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

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


1 Answer 1


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.