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Let $f$ be a function of bounded variation on $[0,1]$. Let $\{[a_n,b_n]\}_{n=1}^\infty$ such that $(a_n,b_n)$ are pairwise disjoint and $\cup_{n=1}^\infty [a_n,b_n]=[0,1]$. (for example, $[1/2, 1], [0,1/3], [1/3,1/3+1/3^2], \cdots$ ) Can we write

$$ \operatorname{Var}_{[0,1]} f=\sum_{k=1}^\infty \operatorname{Var}_{[a_k,b_k]} f? $$

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1 Answer 1

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For your example intervals, suppose that $f$ is $0$ on $[0,1/2)$ and $1$ on $[1/2,1]$.

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  • $\begingroup$ I'm curious as to why this answer got downvoted. It gives a function that has variation equal to 1 on [0,1] and whose variation on each interval of the specified partition is 0. This seems to provide a valid counterexample to the question being asked... so what have I missed that makes this worth a downvote? $\endgroup$
    – postmortes
    Jul 22, 2014 at 7:55
  • $\begingroup$ I also believe this is a good counterexample. But it looks somehow fragile, because for $C^1$ functions we know that $\operatorname{Var}_{[a,b]}f = \int_a^b |f'(x)|\, dx$, and the conjecture seems to be true. $\endgroup$
    – Siminore
    Jul 22, 2014 at 7:59
  • $\begingroup$ Granted, but the OP hasn't placed any restrictions on their function f so I don't think the answer was worth downvoting. However, perhaps you could add an answer that addresses the issue of the function-class that f belongs to? $\endgroup$
    – postmortes
    Jul 22, 2014 at 8:04
  • $\begingroup$ I suspect it is true for continuous functions but haven't thought it through. $\endgroup$ Jul 22, 2014 at 8:04

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