given counterexample in the linked post, and it is actually shows perfectly what fails in the countable case. Thus I will be flagging this post as a duplicate, as it adds nothing new.
I don’t understand why you aren’t satisfied with the proof of the post you refer to.
Take $a=0$ and $b=1$, $a_1=1/2$, $b_1=1$, $a_k=\sum\limits_{l=2}^k \frac{1}{2^{l-1}}-1/2$ and $b_k = a_k +1/2^k$ for $k\ge 2$. The intervals $[a_k, b_k]$ satisfy the hypothesis.
And $$f(x) =\begin{cases} 0 & 0 \le x <1/2\\ 1 & 1/2 \le x \le 1 \end{cases}$$
However $$0 = \sum_{k=1}^\infty V_{[a_k,b_k]}(f)\neq V_{[0,1]}(f)=1.$$