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Suppose $f$ is a bounded function on $[a,b]$, its total variation is defined to be $$ \mathrm{Var}(f) = \sup_{\tau} \sum_{i=1}^{n} |f(x_{i})-f(x_{i-1})|. $$ Furthermore, if $f$ is continuous on $[a,b]$, it can be proved that $$ \mathrm{Var}(f) = \lim_{|\tau|\to 0}\sum_{i=1}^{n} |f(x_{i})-f(x_{i-1})|, $$ where $|\tau|$ denotes the mesh of the partition $\tau$.

When $f$ is discontinuous, can anyone give a counterexample to the above equality, that is the total variation cannot always be calculated by letting the mesh approaching zero?

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    $\begingroup$ If the limit exists then it is equal to the total variation. I don't know if there is an example where the limits does not exist. $\endgroup$ Apr 5, 2019 at 12:07
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    $\begingroup$ The question asked in the title is different that the one in the main formulation. $\endgroup$
    – Hayk
    Apr 17, 2019 at 15:59

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To see that the limit and supremum in question might produce difference results define $f:[0,1]\to[0,1]$ as $$ f(x) = \begin{cases} 0, &\text{ if } \ x \neq 1/2, \\ 1, &\text{ if } \ x = 1/2. \end{cases} $$ Clearly $\mathrm{Var}(f) = 2$, but any partition of $[0,1]$, with diameter however small, will result in $0$ variance-sum unless it includes the point $1/2$ in the partition of $[0,1]$.

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  • $\begingroup$ Thanks a lot for this simple example. But the limit here is mesh-dependent, how about mesh-independent counterexample? Is there such a function : (1) for any mesh appproaching zero, its restricted variation exists. (2) But its true total variation is quite different from the restricted variaton in (1). $\endgroup$
    – C. Davide
    Apr 20, 2019 at 6:58
  • $\begingroup$ @C.Davide, if I understand you correctly, there cannot be a counterexample that is partition independent. This is because the $\sup$ itself is realized on a partition with diameter approaching to $0$. Indeed, notice that $\sup$ is a limit of some sequence of partitions. But for any partition of $[a,b]$ adding more points to it can only increase the variation-sum. Hence, WLOG, we may assume that the sequence of partitions realizing the $\sup$ has diameter converging to $0$. Thus, there is at least one sequence of partitions (meshes) with decaying diameter, that coincides with the $\sup$. $\endgroup$
    – Hayk
    Apr 20, 2019 at 7:28

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